students’ understanding of related rates ...carlson/dissertations/ne...students’ understanding...
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STUDENTS’ UNDERSTANDING OF RELATED RATES PROBLEMS IN
CALCULUS
by
Nicole Marie Engelke Infante
A Dissertation Presented in Partial Fulfillment of the Requirements for the Degree
Doctor of Philosophy
ARIZONA STATE UNIVERSITY
May 2007
STUDENTS’ UNDERSTANDING OF RELATED RATES PROBLEMS IN
CALCULUS
by
Nicole Marie Engelke Infante
has been approved
March 2007
Graduate Supervisory Committee:
Marilyn P. Carlson, Chair Michael Oehrtman
Helene Barcelo Alfinio Flores Donald Jones
ACCEPTED BY THE DIVISION OF GRADUATE STUDIES
iii
ABSTRACT
There is little research that has been conducted on how students understand and
solve related rates problems in calculus. The research to date has focused on classifying
each step that may be used to solve a problem as either procedural or conceptual.
Students’ success has been tied to their ability to effectively complete the conceptual
steps. However, there is little known about the mental model which supports a conceptual
approach to solving related rates problems. The purpose of this study is to address this
gap in the research.
Three mathematicians were observed solving three related rates problems. This
data was analyzed to develop a framework for solving related rates problems. It was
found that the mathematicians identified the problem type as a “related rates problem”
and then engaged in a series of phases to generate pieces of their solution. These phases
were identified as: draw and label a diagram, construct a meaningful functional
relationship, relate the rates, solve for the unknown rate, and check the answer. To
complete each phase and construct a piece of the solution, the mathematicians built and
refined a mental model of the problem situation. The framework captured how the
solution to the problem emerged from the mathematicians’ thinking as they responded to
related rates problems. This framework was then used to analyze the results of a teaching
experiment.
A teaching experiment consisting of six teaching episodes was conducted with
three first semester calculus students. It was expected that the students would develop
covariational reasoning abilities which would foster their ability to construct mental
models of related rates problem situations. By using dynamic computer simulations of
iv
related rates problem situations, the students explored the concept of rate and developed
language and notation to talk about rates. The students developed an approach to solving
related rates problems which was centered on the idea of relating the rates by creating a
“delta equation.” The delta equation was a restatement of the problem by applying the
chain rule and figuring out what each piece of the equation represented. This approach
was also observed in one of the mathematician’s solutions.
v
To my husband, Niel. You are my best friend and biggest champion.
I could not have done it without you.
To my parents, Doug and Mary Engelke. You always encouraged me to pursue my dreams.
I have done just that with your unending love and support.
vi
ACKNOWLEDGEMENTS
I could not have completed this dissertation without the guidance and support of
some wonderful people. First, I would like to thank my advisor, Dr. Marilyn Carlson, for
her encouragement and support. My many discussions with you enriched my knowledge,
directed this dissertation, and made me a better researcher. Many thanks also go to Dr.
Patrick Thompson for his guidance in refining my instructional materials; they would not
be what they are today without your insights.
I also wish to thank my committee members Dr. Michael Oehrtman, Dr. Alfinio
Flores, Dr. Helene Barcelo, and Dr. Don Jones. I appreciate your patience and helpful
suggestions. My conversations with you helped me to develop as a mathematician and an
educator.
My greatest thanks are to my family, Mom, Dad, Sarah, Kira, Kirk, and Zach.
Thanks for believing in me and encouraging me along the way.
I could not have done this without the love, support, and programming skills of
my husband, Niel. The computer simulations you created are wonderful and bring life to
the instructional activities. Thank you for reading this document again and again and
providing valuable insights. You always encouraged me when I needed it and celebrated
the small victories along the way with ice cream.
Thanks to my many friends, especially the UConn crowd: Jonathan, Jack, Jenn,
Mark, Dave, and Jason. You are the best friends a person could ask for, and you never
doubted me. Then there are the ASU people: Amylou, Rachel, Ryan, Phil, Jessica, and
Vicki. Last but not least, my friends from VOTS – there are too many of you to mention
by name. We’ve had some great times folks, and I am sure more will follow.
vii
TABLE OF CONTENTS Page
LIST OF TABLES............................................................................................................ xii
LIST OF FIGURES ......................................................................................................... xiv
CHAPTER 1: INTRODUCTION........................................................................................1
1.1 Background ....................................................................................................... 1
1.2 Research Questions........................................................................................... 7
1.3 Overview of the Study ...................................................................................... 8
CHAPTER 2: LITERATURE REVIEW ...........................................................................10
2.1 The Concept of Variable................................................................................. 10
2.2 The Concept of Function ................................................................................ 11
2.3 Function Representations................................................................................ 13
2.4 The Concept of Derivative.............................................................................. 19
2.4.1 A Derivative Framework ................................................................. 20
2.4.2 The Derivative in Related Rates Problems ...................................... 22
2.5 Related Rates .................................................................................................. 23
CHAPTER 3: THEORETICAL PERSPECTIVE..............................................................30
3.1 Transformational Reasoning........................................................................... 30
3.2 Covariational Reasoning................................................................................. 31
3.3 Mental Models ................................................................................................ 36
3.4 Problem Solving.............................................................................................. 38
CHAPTER 4: PILOT STUDIES .......................................................................................44
4.1 Overview......................................................................................................... 44
viii
Page
4.2 Initial Results .................................................................................................. 53
4.3 Development of Classroom Activities ............................................................ 61
4.4 Limitations ...................................................................................................... 69
4.5 Summary......................................................................................................... 69
CHAPTER 5: METHODS.................................................................................................71
5.1 Data Collection ............................................................................................... 72
5.1.1 Subjects ............................................................................................ 72
5.1.2 Individual Interviews ....................................................................... 74
5.1.3 The Teaching Experiment................................................................ 75
5.2 Data Analysis .................................................................................................. 81
CHAPTER 6: AN EMERGENT FRAMEWORK.............................................................85
6.1 Mathematicians Solve Related Rates Problems.............................................. 86
6.1.1 The Trough Problem........................................................................ 86
Dan’s solution. .............................................................................. 87
Adam’s solution. ........................................................................... 96
Bob’s solution. ............................................................................ 101
Summary. .................................................................................... 108
6.1.2 The Plane Problem......................................................................... 109
6.1.3 The Coffee Cup Problem ............................................................... 115
6.1.4 Summary of the Mathematicians’ Solution Process ...................... 125
6.2 A Framework for Solving Related Rates Problems...................................... 129
6.3 Using the Framework.................................................................................... 141
ix
Page
CHAPTER 7: TEACHING EXPERIMENT RESULTS .................................................143
7.1 Students’ Individual Interviews .................................................................... 143
7.1.1 Initial Understandings of Variable................................................. 143
7.1.2 Initial Understandings of Function ................................................ 147
7.1.3 Evidence of Engagement in Covariational Reasoning................... 162
7.2 Addressing Time as a Variable and Rate of Change .................................... 170
7.3 Revisiting the Chain Rule ............................................................................. 180
7.4 Students Solve Related Rates Problems ....................................................... 186
7.4.1 The Plane Problem......................................................................... 186
7.4.2 The Trough Problem...................................................................... 205
7.4.3 The Coffee Cup Problem ............................................................... 228
7.5 Summary....................................................................................................... 242
CHAPTER 8: DISCUSSION AND CONCLUSIONS....................................................246
8.1 Discussion of the Framework ....................................................................... 246
8.1.1 The Role of the Mental Model....................................................... 247
8.1.2 The Role of Content Knowledge ................................................... 249
8.1.3 The Role of Time ........................................................................... 251
8.1.4 The Phases ..................................................................................... 252
8.1.5 The Revised Framework ................................................................ 256
8.1.6 Generalization of the Framework .................................................. 262
8.2 Conclusions................................................................................................... 264
8.2.1 General Conclusions ...................................................................... 265
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Page
8.2.2 Answering the Research Questions ............................................... 266
Question 1. .................................................................................. 267
Question 2. .................................................................................. 268
Question 3. .................................................................................. 271
Question 4. .................................................................................. 271
8.3 Limitations of the Study................................................................................ 272
8.4 Directions for Future Research ..................................................................... 273
8.4.1 Refinement of the Classroom Materials ........................................ 274
8.4.2 The Role of Affect ......................................................................... 274
8.4.3 Use of the Word “Like” ................................................................. 275
8.4.4 Dealing with Time as a Variable ................................................... 276
REFERENCES ................................................................................................................277
APPENDIX
A HUMAN SUBJECTS MATERIALS...................................................................286
Recruitment - Verbal Script ................................................................................ 287
Letter of Consent................................................................................................. 288
B PILOT STUDY MATERIALS ............................................................................290
Review of Similar Figures for Calculus Students Using GSP............................ 291
Pilot Study Version - Related Rates Activity ..................................................... 292
Pilot Study Version - Related Rates Activity 2 .................................................. 294
Related Rates Puzzle........................................................................................... 295
C TEACHING EXPERIMENT MATERIALS .......................................................299
xi
Page
Individual Interview Protocol ............................................................................. 300
Instructional Sequence Lesson Logic ................................................................. 302
Student Activity Sheets....................................................................................... 315
xii
LIST OF TABLES
Table Page
1 Common Related Rates Problems .......................................................................... 3
2 Steps to Solve a Related Rates Problem ................................................................. 6
3 A Derivative Framework ...................................................................................... 21
4 Martin’s Step Classification.................................................................................. 24
5 Steps to Solve a Related Rates Problem ............................................................... 28
6 A Covariational Framework ................................................................................. 34
7 Carlson and Bloom’s Multidimensional Problem Solving Framework................ 41
8 Pilot Study Data Codes ......................................................................................... 47
9 Example of Pilot Study Coding ............................................................................ 51
10 Hypothetical Learning Trajectory......................................................................... 78
11 Data Sources ......................................................................................................... 83
12 Summary of the Phase: Draw a Diagram............................................................ 132
13 Summary of the Phase: Construct Meaningful Functional Relationships .......... 133
14 Summary of the Phase: Relate the Rates ............................................................ 134
15 Summary of the Phase: Solve for the Unknown Rate......................................... 135
16 Summary of the Phase: Check the Answer for Reasonability ............................ 136
17 Sample Summary Table...................................................................................... 142
18 Summary Table for Draw a Diagram - Plane Problem....................................... 190
19 Summary Table for Construct a Functional Relationship - Plane Problem........ 194
20 Summary Table for Relate the Rates - Plane Problem ....................................... 199
21 Summary Table for Find the Unknown Rate - Plane Problem ........................... 201
xiii
Table Page
22 Summary Table for Relate the Rates, Alternate Approach – Plane Problem ..... 204
23 Summary Table for Draw a Diagram - Trough Problem.................................... 211
24 Summary Table for Construct a Functional Relationship - Trough Problem..... 216
25 Summary Table for Relate the Rates - Trough Problem..................................... 221
26 Summary Table for Find the Unknown Rate - Trough Problem ........................ 223
27 Summary Table for Draw a Diagram - Cup Problem......................................... 229
28 Summary Table for Relate the Rates Part 1 - Cup Problem ............................... 231
29 Summary Table for Construct a Functional Relationship - Cup Problem.......... 237
30 Summary Table for Relate the Rates Part 2 - Cup Problem ............................... 240
31 Summary Table for Find the Unknown Rate - Cup Problem ............................. 242
32 A Revised Framework for the Solution Process for Related Rates Problems .... 258
33 General Framework Template ............................................................................ 262
34 Possible Framework for Extreme Value Problems............................................. 264
35 Lesson Logic for Plane Problem Using Computer Program .............................. 302
36 Lesson Logic for Snowball Problem Using Computer Program ........................ 304
37 Lesson Logic for Chain Rule - Car Problem ...................................................... 305
38 Lesson Logic for Chain Rule - Hiking Problem ................................................. 306
39 Lesson Logic for Chain Rule - Ball Problem...................................................... 306
40 Lesson Logic for Related Rates - Plane Problem ............................................... 308
41 Lesson Logic for Related Rates - Snowball Problem ......................................... 310
42 Lesson Logic for Related Rates - Trough Problem ............................................ 311
xiv
LIST OF FIGURES
Figure Page
1 Visualization schema ............................................................................................ 38
2 The solution process for related rates problems ................................................. 138
3 Ali’s solution to the genie problem..................................................................... 151
4 Ann’s solution to the genie problem................................................................... 156
5 Ben’s solution to the genie problem ................................................................... 160
6 Computer simulation of the plane problem 1 ..................................................... 171
7 Computer simulation of the plane problem 2 ..................................................... 174
8 Computer simulation of the plane problem 3 ..................................................... 176
9 Ann’s solution to the trough problem ................................................................. 225
10 Ali’s solution to the trough problem................................................................... 227
11 The revised solution process for related rates problems..................................... 261
Chapter 1: Introduction
This dissertation describes a framework that emerged from a study that
investigated the reasoning patterns, knowledge, and general problem solving approaches
for solving related rates problems. It also describes the results and implications of a
teaching experiment that was analyzed using this framework.
1.1 Background
Related rates problems emerged historically as a means to reform the manner in
which calculus was taught. Rev. William Ritchie (1790-1837) was a pioneer in
mathematics education; it was his desire to write a reform calculus text that was more
accessible to ordinary, non-university students (Austin, Barry, & Berman, 2000). In
Ritchie’s text, related rates problems were meant to be fundamental, explanatory
problems that illustrate the power of calculus. Much of his text focused on investigating
the change in a magnitude over time. After an intuitive introduction to limits, he used an
expanding square to illustrate the concept of function and how uniform increases to the
independent variable may cause the dependent variable to increase at an increasing rate.
It was Ritchie’s belief that “The object of the differential calculus, is to determine the
ratio between the rate of variation of the independent variable and that of the function
into which it enters.” (Ritchie, 1836, p. 11) Related rates problems also require students
to relate mathematical concepts from geometry, algebra, trigonometry, and calculus. The
idea of relating changing quantities through a third variable (usually time) is important
beyond the calculus classroom, particularly in physics. In today’s calculus classroom,
related rates problems are often given only a cursory treatment and are sometimes even
omitted from the curriculum.
2
Related rates problems require the student to investigate the relationship(s)
between two or more changing quantities, one of which is unknown and needs to be
found. The relationships between the variables tend to fall into two categories: 1) the
variables are related through function composition (the trough problem in Table 1) or 2)
the variables are related parametrically through a third variable, usually time (the plane
problem in Table 1). The problem solver needs to be able to reconceptualize the static
variables that occur in the geometric formulas that represent these problem situations as
functions of time in order to successfully solve the problem. A list of some common
related rates problems is provided in Table 1 below.
3
Table 1: Common Related Rates Problems
Short Name Related Rates Problem
Trough Problem
A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3 feet across at the top and have a height of 1 foot. If the trough is filled with water at a rate of 12 ft3/min, how fast does the water level rise when the water is 6 in. deep? (Stewart, 1991, p.164)
Plane Problem
A plane flying horizontally at an altitude of 3 miles and a speed of 600 mi/hr passes directly over a radar station. When the plane is 5 miles away from the station, at what rate is the distance from the plane to the station increasing?
Ladder Problem
A 17 ft. ladder is leaning against a wall. If the bottom of the ladder is pulled along the ground away from the wall at a constant rate of 5 ft/sec, how fast will the top of the ladder be moving down the wall when it is 8 ft above the ground? (Anton, 1995, p. 183)
Snowball Problem
A spherical snowball is melting in such a way that its volume is decreasing at a rate of 1 cm3/min. Let d represent the diameter of the snowball in cm. At what rate is the diameter decreasing with respect to time?
Coffee Cup Problem
Coffee is poured at a uniform rate of 20 cm3/sec into a cup whose inside is shaped like a truncated cone. If the upper and lower radii of the cup are 4 cm and 2 cm, respectively, and the height of the cup is 6 cm, how fast will the coffee level be rising when the coffee is halfway up the cup? (Anton, 1995, p.185)
Boat Problem Two boats are racing with constant speed toward a finish marker, boat A sailing from the south at 13 mph and boat B approaching from the east. When equidistant from the marker the boats are 16 miles apart and the distance between them is decreasing at the rate 17 mph. Which boat will win the race?
Thin Lens Problem The thin lens equation in physics is
fSs
111=+ where s is the object
distance from the lens, S is the image distance from the lens, and f is the focal length of the lens. Suppose that a certain lens has a focal length of 6 cm and that an object is moving toward the lens at the rate of 2 cm/sec. How fast is the image distance changing at the instant when the object is 10 cm from the lens? Is the image moving away from the lens or toward the lens? (Anton, 1995, p.185)
Balloon Problem
A spherical balloon is to be deflated so that its radius decreases at a constant rate of 15 cm/min. At what rate must the air be removed when the radius is 9 cm? (Anton, 1995, p. 183)
4
To an experienced mathematician, solving a related rates problem may seem
routine. However, the task is often a source of frustration for the calculus student. At the
beginning of the section on related rates in his calculus textbook, Stewart (1991) stated:
In a related rates problem the idea is to compute the rate of change of one quantity
in terms of the rate of change of another quantity (which may be more easily
measured). The procedure is to find an equation that relates the two quantities and
then use the Chain Rule to differentiate both sides with respect to time. (p. 160)
Stewart’s description implies that these problems are easily solved in two quick steps.
Two pages later, he outlines a strategy for solving related rates problems that involves
seven steps:
1) Read the problem carefully.
2) Draw a diagram if possible.
3) Introduce notation. Assign symbols to all quantities that are functions of time.
4) Express the given information and the required rate in terms of derivatives.
5) Write an equation that relates the various quantities of the problem. If
necessary, use the geometry of the situation to eliminate one of the variables
by substitution.
6) Use the Chain Rule to differentiate both sides of the equation with respect to t.
7) Substitute the given information into the resulting equation and solve for the
unknown quantity. (p. 162)
He follows this recommendation with the warning that a common error is to substitute the
given numerical information too early. If solving a related rates problem was as simple as
5
following these steps with a procedural orientation, students should have little difficulty
solving such problems. Unfortunately, this is not the case. This leads to the question:
What makes solving a related rates problem so difficult for students?
To solve most related rate problems, a student must incorporate recent calculus
knowledge with knowledge from previous mathematics courses such as geometry and
precalculus. The trough problem as stated in Table 1 above is a typical related rates
problem from a textbook.
A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3
feet across at the top and have a height of 1 foot. If the trough is filled with water
at a rate of 12 ft3/min, how fast does the water level rise when the water is 6 in.
deep? (Stewart, 1991, p.164)
The solution requires the student to work through what appears to be a minimum of 11
steps which rely on the student’s knowledge of many concepts such as triangle geometry,
variable, function, derivative (including the use of the chain rule to account for the
implicit variable of time), and basic algebraic manipulations. Table 2 below outlines a
possible solution process for this specific problem. It also includes my initial description
of a general approach to solving related rates problems.
6
Table 2: Steps to Solve a Related Rates Problem
Steps Taken to Solve the Trough
Problem
Generalized Step
Draw a picture of an isosceles triangle and label the known and unknown quantities.
Interpret the words in the problem to form a mental image of the situation, draw a diagram of the situation, and label the known and unknown quantities.
Define what is known:
bhlVldt
dV
2
1,10,12 ===
Define what is known.
Determine what rate is asked for in the
problem: dt
dh
Define what rate is asked for in the problem.
Note that the solution requires a formula that defines volume in terms of height:
________ honlyV =
Identify which two of the varying quantities need to be related in the formula: one known and one unknown.
Imagine the trough filling with water. Notice that other quantities that are varying in the situation.
Imagine the system changing (engaging in covariational reasoning, to be defined later).
Notice that the height and width of the water are changing.
Note the quantities that are changing.
Define which quantities are changing: b and h
Define which quantities are changing with variables.
Define the volume of the water in the trough in terms of the changing quantities:
bhV 5=
Write a formula that relates the changing quantities in the situation.
Eliminate b in the volume formula by defining b in terms of h: bhV 5=
Is needed: 1
3=
h
b
Substitute hb 3= into the volume formula
to obtain: ( )( ) 21535 hhhV ==
Eliminate unnecessary variables if needed. If needed, determine the relationship between 2 changing quantities. Make the substitution (employ function composition).
Differentiate with respect to time:
dt
dhh
dt
dV30=
Differentiate
Plug in values and solve for the unknown rate.
Substitute in values and solve for the unknown rate.
7
There is very little literature that pertains to how students understand and solve
related rates problems. However, there have been numerous investigations how students
understand related concepts such as variable, function, and derivative which may provide
some insight. It has been suggested that students have a weak understanding of the
concepts of variable and function (Breidenbach, Dubinsky, Hawks, & Nichols, 1992;
Carlson, 1998; Clark, Cordero, Cottrill, Czarnocha, Devries, St. John, Tolias, &
Vidakovic, 1997; Dubinsky & Harel, 1992; Frid, 1992; Lithner, 2003; Orton, 1983;
Sfard, 1992; Vinner & Dreyfus, 1989; White & Mitchelmore, 1996). The research also
suggests that students learn to manipulate symbols without using them to represent
concepts and ideas (White & Mitchelmore, 1996). Lithner (2003) reported that when
calculus students are solving textbook exercises, they are focused on mimicking
previously worked examples rather than engaging in genuine problem solving processes.
It is the purpose of this study to investigate what type of mental model(s) and conceptual
understandings facilitate the successful completion of related rates problems.
1.2 Research Questions
Since related rates problems arise in situations beyond the mathematics
classroom, it would appear that further investigation of how students understand and
solve these problems would be beneficial not only for calculus students but physics and
engineering students, too. It is the purpose of this study to investigate the process by
which students understand and solve related rates problems. This study will also seek to
determine whether instructional materials may aid students in understanding and solving
related rates problems. I investigated the following research questions:
8
1. What is involved in solving a related rates problem? What problem solving
processes are necessary? What cognitive constructions are necessary for the
student to have an effective mental model? What formal mathematical
knowledge is necessary?
2. How do students’ function conception and covariational reasoning abilities
influence their understanding of and ability to complete related rates
problems?
3. How do students’ understandings of the chain rule influence their ability to
solve related rates problems?
4. Does the use of a computer generated dynamic animation of related rates
problem situations impact students’ ability to solve and understand related
rates problems? If so, how?
1.3 Overview of the Study
This study consisted of two parts. In the first part, a teaching experiment was
conducted with three first semester calculus students. The purpose of the teaching
experiment was to evaluate the effectiveness of some in-class activities that were
developed as a result of the pilot studies.
In the second part of the study, three mathematicians were observed solving
related rates problems in a think aloud setting. This data allowed for the development of a
framework which would characterize what is likely involved in understanding and
solving a related rates problem. This framework would then be used to analyze the
teaching experiment data.
9
This dissertation provides a review of the relevant literature and the theoretical
perspective I used to analyze my data. I then provide an overview of the results of my
pilot study and how that guided the development of the related rates instructional
materials. A description of the methods I used to investigate these questions is followed
by the analysis of the mathematicians’ data which led to the emergent framework. The
results of the teaching experiment are then provided. Lastly, I present a summary of the
results, general conclusions, and directions for future research.
Chapter 2: Literature Review
There is very little literature that investigates students’ understanding of related
rates problems. Martin’s (1996, 2000) summarized dissertation work and White and
Mitchelmore’s (1996) study of introductory calculus concepts are the most recent
literature that directly address related rates problems. However, related rates problems
utilize many mathematical concepts such as variable and function which have been
studied more extensively. I will begin with a review of the literature on the concept of
variable, the concept of function, and the concept of derivative. I will conclude with how
these concepts relate to the literature on related rates and to the student’s ability to
successfully solve a related rates problem.
2.1 The Concept of Variable
Related rates problems require the student to be able to represent changing
quantities with variables. Research suggests that students have difficulty with the concept
of variable from when they begin to learn algebra and even into calculus (Jacobs, 2002;
Kieran, 1992; Orton, 1983; White & Mitchelmore, 1996). Jacobs (2002) reported that
students who had a robust understanding of variable thought about it as a tool for
expressing relationships rather than just a computational tool (a letter that represents a
number that will be solved for). In a related rates problem, being able to correctly
represent a relationship between changing quantities is a critical part of the solution
process. Jacobs also concluded that a dynamic conception of variable (one in which the
student can imagine how two or more variables relate to each other) supports deeper
understandings of other calculus conceptions such as limit and derivative.
11
2.2 The Concept of Function
Related rates problems require students to have a working knowledge of the
concept of function, particularly function composition. An understanding of the concept
of function that goes beyond the ability to evaluate a function at a specific point allows
the student to easily construct a formula that relates the appropriate quantities.
Specifically, being able to compose two or more functions to achieve a function of one
variable in terms of another variable (where the variables are the appropriate independent
and dependent variables) is a critical skill necessary for being able to solve a related rates
problem.
Breidenbach, Dubinsky, Hawks, and Nichols (1992) and Dubinsky and Harel
(1992) proposed four levels of understanding for the concept of function: pre-action
(little, if any, understanding of the concept of function), action (repeatable physical or
mental operations on objects), process (dynamic transformation of quantities according to
repeatable means and is a complete activity, for example able to compute composition
and inverses), and object (ability to perform actions on the function itself). It should be
noted that frequently it is useful to move between a process and object view when
attempting to solve a problem. These levels of understanding are one instantiation of
what is known as APOS (Action, Process, Object, Schema) theory which has been used
by many authors to describe how students understand various mathematical concepts
(Asiala, Brown, Devries, Dubinsky, Matthews, & Thomas, 1996; Breidenbach et al.,
1992; Dubinsky & Harel, 1992).
Carlson (1998) investigated and described what is required for students to gain a
mature understanding of the concept of function. The study used students from college
12
algebra, second semester calculus, and first semester graduate students (complex analysis
or abstract algebra) – each had earned an A in their course. She found that even strong
precalculus students may possess an action view of function and could not easily access
newly acquired information. Both second semester calculus and graduate students in the
study also exhibited difficulty. Carlson concluded that the function concept is slow to
develop, even in strong students.
College algebra students frequently did not refer to the input-output nature of the
concept of function and expected them to be “nice,” that is smooth, continuous, and 1-1.
Misconceptions about the role of the independent and dependent variables were also
identified in college algebra students as they found it difficult to interpret statements such
as “express A in terms of r.” They also had difficulty computing ( )axf + , and struggled
to verbally express the meaning of algebraic symbols. This weak function conception
presented difficulty when composing and inverting functions, and when using functions
to describe relationships in word problems.
Carlson (1998) used a bottle problem (given a particular bottle, construct a graph
of height versus volume) to investigate students’ ability to coordinate the changes in one
variable with the changes in another variable and found that the population ran the gamut
from no evidence of covariational reasoning to clear, articulate reasoning which included
calculus concepts. Using Monk’s (1992) car problem [the speed vs. time graph for two
cars, one concave up, one concave down, is given and students are asked to determine the
position of the cars at time t = 1], she identified iconic translation, the tendency for a
student to interpret a graph as a much more literal picture of the situation than it really is,
as a source of difficulty for students (Hitt, 1998; Monk, 1992).
13
Sfard (1991, 1992) discussed the historical development of the concept of
function and its roots in computational processes. She described three steps required to
move from what she calls an operational approach (similar to an action/process view) to a
structural approach (similar to a process/object view). Her three steps are: 1)
interiorization – process performed on already familiar objects, 2) condensation – turning
this process into a compact, self-contained unit, and 3) reification – the ability to view
this new entity in its own right. This approach is similar to Dubinsky et al’s approach in
that to solve complex function tasks, one must move flexibly between an operational and
structural mode of thinking. The difference in terminology here highlights the still
emerging language of mathematics education research.
2.3 Function Representations
Function notation has been shown to be problematic for students (Carlson, 1998;
Even, 1998). The concept of variable is embedded into the concept of function and there
is evidence that even calculus students have a rigid conception of variable (Jacobs, 2002;
Orton, 1983; Vergnaud, 1998; White & Mitchelmore, 1996). That x represents the
independent variable (input) and ( )xf represents the dependent variable (output) of a
function is difficult for students to grasp. White and Mitchelmore (1996) reported that
students’ weak concept of variable, particularly in the case of function, severely limits
their understanding of calculus concepts. They identified this issue when they studied
how students interpreted and solved word problems, some of which were related rates
problems (see section 2.5.2 for a more complete discussion of this topic). Orton (1983)
indicated that basic algebraic errors such as improper factorization and “losing the middle
term” when expanding a quadratic were very common when working with functions. He
14
argued, “algebraic difficulties could be obscuring the ideas of calculus,” noting that some
students appeared to have a reasonable understanding of algebraic situations, but could
not execute the algebraic procedures without error.
Composition of functions, denoted by ( )( )xgf or ( )( )xgf o , is often confused
with function multiplication. Engelke, Oehrtman, and Carlson (2005) reported that when
faced with composition, 95% of students can compute a value when given an explicit
formula. When faced with a table or graph, that number dropped to 35%, and when given
the problem with a context only 20% of the same precalculus students provided a correct
response. This indicates that students have an action view of function and have a difficult
time moving between function representations. More evidence of this comes from Sfard
(1992), as she found that students had particular difficulty solving composition problems
that required them to use unspecified functions, for example, ( ) ( ) ( )( )21 , xgxgfxh = . This
difficulty in dealing with more abstract levels of function composition indicates that
students do not see functions as objects, the building blocks for other functions (Carlson,
1998; Dubinsky & Harel, 1992; Engelke et al., 2005; Even, 1998).
The notation for the inverse of a function, ( )xf 1− , is often confused with the
notation for negative exponents (Engelke et al., 2005; Even, 1992; Vidakovic, 1996). As
a result, students often interpret ( )xf 1− as ( )xf
1. When given an explicit formula, some
students are able to carry out the procedure of switching the x and y values and solving
for y to obtain the formula for ( )xf 1− , but this likely only demonstrates the ability to
carry out a memorized algorithm rather than understanding that the inverse reverses the
functions process. When asked to compute an inverse given a graph or table (more
15
conceptual questions), students reveal more difficulties. Even’s (1992) work with
preservice teachers indicates that a naïve conception of “undoing” is not sufficient to
understand the concept of inverse function. Her subjects could not differentiate between
exponential ( x2 ) and power functions ( 2x ) and thought that taking the log and taking the
root were the same thing.
Vidakovic (1996) claimed that
subjects with schemas for composition of functions and inverse function can
coordinate them (coordination of a higher order) to obtain a new process. For
example, this coordination explains why the inverse of the composition of two
functions is the composition of the inverses of two functions in reverse order. (p.
310)
Asiala, Cottrill, Dubinsky, and Schwingendorf (1997) provided evidence that
these problems with the concept of function carry over to calculus. Using the APOS
framework, they detailed students’ difficulties with understanding the derivative and its
graphical representation. The students were presented with the graph of a curve and a
tangent to the point ( )5f and asked to find ( )5f and ( )5f ′ . Most students attempted to
construct a formula for the given graph so they could find a formula for the derivative
using known rules; however, the slope could be calculated directly from the graphical
representation. The students did not readily make the connection between the slope of the
tangent line and the derivative.
The difficulty with notation and representation is that students learn to use
symbols without any regard for their meaning; the symbol is merely an abbreviation for a
word, not a designate for the concept (Ferrini-Mundy & Geuther-Graham, 1991; Sajka,
16
2003; White & Mitchelmore, 1996). Given that students have such a difficult time with
the notation of functions; it is not surprising that this fosters problems with
representation, particularly in word problems.
Janvier (1987) defines representation as a combination of: symbols (written), real
objects, and mental images. He argues that there is also a verbal/semantic component to
understanding a concept. An algebraic formula (probably the most common and compact
notation), graph, table, or contextual situation can be used to represent a function. To be
effective problem solvers, students must be able to flexibly move between the various
function representations, sometimes taking a global view of the problem and sometimes a
pointwise view. Research suggests that precalculus level students and some secondary
mathematics teachers have difficulty moving among function representations
(Breidenbach et al., 1992; Carlson, 1998; Even, 1998). To encourage this flexibility,
Thompson (1994c) suggested that teachers need:
to find situations that are sufficiently propitious for engendering multitudes of
representational activity and (2) to orient students toward drawing connections
among their representational activities in regard to the situation that engendered
them. The situation being represented must be paramount in students’ awareness,
for if they do not see something remaining the same as they move among tables,
graphs, and expressions, then it increases the probability that they will see each as
a “topic” to be learned in isolation of the others. (p. 23)
Related rates problems in calculus require the student to be adept at moving between
function representations, particularly between the contextual situation, a diagram of the
situation, and the algebraic formula.
17
With regard to word problems, problem representation is directly related to
success (Cifarelli, 1998; Even, 1998; Hegarty, Mayer, & Monk, 1995; Hitt, 1998;
Janvier, 1987; Silver & Marshall, 1989; Vergnaud, 1998). In solving problems, many
unstable representations are often created before reaching a stable one that is helpful
(Silver & Marshall, 1989). These representations are often based on heuristics and
knowledge of previous examples and problems.
There is cognitive research that supports the idea that categories and concepts are
frequently represented by prototypes (Medin, 1998; Rosch & Mervis, 1975). A prototype
is something that represents a category. For example, a robin may be a prototype for
bird. While an ostrich is also a bird, it is less of a bird than a robin to most people and
thus is not a prototype for bird. In math, students may have prototypes for categories of
word problems. The prototype for a related rates problem is usually a textbook example.
When a problem does not align with this prototype, students encounter difficulties
solving the problem and are more likely to make errors. Johnson-Laird (1983) asserts
that prototypes “are procedures that specify by default the values of certain variables in
mental models.” (p. 446) The use of a prototype to solve problems is also known as
transfer or analogical problem solving.
There is a substantial amount of research that has been conducted on transfer,
primarily by cognitive psychologists (Bassok, Chase, & Martin, 1998; Bernardo, 2001;
Campione, Brown, & Connell, 1989; Catrambone, 1994, 1995, 1996; Kapa, 2001; Kulm
& Days, 1979; Lithner, 2003; Robertson, 2000; Schoenfeld, 1989; Sierpinska, 1995;
Silver & Marshall, 1989). While students may prefer to learn from examples, there is
evidence that this method is not as effective as one might hope (Bernardo, 2001; Kulm &
18
Days, 1979; Robertson, 2000; Schoenfeld, 1989; Silver & Marshall, 1989). Silver and
Marshall (1989) indicated that while experts focus on structure, novices tend to focus on
superficial features when attempting to solve a word problem resulting in poor transfer.
Robertson (2000) indicated that mapping errors (not being able to properly choose and
align similar problems), over-transfer/matching errors (not being able to properly assign
values to variables in the problem) and frame errors (inability to generate and adapt the
relevant equation) are extremely common. Kulm and Days (1979) proposed that solving
more complex problems before solving simpler ones may increase the chances of
transfer, and Catrambone (1994, 1995, 1996) proposed a subgoal structure to promote
transfer.
Catrambone (1994, 1995, 1996) studied how groups of psychology students
learned to solve statistics problems by studying previously worked examples. His
experiments determined that when the worked example had related steps grouped
together and labeled (a subgoal), students were more successful in transferring their
knowledge to new, novel problem situations. He also indicated that the labels do not need
to be meaningful, and that unspecific labels may be beneficial as the student is then
forced to construct the structure associated with the label. The idea is that the problem
has a general goal (solution) and that to attain this goal, the problem may be broken down
into subgoals, smaller problems to be solved. The primary purpose of subgoals is to
restrict the solution space the student must search mentally. The subgoal structure may be
more easily adapted to novel problems as it creates a structural hierarchy for solving
problems.
19
Analogical problem solving strategies are used extensively by students solving
textbook word problems (Lithner, 2003; White & Mitchelmore, 1996). This appears to be
true of students solving related rates problems. As was reported in previous studies and
will be seen in my pilot study results, students are actively looking for previously solved
problems and examples that they may follow line by line to solve a similar problem
rather than construct a new representation for the current problem situation (Engelke,
2004; Lithner, 2003; White & Mitchelmore, 1996). Construction of a representation of
the problem situation may rely on the student’s ability to engage in covariational
reasoning which will be discussed in chapter three.
2.4 The Concept of Derivative
What does it mean to understand the concept of derivative? There have been a
number of studies that attempt to answer this question (Asiala et al., 1997; Ferrini-Mundy
& Gaudard, 1992; Ferrini-Mundy & Geuther-Graham, 1991; Kaput, 1994; Lauten,
Graham, & Ferrini-Mundy, 1994; Orton, 1983; Zandieh, 2000). “It is known that some
students are introduced to differentiation as a rule to be applied without much attempt to
reveal the reasons for and justifications of the procedure.” (Orton, 1983, p. 242) In fact,
many first semester calculus students survive without ever achieving a conceptual
understanding of the derivative. They can use the rules to find the derivative function and
use this to compute the desired answer. For example, when asked about the chain rule,
most students will simply provide an example of what it is rather than explain how it
works (Clark et al., 1997; Cottrill, 1999). “Calculus has come to be regarded as a course
(or sequence of such) rather than a collection of ideas that can be approached gradually,
involving many different perspectives, levels, and representational tools, as is the case
20
with geometry, for example.” (Kaput, 1994, p. 78) I will now attempt to briefly describe
what is known about how students understand the concept of derivative and the role of
the concept of derivative in solving related rates problems.
2.4.1 A Derivative Framework
Zandieh (2000) constructed a framework that may be used to assess what students
understand about the concept of derivative. Her framework consists of two main
components: multiple representations/contexts and layers of process-object pairs. This
framework is intended to be a “map of the territory” which allows one to see where
students are in their understanding of the derivative. It is not intended to be
developmental or hierarchal in nature. She examined the multiple representations of the
derivative using the following categories: graphical (slope), verbal (rate), paradigmatic
physical (velocity), symbolic (difference quotient), and other. Her three layers of
understanding are ratio, limit, and function. These layers consist of process-object pairs
and are used to describe what a student understands about the concept of derivative in
terms of the multiple representations. The process-object pairs come from using Sfard’s
(1992) framework of operational and structural understandings. She described the
following chain of understanding for the derivative:
The ratio process takes two objects (two differences, two lengths, etc.) and acts by
division. The reified object (the ratio, slope, velocity, etc.) is used by the next
process, that of taking a limit. The limiting process “passes through” infinitely
many of the ratios approaching a particular value. The reified object, the limit, is
used to define each value of the derivative function. The derivative function acts
as a process of passing through (possibly) infinitely many input values and for
21
each determining an output value given by the limit of the difference quotient at
that point. The derivative function may also be viewed as a reified object, just as
any function may. (Zandieh, 2000, p. 107)
Thus, Zandieh created the following chart that may be used to asses what an individual
student understands about the concept of derivative.
Table 3: A Derivative Framework
Graphical (Slope)
Verbal (Rate)
Physical Paradigmatic (Velocity)
Symbolic (Difference Quotient)
Other
Ratio
Limit
Function
If a student uses an object without regard for its internal structure, this is what Zandieh
calls a pseudo-object. For example, if a student refers to the limit value without
consideration for the limiting process and the epsilon-delta definition, the student is
operating with a pseudo-object. In coding student data, she used an open circle to denote
that a student had demonstrated at least a pseudo-object understanding and a filled in
circle to denote that the student demonstrated an understanding of the process involved in
the layer and context. Other researchers have investigated smaller pieces of this
framework in their attempts to describe what students understand about the concept of
derivative and how this may inform future teaching strategies for the concept of
derivative (Asiala et al., 1997; Cornu, 1991; Davis & Vinner, 1986; Ferrini-Mundy &
Gaudard, 1992; Ferrini-Mundy & Geuther-Graham, 1991; Orton, 1983; Tall & Vinner,
1981; Thompson, 1994a, 1994b; Williams, 1991; Zandieh, 2000).
22
2.4.2 The Derivative in Related Rates Problems
Related rates problems often require the student to move flexibly between the
verbal, physical paradigmatic, and the symbolic columns of Zandieh’s (2000) framework.
These problems are typically presented in a verbal context that must be interpreted in
terms of a physical situation and represented by symbols that are to be manipulated to
obtain a solution. However, White and Mitchelmore (1996) found that students appear to
focus on the symbolic column when they solve related rates problems.
White and Mitchelmore (1996) studied students’ understanding of related rates
and extrema problems. From direct instruction and the use of examples, students often
find they can solve problems with what White and Mitchelmore term a
…manipulation focus, in which students base decisions about which procedure to
apply on the given symbols and ignore the meaning behind the symbols. […]
manipulation focus errors were not just bad luck, but that students were actively
looking for symbols to which they could apply known manipulations. (p. 88)
This was based on their work with students solving calculus word problems that
involved rates of change, related rates, and maximization/minimization. Students were
asked to solve one of four variations of the same four problems. The variations were on
the amount of information students had to extract from the problem context. The
extremes being, no information had to be extracted (the problem was presented in a
strictly symbolic format) and having to extract all the information (no symbols, only a
contextually worded problem). The results of their study report that students have weak
conceptions of variable and function which make it difficult for them to accurately
23
represent the problem situation when they must extract all the information from the
contextually worded problem.
Two other forms of manipulation focus emerged. One other form of the
manipulation focus White and Mitchelmore (1996) describe is that of the x,y syndrome in
which students remember a procedure in terms of the symbols first used to introduce the
concept without understanding the meaning of the symbols. A third form of manipulation
focus they identify is that students fail to distinguish a general relationship from a
specific value. They also note that students with a “manipulation focus have a concept of
variable that is limited to algebraic symbols; they have learned to operate with symbols
without any regard to their possible contextual meaning.” (p. 91) Thus, it is possible for
students to be adept at calculating derivative functions and using them to compute
answers without any deep understanding of the concept of derivative.
2.5 Related Rates
Problems that incorporate the concepts of calculus such as those of related rates
emerge as a source of frustration for students and pedagogical complexity for instructors
(Martin, 1996, 2000; Orton, 1983; Selden, Selden, & Mason, 1994; White &
Mitchelmore, 1996). Related rates problems require the student to investigate the
relationship(s) between two or more changing quantities, one of which is unknown and
needs to be found. The relationships between the variables tend to fall into two
categories: 1) the variables are related through function composition, as in the case of the
trough problem, or 2) the variables are related parametrically through a third variable,
usually time, as in the case of the plane problem. The ability to reconceptualize the static
24
variables that occur in the geometric formulas that represent these problem situations as
functions of time appears to be crucial to the student’s success.
Martin (2000) conducted a study investigating students’ difficulties with
geometric related rates problems. In attempting to understand students’ difficulties with
these problems, Martin broke down the procedure for solving them into seven steps. She
then classified these steps as either conceptual or procedural as outlined in Table 4.
Table 4: Martin’s Step Classification
Step Description Classification
1 Sketch the situation and label variables Conceptual
2 Summarize the problem and identify given and requested
information
Conceptual
3 Identify the relevant geometric formula Procedural
4 Implicitly differentiate the geometric equation Procedural
5 Substitute specific values and solve Procedural
6 Interpret and report results Conceptual
7 Solve an auxiliary problem (e.g. solve a similar triangles
problem before attempting to use the volume-of-a-cone
formula to relate the variables)
Varies
In her study, Martin found that the problems that appeared to be the easiest for
students were the ones that required only the selection of the appropriate geometric
formula, differentiation, substitution, and algebraic manipulation. The most difficult
25
questions were those that required Step 7, solving an auxiliary problem. She also
indicated that the conceptual steps are more difficult for students than the procedural
ones. However, Martin concluded that students’ poor performance on these types of
problems was linked to difficulties with both procedural and conceptual understandings.
Simon (1996) indicated that transformational reasoning is a critical component of
mathematics learning and understanding. He states, “Central to transformational
reasoning is the ability to consider, not a static state, but a dynamic process by which a
new state or a continuum of states are generated.” (p. 201) Transformational reasoning
applied to related rates problems allows students to create a mental model that can be
mentally manipulated to see and understand relationships between changing quantities in
the problem, thus allowing them to construct a meaningful diagram of the situation (Step
1).
While Step 3 is listed as procedural, it embeds the conceptual component of
function composition. Sfard (1992), Engelke, Oehrtman, and Carlson (2005), and Carlson
(1998) provide evidence that first-semester calculus students usually do not have an
object view of function composition and experience frustration when confronted with
composition problems. (See section 2.3 for more detail about function composition.)
Step 4 also has an embedded conceptual component, that of the chain rule, which
Clark, et al. (1997) have investigated. Clark et al. sought to understand how students
construct their understanding of the chain rule, what component concepts are needed for
this construction, and how students recognize and apply the chain rule in various
mathematical situations. They classified the students’ understandings into three stages:
Intra, Inter, and Trans. This framework was based on Piaget and Garcia’s work,
26
Psychogenesis and the History of Science (1989). The Intra stage is characterized by the
student focusing on a single object in isolation from others (each instance of the chain
rule is unique); the Inter stage is distinguished by recognizing the relationships between
different actions, processes, objects, and/or schemas (recognizing that there are
similarities between instances of the chain rule); and the Trans stage is described by the
construction of a coherent structure that includes the link between function composition,
decomposition, and inverse (knowing how composition and the chain rule are related).
In an 11-task questionnaire administered in Clark et al.’s (1997) study, one
question involved implicit differentiation and another was a related rates problem. When
asked to find the derivative of Axyyx =+ , only 39% of the students were
successful, and only 34.1% of students were able to correctly solve the following ladder
problem (similar to the one described in section 2.4): A ladder A feet long is leaning
against a wall, but sliding away at the rate of 4 ft/sec. Find a formula for the rate at
which the top of the ladder is moving down the wall. In addition to demonstrating a weak
function conception, the students in their study exhibited characteristics similar to what
White and Mitchelmore (1996) had described as a manipulation focus when solving these
problems. The students were much more focused on manipulating symbols to get an
answer rather that attending to what the symbols represented.
As has been elaborated in the above literature review, solving related rates
problems is complex. It requires an ability to construct a mental model of a real situation
and represent that situation using the various function representations. It also requires one
to have a robust and flexible understanding of the concepts of variable and function.
Understanding the concept of the derivative, especially the chain rule, in conjunction with
27
being able to think about the variables as functions of time is necessary for the student to
understand where the dt comes from in the solution. Recall the trough problem and its
solution:
A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3
ft across at the top and have a height of 1 ft. If the trough is filled with water at a
rate of 12 ft3/min, how fast does the water level rise when the water is 6 in. deep?
(Stewart, 1991, p.164)
28
Table 5: Steps to Solve a Related Rates Problem
Steps Taken to Solve the Trough
Problem
Generalized Step
Draw a picture of an isosceles triangle and label the known and unknown quantities.
Interpret the words in the problem to form a mental image of the situation, draw a diagram of the situation, and label the known and unknown quantities.
Define what is known:
bhlVldt
dV
2
1,10,12 ===
Define what is known.
Determine what rate is asked for in the
problem: dt
dh
Define what rate is asked for in the problem.
Note that the solution requires a formula that defines volume in terms of height:
________ honlyV =
Identify which two of the varying quantities need to be related in the formula: one known and one unknown.
Imagine the trough filling with water. Notice that other quantities that are varying in the situation.
Imagine the system changing (engaging in covariational reasoning, to be defined later).
Notice that the height and width of the water are changing.
Note the quantities that are changing.
Define which quantities are changing: b and h
Define which quantities are changing with variables.
Define the volume of the water in the trough in terms of the changing quantities:
bhV 5=
Write a formula that relates the changing quantities in the situation.
Eliminate b in the volume formula by defining b in terms of h: bhV 5=
Is needed: 1
3=
h
b
Substitute hb 3= into the volume formula
to obtain: ( )( ) 21535 hhhV ==
Eliminate unnecessary variables if needed. If needed, determine the relationship between 2 changing quantities. Make the substitution (employ function composition).
Differentiate with respect to time:
dt
dhh
dt
dV30=
Differentiate
Plug in values and solve for the unknown rate.
Substitute in values and solve for the unknown rate.
29
A students’ understanding of the concepts of variable, function, and derivative
may directly impact their solution procedure. The ability to engage in covariational
reasoning also appears to be emerging as an integral part of their ability to successfully
solve a related rates problem. Solving a related rates problem requires that the student
engage in covariational reasoning to understand how the problem works, construct a
mental model that allows them to recognize which variables are changing, construct a
meaningful relationship between the changing quantities (create an appropriate
formula),and reconceptualize the variables in their formula as functions of time. Only
then may they use the chain rule to correctly differentiate their formula with respect to
time and solve for the desired variable.
Chapter 3: Theoretical Perspective
From the literature review, it appeared that some of the major obstacles students
encounter in solving related rates problems are: immature conceptions of variable and
function; an inability to construct an accurate mental model; not understanding the role of
the chain rule (Clark et al., 1997; Orton, 1983); and manipulating symbols that have no
meaning (Sfard, 1992; White & Mitchelmore, 1996). These obstacles formed the draft of
a framework around which instructional activities were developed, intending to enhance
the introduction and instruction of related rates. My dissertation further develops this
framework and the instructional activities as the result of a teaching experiment.
3.1 Transformational Reasoning
Most mathematicians are aware that they use inductive and deductive reasoning
strategies to solve mathematics problems. There is another type of reasoning that they
likely engage in without a conscious effort, transformational reasoning. To successfully
solve mathematics problems, Simon (1994, 1996) argued that transformational reasoning
is an integral part of the problem solving process. He described transformational
reasoning as:
Transformational reasoning is the mental or physical enactment of an operation or
set of operations on an object or set of objects that allows one to envision the
transformations that these objects undergo and the set of results of these
operations. Central to transformational reasoning is the ability to consider, not a
static state, but dynamic process by which a new state or a continuum of states are
generated. (Simon, 1996, p. 201)
This type of reasoning may range from trivial to extremely powerful in nature and
tends to be most powerful when a range of potential situations is considered. Simon
31
proceeds to state that recognizing the appropriateness of engaging in such a process is an
essential part of transformational reasoning. Not engaging in transformational reasoning
makes it difficult for students to visualize the situation, identify relevant variables, and
thus construct meaningful relationships. Simon has also noted that a primary affective
consequence of transformational reasoning is that it gives one a “sense of understanding
how it works.”
Simon gave the following example of a student engaged in transformational
reasoning. In a lesson on isosceles triangles, the teacher asked Mary to make an isosceles
triangle by specifying two angles and the included side. Mary responded, “Well, I know
that if two people walked from the ends of this side at equal angles towards each other,
when they meet, they would have walked the same distance.” When questioned about
what would happen if one person walked at a smaller angle, Mary replied, “Then that
person would walk further before they meet.” Thus, Mary has demonstrated that she
understands the relationship between the legs of an isosceles triangle and its base angles.
It appears that Mary has constructed a mental model of the situation. She views it as a
dynamic process that generates triangles from the ends of a base segment and she has
shown that she has an understanding of “how it works.” One could also argue that Mary
used covariational reasoning because she is able to coordinate two changing quantities,
the lengths of the triangle’s legs. Transformational and covariational reasoning are
closely related and both appear to be necessary for solving a related rates problem.
3.2 Covariational Reasoning
Covariational reasoning is employed when making sense of dynamic functional
relationships, a coordination of the inputs and outputs over some interval (Carlson,
32
Jacobs, Coe, Larsen, & Hsu, 2002; Hitt, 1998; Saldanha & Thompson, 1998). Carlson et
al. (2002) suggested that covariational reasoning involves similar cognitive actions as
transformational reasoning.
This type of reasoning has been shown to be employed when constructing meaningful
graphical representations of functional relationships (Carlson et al., 2002; Saldanha &
Thompson, 1998). When asking students to interpret graphical data, misconceptions
about the dynamic nature of the concept of function emerged (Carlson, 1998; Carlson et
al., 2002; Monk, 1992; Monk & Nemirovsky, 1994; Saldanha & Thompson, 1998).
Students can usually interpret a single point on a graph correctly, but have difficulty
when asked to consider a range of values for the input and the corresponding output. The
action view of function exhibited by the subjects in these studies suggests that students
view functions as static, not dynamic, entities (Breidenbach et al., 1992; Carlson, 1998;
Carlson et al., 2002; Dubinsky & Harel, 1992; Monk, 1992; Monk & Nemirovsky, 1994).
Using a typical related rates problem, Monk (1992) investigated how students
interpreted graphical representations of function by asking students two types of
questions. Pointwise questions ask students to interpret the input and output of a
particular point on the graph. Across-time questions ask students to interpret the graph’s
behavior over an interval. As one example, Monk asked students a series of questions
about a ladder sliding down a wall. Building up to something similar to the traditional
ladder problem of related rates: A 17 ft. ladder is leaning against a wall. If the bottom of
the ladder is pulled along the ground away from the wall at a constant rate of 5 ft/sec,
how fast will the top of the ladder be moving down the wall when it is 8 ft above the
ground? Pointwise questions were easier for students than across-time questions.
33
There is evidence that when students interact with physical models that challenge
their beliefs, misconceptions can be improved and/or rectified (Bowers & Doerr, 2001;
Monk, 1992; Monk & Nemirovsky, 1994; Pea, 1985; Saldanha & Thompson, 1998).
Monk found that when students interacted with a physical model to assist in visualizing
problem situations such as the ladder problem stated above or to create a flow rate graph
for air in a vacuum tube, their graphical interpretations of the situation were improved
(Monk, 1992; Monk & Nemirovsky, 1994). Saldanha and Thompson (1998) had similar
results when they investigated how a student would interpret a graphical situation after
having the student interact with a computer program that allowed him to track and
describe the behavior of distances between a car and each of two cities as the car moves
along a road.
Saldanha and Thompson’s (1998) work on covariational reasoning provided a
three phase model: engagement, move to representation, and move to reflection. Their
model made more explicit Monk’s (1992) pointwise and across-time definitions to
describe students’ graphical understanding of functions. Carlson et al. (2002) extended
Saldanha and Thompson’s (1998) work on covariational reasoning and provided a
framework which has five levels of understanding and is based on five observable mental
actions. Mental actions 1-2 are an extension of Saldanha and Thompson’s engagement
phase and are similar to what Monk described as the reasoning needed to answer a
pointwise question. Mental actions 3-5 are an elaboration of what Saldanha and
Thompson described as the move to representation phase and are similar to what Monk
termed necessary to answer an across-time question. Table 6 below provides a brief
34
summary of the mental actions and how one would likely be thinking about the above
ladder problem:
Table 6: A Covariational Framework
Mental
Action
Description
MA 1 Coordinate the value of one variable with changes in the other. (As the
base of the ladder is moved, the top of the ladder moves.)
MA 2 Coordinate the direction of change of one variable with changes in the
other. (As the base of the ladder is pulled away from the wall, the top of
the ladder slides down the wall.)
MA 3 Coordinate the amount of change of one variable with changes in the
other variable. (If the base of the ladder is moved x feet, the top of the
ladder slides down y feet.)
MA 4 Coordinate the average rate-of-change of the function with uniform
increments of change in the input variable. (As the base of the ladder
slides away from the wall every x feet, the top of the ladder slides down
by r ft/sec.)
MA 5 Coordinate the instantaneous rate of change of the function with
continuous changes in the independent variable for the entire domain of
the function. (As the base of the ladder moves away from the wall at x
ft/sec, the top of the ladder slides down the wall at r ft/sec.)
35
Note that a student who has a higher-level understanding, say level 5, is not only able to
engage in MA-5 level covariational reasoning but also has all the previous mental actions
as well and may unpack them to use as necessary in solving a particular problem.
Covariational reasoning plays a fundamental role in solving mathematics
problems, particularly in calculus (Carlson et al., 2002; Ferrini-Mundy & Geuther-
Graham, 1991; Hitt, 1998; Monk, 1992). Understanding and coordinating rate of change
over intervals is essential for understanding the concepts of limit, derivative, and
accumulation. To understand the concept of limit, a student must recognize the dynamic
process of the “…values of a function approaching its limiting value as the values in the
domain approach some quantity.” (Cottrill, Dubinsky, Nichols, Schwingendorf, Thomas,
& Vidakovic, 1996 p. 6) The derivative can be viewed as a function that runs “through
infinitely many input values and for each determining an output value given by the limit
of the difference quotient at that point.” (Zandieh, 2000, p. 107). The fundamental
theorem of calculus hinges on the idea that there is an accumulation of a quantity (say
total distance) and the rate of change of its accumulation (speed and amount of time
passed) are directly related (Carlson et al., 2002; Thompson, 1994c).
More specifically, covariational reasoning plays an integral role in the solving of
word problems. Related rates problems require students to envision and play out what
happens given a particular situation. The situations in related rates problems usually
involve a geometric figure that is changing over time. A similar argument could be made
for maximization and minimization problems, such as the box problem. Students need to
be encouraged to use covariational reasoning in these problem situations.
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3.3 Mental Models
The prototype approach to problem solving as described in Section 2.3 does not
encourage students to engage in covariational reasoning or model construction. Research
suggests that successful problem solvers engage in covariational reasoning by
constructing mental models that allow them to play out multiple scenarios (Carlson &
Bloom, 2005; Greeno, 1991; Johnson-Laird, 1983; Newell & Simon, 1972; Polya, 1945;
Schoenfeld, 1992; Simon, 1996). An ability to consider multiple perspectives and
representations in conjunction with the ability to easily move between them also
facilitates the construction of useful mental models.
Johnson-Laird (1983) presented three types of mental representation:
“propositional representations which are strings of symbols that correspond to natural
language, mental models which are structural analogues of the world, and images which
are the perceptual correlates of models from a particular point of view.” (p. 165) Glasgow
and Papadias (1998) describe mental images as having a deep representation, a visual
representation, and a spatial representation. The deep representation is stored in long term
memory and contains all relevant information. Visual representation allows one to
recognize an object while spatial representation allows one to relate an object to other
objects. It is the mental models and the manipulation of them that corresponds to Simon’s
transformational reasoning. When solving word problems in mathematics, it is often
useful to create a mental model that may be manipulated to understand “how it works.”
We encourage our students to draw a diagram on their paper: an image that represents
one particular point of view.
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Visualization and representation are dynamic in nature and are an important part
of being able to solve word problems (Booth & Thomas, 2000; Carlson & Bloom, 2005;
Cifarelli, 1998; Gravemeijer, 1997; Greer, 1997; Hegarty et al., 1995; Lucangeli,
Tressoldi, & Cendron, 1998; Silver & Marshall, 1989; Simon, 1996; Vergnaud, 1998).
Students generally construct representations in one of two ways: a short-cut way based on
key words and phrases or a meaningful way based on a mental model (Hegarty et al.,
1995). The representation of mathematical ideas may carry valuable information or they
may be only of superficial value. Booth and Thomas (2000) argued that visualization
plays a valuable, integral role in solving word problems but that reliance on a concrete
image or diagram may cause a student to attach irrelevant details or false data to it.
Students need to be able to flexibly move from lower (visuo-spatial) to higher
(verbal/logical) cognitive schemas and vice versa. This is what they defined as cognitive
integration. It was stated that it was necessary, “to have developed good spatial skills in
order to analyze a picture, abstract from it the essential elements of the problem, mentally
convert these copies of reality into theoretical objects, and at each stage relating these to
the verbally stored conceptual ideas (p. 185).”
Booth and Thomas (2000) further elaborated that problem solvers need to be able
to move between a picture, a drawing, a diagram, and a theoretical figure (a drawing is
more easily related to reality – drawing actual tree like structures, while a diagram is
more stylized – trees represented by sticks). They represented the dynamic process of
moving from a perceived reality to a picture of reality to a diagram or drawing to a
theoretical mathematical figure with Figure 1 (p. 183).
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Figure 1: Visualization schema.
Booth and Thomas related this to Mason’s (1992) distinction between looking at a
diagram and working on a diagram. They quoted Mason:
…to get much educational benefit, students need to be active in processing
images; they need to work on the images, not just look at them…The point about
processing diagrams is that it is not just a matter of looking at them, but rather
looking through them. (p. 185)
It was further noted that students should have two abilities to be successful in solving
word problems: 1) the ability to translate the word problem into a diagrammatic
representation; and 2) the ability to interpret the relevance of a diagram to the problem
situation. (p. 185) This is particularly true when considering related rates problems in
calculus that involve a geometric situation that can be represented by a diagram.
3.4 Problem Solving
Working through a related rates problem often requires students to make multiple
attempts, necessitating the use of various problem solving behaviors and strategies. In
1992, Schoenfeld (1992) developed a problem solving framework based on five
categories: the knowledge base, problem-solving strategies, monitoring and control,
beliefs and affects, and practices. The knowledge base consists of what a person knows
(formal and intuitive: facts, definitions, algorithmic procedures, etc.), how it is
cognitively organized (categorization of problems by type, etc.), and how it is deployed
Perceived Reality
Picture of Reality
Diagram or Drawing
Theoretical Mathematical Figure
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(heuristics, guess, etc.). Problem-solving strategies, or heuristics, are rooted in Polya’s
(1945) How to Solve It and include things like analogy, auxiliary elements, decomposing
and recombining, induction, specialization, variation, and working backwards.
Monitoring and control are those behaviors that cause a person to plan, check and
reevaluate their problem solving activity. Essential to control is being able to recognize
when a strategy is failing. Students often choose a strategy quickly and pursue it “come
hell or high water.” Schoenfeld attempted to overcome this in his problem-solving course
by reserving the right at any time to ask the questions: What are you doing? Why are you
doing it? How does it help you? Beliefs and affect involve the beliefs held by students,
teachers, mathematicians, and “just plain folks” and the impact that has on one’s
approach to problem solving. Practices focuses on what is actually happening in the
classroom; how is problem solving being taught? He provides examples of classroom
environments that were “designed to be consonant with the instructors’ epistemological
sense of mathematics as an ongoing, dynamic discipline of sense making through the
dialectic of conjecture and argumentation (p. 363).”
Building on this and other work, Carlson and Bloom (2005) studied twelve
mathematicians and identified four phases of problem solving that were repeated in a
cyclic fashion: Orienting, Planning, Executing, and Checking, but they also identified a
sub-cycle in the planning phase, that of conjecture-imagine-evaluate. In the orienting
phase, one makes sense of the problem and organizes the data. The planning phase
requires that one use transformational reasoning abilities to make a conjecture, play it out
in the mind, evaluate it for reasonableness, and then adjust as necessary. Completion of
the problem occurs in the executing phase. Finally, at the checking phase the work is
40
checked for correctness and reasonableness. These data also supported that
mathematicians utilize resources, heuristics, monitoring, and control as an integral part of
their problem solving process. This resulted in the emergence of the multidimensional
problem solving framework in Table 7 below (Carlson & Bloom, 2005, p. 67).
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Table 7: Carlson and Bloom’s Multidimensional Problem Solving Framework
Phase
• Behaviors
Resources Heuristics Affect Monitoring
Orienting
• Sense making
• Organizing
• Constructing
Mathematical concepts, facts and algorithms were accessed when attempting to make sense of the problem.The solver also scanned her/his knowledge base to categorize the problem.
The solver often drew pictures, labeledunknowns and classifies the problem. (Solvers were sometimes observed saying, “this is an X kind of problem.”)
Motivation to make sense of the problem was influenced by their strong curiosity and high interest. High confidence was consistently exhibited, as was strong mathematical integrity.
Self-talk and reflective behaviors helped to keep their minds engaged. The solvers were observed asking “What does this mean”; “How should I represent this?”; “What does that look like?”
Planning
• Conjecturing
• Testing
• Strategizing
Conceptual knowledge and facts were accessed to construct conjectures and make informed decisions about strategies and approaches.
Specific computational heuristics and geometric relationships were accessed and considered when determining a solution approach.
Beliefs about the methods of mathematics and one’s abilities influenced conjectures and decisions. Signs of intimacy, anxiety, and frustration were also displayed.
Solvers reflected on the effectiveness of their strategies and plans. They frequently asked themselves questions such as “Will this take me where I want to go?”, “How efficient will approach X be?”
Executing
• Computing
• Constructing
Conceptual knowledge, facts and algorithms were accessed when executing, computing and constructing. Without conceptual knowledge, monitoring of constructions was misguided.
Fluency with a wide repertoire of heuristics, algorithms and computational approaches were needed for the efficient execution of a solution.
Intimacy with the problem, integrity in constructions, frustration, joy, defense mechanisms and concern for aesthetic solutions emerged in the context of constructing and computing.
Conceptual understandings and numerical intuitions were employed to reflect on the sensibility of the solution progress and products when constructing solution statements.
Checking
• Verifying
Resources, including well-connected conceptual knowledge informed the solver as to the reasonableness or correctness of the solution attained.
Computational and algorithmic shortcuts were used to verify the correctness of their answers and to ascertain the reasonableness of thecomputations.
As with the other phases, many affective behaviors were displayed. It is at this phase that frustration sometimes overwhelmed the solver.
Reflections on the efficiency, correctness and aesthetic quality of the solution provideduseful feedback to the solver.
In Lithner’s (2003) study of how calculus students reason through textbook
exercises, he proposed a framework to evaluate how students are reasoning about their
textbook exercises. Students engage in a reasoning structure that has the following four
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components: 1) problematic situation, 2) a strategy choice, 3) a strategy implementation,
and 4) conclusion. He also identified three reasoning types that play a role in parts 2 and
3 of the reasoning structure. He has labeled these types of reasoning as: plausible
reasoning, established experiences, and identification of similarities.
Plausible reasoning (PR) is:
i. Founded on intrinsic mathematical properties of the components involved in
the reasoning, and
ii. Meant to guide towards what probably is the truth, without necessarily having
to be complete or correct. (p. 33)
Reasoning based on established experiences (EE) is:
i. Founded on notions and procedures established on the basis of the
individual’s previous experiences from the learning environment, and
ii. Meant to guide towards what probably is the truth, without necessarily having
to be complete or correct. (p. 34)
Reasoning based on identification of similarities (IS) is:
i. The strategy choice is founded on identifying similar surface properties in an
example, theorem, rule, or some other situation described earlier in the text.
ii. The strategy implementation is carried through by mimicking the procedure
from the identified situation. (p. 35)
Lithner concluded that students spend a disproportionate amount of time on EE and IS
reasoning strategies without understanding the big ideas and relevant relationships. In
using EE and IS strategies, students focus on superficial features of the problem that do
not lead them to construct meaningful knowledge. It is not that we expect them to behave
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as mathematicians, but without plausible reasoning their progress toward understanding
mathematical concepts is impeded. For example, when students are engaged only in EE
and IS reasoning they may not recognize that ( )3sin x and xln are both composite
functions that are differentiated under the same mathematical principle, the chain rule.
Rather they see the functions as different because one involves the natural log and one
involves sine causing them to search for different procedural information. In solving a
related rates problem such as the ladder problem described previously, students tend to go
back to a ladder problem that was presented in class or in the textbook and attempt to
mimic each step rather than constructing their own mental model of the situation and
attending to its mathematical properties.
When calculus students are presented with related rates problems, the extent to
which they engage in these problem solving behaviors may play an integral role in their
success. In this study, I investigated the extent to which students engage in covariational
reasoning to construct mental models of the problem situation and to support their
problem solving processes. I also investigated how students construct their
representations and whether the use of a computer program to model the situations in
related rates problems has any impact on the students’ ability to model and solve these
problems. Students’ conceptions of relevant mathematical concepts such as variable,
function, derivative, and the chain rule were also studied in this context.
Chapter 4: Pilot Studies
The pilot study began with two rounds of individual interviews conducted in the
fall 2002 and spring 2003 semesters. These interviews were designed to uncover
conceptual difficulties students encounter when solving related rates problems. The
results of these interviews, described in sections 4.1 and 4.2, informed the design of my
instructional materials. The instructional materials were piloted in the fall semester of
2004 with students enrolled in a calculus seminar. Both first and second semester
calculus students were enrolled in this course. The calculus seminar was a one credit hour
course designed to supplement the students’ regular section of calculus. Section 4.3
covers the results of the pilot of the instructional materials.
4.1 Overview
The pilot study was designed with three open ended interview questions. There
were four students in the first round of interviews conducted in the fall 2002 semester.
These students were enrolled in a conceptually based calculus course. Following an
initial analysis of the data after two interviews, the interview protocol was refined to
include a problem to assess whether this situation was easier for students to visualize and
solve. There were four students who participated in a second round of interviews
conducted in the spring 2003 semester. These students had volunteered to participate in a
series of interviews throughout the semester. They were asked to solve the following
problems:
1. A plane flying horizontally at an altitude of 3 miles and a speed of 600 mi/hr passes
directly over a radar station. When the plane is 5 miles away from the station, at what
rate is the distance from the plane to the station increasing?
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2. A spherical balloon is to be deflated so that its radius decreases at a constant rate of
15 cm/min. At what rate must the air be removed when the radius is 9 cm?
3. Coffee is poured at a uniform rate of 20 cm3/sec into a cup whose inside is shaped
like a truncated cone. If the upper and lower radii of the cup are 4 cm and 2 cm,
respectively, and the height of the cup is 6 cm, how fast will the coffee level be rising
when the coffee is halfway up the cup?
The students were videotaped while working the problems, and the interviewer asked
questions to determine why the student chose to perform certain operations. The
interviewer also prompted the subject to provide clarification about their reasoning as
deemed necessary by the interviewer. The videotapes were transcribed and analyzed
using Strauss and Corbin’s (1990) open and axial coding techniques.
Strauss and Corbin’s (1990) grounded theory approach provides a detailed
description of how qualitative research can be scientific and rigorous. The primary
principle of grounded theory is that the theory will emerge from the systematic collection
and analysis of data. They outline several steps for the researcher to follow including:
open coding, axial coding, selective coding and theoretical sampling. Through these
steps, one can analyze the interview data that has been collected.
Open coding is the first step and requires the researcher to look for distinctive or
repeated traits in the transcripts. Initially, one should conduct a microanalysis on a small
portion of the data, examining closely each word and phrase that the student uses, and
considering why they were chosen. As the data is read and reread, the researcher assigns
codes to these phenomena. Codes may be terminology used by the individual or made up
by the researcher to describe the phenomena. The researcher begins to organize the codes
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around central themes and categories arise. These categories can then be described in
terms of properties and dimensions that describe the phenomena.
After open coding and the emergence of categories, axial coding takes place to
assemble the categories in a manner that describes the big picture. Actions and
interactions of the categories are explored, while relationships between the categories are
formed. Strauss and Corbin (1990) suggest attempting to diagram the relationships
between categories. From this, a central category should emerge.
Selective coding involves the researcher comparing his data to the data and results
of other researchers. At this point, the results should have a certain level of internal
consistency, and the theory should make sense.
Theoretical sampling occurs as a result of wanting to fill in any holes that may
exist in the data. Targeted interviewing, identifying outliers, and revisiting the original
data all fall under this umbrella. The goal is to be able to have categories that are
saturated, with no holes and explanations for any outliers. After theoretical sampling and
ensuring there are no holes in your theory, you are ready to write a complete, detailed
assessment of your research and results. While grounded theory may be time consuming,
the results it generates are soundly based on empirical evidence and are scientifically
rigorous.
The transcribed data of my pilot study were coded using the NVivo program.
NVivo allows one to assign codes to individual words or larger segments of text. These
codes may then be organized into tree structures, allowing the researcher to organize the
codes in a manner that may help to identify relationships. As a result of my open coding,
categories began emerging from how I structured my tree nodes. The categories that
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emerged were: diagramming (includes ideas related to transformational/covariational
reasoning), knowing the process (problem solving abilities and heuristics), algebraic
proficiency (includes ideas related to the concepts of variable and function), geometric
proficiency, knowledge of derivatives, and mathematical problem solving
attitudes/beliefs. The following table summarizes my codes by their emergent categories
and their descriptions:
Table 8: Pilot Study Data Codes
Code Description
Diagram The student draws a diagram of some kind. Is it appropriate?
Labeling Does the student label the diagram? If so, how does the student label the diagram?
Know the process Trying to identify how much the student knows about the process of solving for related rates. Do they know or use the steps as outlined in the book?
Want to find Are students able to identify what the problem is asking for?
Known Can the student identify what is known/given in the problem?
Unknown Can the student identify unknowns? How do they interpret variables?
Don’t want Does the student recognize irrelevant information? Does the student realize when he has found something that is of no consequence?
Identifies sub-problem The student identifies a sub-problem to be solved.
Irrelevant steps The student is performing irrelevant operations on a relevant or irrelevant formula.
Correct reasoning The student has figured out how to account for the shape of the cup either by subtraction, or figuring in the additional height.
Not knowing when problem solved
The student has calculated a number, but is not sure whether the question has been answered.
Algebra Identifying strengths and weaknesses in basic algebra skills.
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Fractions Identifying strengths and weaknesses relating to fractional computations.
Function Identifying if a student has a process view of function. Do they just remember certain catch phrases and graphs from the review?
Incorrect Input A student has a difficult time identifying the correct inputs for a function.
In terms of Identifying if a student knows that they want a formula that has one thing in terms of one other. Do they recognize to relate r and h?
Formula How does a student deal with an equation? Do they feel they need a formula for everything? Can they construct their own formula?
Plug in The student wants to plug into the formula, frequently too early in the process.
Derivative Identifies how often students use the term derivative (or a variation) and looking for understanding about derivative properties.
Chain rule Identifying if students understand how the chain rule fits into the problem.
Product rule Identifying when students feel the product rule is necessary.
Quotient rule Students feel the quotient rule is necessary.
Implicit differentiation Identifying how students view implicit differentiation.
Rate of change Identifying how students understand and use rate of change vocabulary.
Rate How do students use the term rate?
Relation Identifying how students identify which relation to use specifically with regard to r and h.
Geometry Identifying how students view geometry and what they know about geometry.
3D objects How students deal with the 3D objects. Does it create visual difficulties?
Dimension Identifying how students use dimension in their reasoning.
Volume Identifying how students understand volume. Specifically with regard to the cone, sphere.
Volume of cone
Volume of sphere
Volume of cup How does the nonstandard shape of the cup give students difficulty? How do they approach it?
Subtract small cone One approach to dealing with the shape of the cup.
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Extra height Can students identify the sub-problem here and solve it? How do they deal with this unknown?
Truncated cone Do students understand truncated?
Cylinder Students try to fit the cup into the context of a cylinder.
Circle The existence of the circular base of the cone causes them to know they have to have a pi somewhere.
Trig Trig is a good way to deal with triangles.
Triangle Triangles are very common. How do students approach them?
Isosceles triangle Isosceles appears to translate to similar.
Pythagorean theorem
Similar triangles Do students see similar triangles? Do they know what it means for triangles to be similar?
Ratio Are students comfortable with the use of ratios? Do they know what a ratio is?
Trapezoid Students view the cup as a trapezoid. Likely due to the swimming pool problem from class.
Rotation of trapezoid Identify the cup as a trapezoid that's been spun around.
Radius How do students deal with the radius of the cone? Of the fact the cup has 2 radii?
Cone What do students know about cones?
Angles The students feel they need to know something about the angles that occur in the cup/cone.
Thoughts about math problems Identifies how students think about math.
Flexible thoughts Does the student believe there is more than one approach to the problem?
Questioning Does the student ask questions of himself? Is he asking questions of the interviewer for clarification or help?
Should be easy Math problems are easy if you see it.
Textbook Students want to consult their textbook for formulas and shortcuts.
Wants to use everything given
Student feels they must use every piece of information given.
Redefine problem Students want to define the problem differently to make it easier.
Feelings
Confidence
Confusion
Monitoring
Persistence
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Units How important are the units in related rates problems?
Checks Does the student have a means to check if his answer is reasonable? Are the units incorporated into this?
Dynamical reasoning The student recognizes that the situation in the problem is dynamic.
For example, the following table illustrates how the transcript of Kim solving the
sphere problem was coded:
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Table 9: Example of Pilot Study Coding
Transcript Codes
Kim: (Reads question 1) So what we’re going to do is have the air be removed. We have a balloon and it’s spherical. So, what we have (draws picture of sphere) the radius is 9 cm and we’re giving the rate of the radius changing over time is 15cm/min. All right. Then, I think we’re just asking is … at what rate must the air be removed so we’re deflating it. So, we’re going to find the rate of the volume, rate of change of the volume over time. Is that what it’s asking?
Geometry - it’s spherical Diagram - draws picture of sphere Known - we have the radius is 9 cm and we’re giving the rate of the radius changing over time is 15cm/min Rate of change - rate of change of the volume over time Want to find - we’re going to find the rate of the volume, rate of change of the volume over time
Kim: So, we have to try to find dv/dt. We know volume of a sphere, dude, I so suck at geometry, but it’s 4/3 pi r^3, right?
Want to find - we have to try to find dv/dt Volume of a sphere - it’s 4/3 pi r^3 Confidence (lack of) – I so suck at geometry
Kim: Yeah. So that’s 4/3 pi r ^3 would be the volume. So, finding dv/dt, do the derivative so it’s 3 times 4/3, 4 pi r^2 dr/dt. Because they’re in different terms so you have to do the dr/dt. Then we know the r at the time given is 9cm. So we can find dv/dt by plugging everything in. (Starts plugging in values.) And dr/dt is given too 15 cm/min. So this is time 15 cm/min. So, dv/dt equals 4 times 81 times 15 pi cm cubed per minute, and that’s how fast it should be removed.
Derivative - do the derivative so it’s 3 times 4/3, 4 pi r^2 dr/dt Want to find - finding dv/dt; we can find dv/dt by plugging everything in Chain rule - Because they’re in different terms so you have to do the dr/dt In terms of - Because they’re in different terms Known - we know the r at the time given is 9cm; dr/dt is given too 15 cm/min Plug in - we can find dv/dt by plugging everything in Algebra - dv/dt equals 4 times 81 times 15 pi cm cubed per minute, and that’s how fast it should be removed
NVivo allows the user to view all the data that has been assigned one or more codes
simultaneously. This fosters the development of categories and allows the researcher to
explore relationships among categories. Again, the categories that emerged from my pilot
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study data were: diagramming (includes ideas related to transformational/covariational
reasoning), knowing the process (problem solving abilities and heuristics), algebraic
proficiency (includes ideas related to the concepts of variable and function), geometric
proficiency, knowledge of derivatives, and mathematical problem solving
attitudes/beliefs.
Diagramming involves the student being able to draw an appropriate picture for
the situation in the problem, labeling known/unknowns, and visualizing what’s
happening. The dimensions range from not being able to draw a picture to drawing an
accurate view of what is presented in the problem and using it appropriately.
Knowing the process refers to how students interpret a related rates problem. Do
they recognize it as such, and consequently, can they solve it appropriately? What is the
role of the book’s seven step procedure? The dimensions range from not understanding
anything about the process, except maybe having memorized the “5 step procedure”
given by the book to knowing and understanding the 5 steps in a general sense that allows
the student to apply them broadly. This would include understandings of the dynamics of
the situation and how the chain rule relates to these problems.
Algebraic proficiency is the ability of the student to manipulate the algebraic
formulas in solving these problems. It includes proficiency with fractions and functions.
The dimensions range from struggling to solve algebraic equations, particularly when
fractions are involved, to readily seeing that you want your formula to involve just one
variable and easily simplifying formulas before simplification.
Geometric proficiency refers to the students being able to visualize the
appropriate situation and determine the correct relation to use in the given situation. The
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dimensions range from the student not being able to tell whether to use a volume or area
formula or even the correct “shape” to being able to determine the correct “shape” and
thus the correct “dimensionality.” It also involves being able to see the subtle sub-
problems that occur as in the similar triangles in the cone.
Knowledge of derivatives indicates how well the student understands the concept
of derivative, particularly with regard to implicit differentiation and the chain rule. The
dimensions range from knowing how to mechanically take a derivative (possibly
forgetting the chain rule) to understanding the derivative as a rate of change and
understanding how the chain rule fits into related rates problems.
Mathematical problem solving attitude/beliefs refers to the students’ attitudes that
they bring to the table. The students frequently commented on the problems being hard
or easy if you see it. They also commented that they hate geometry or are not really good
at math. There were students who were determined to get the right answer. The
dimensions here range from very negative to very positive thoughts about mathematics
and problem solving in general.
4.2 Initial Results
Students had particular problems recognizing when to use the similar triangle
relationship in the cup problem; they did not understand the power of substitution and
function composition; and they were not effective in determining what algebraic
procedures to implement to determine the most appropriate defining relationship. This
was true for at least five of the eight students who participated in the first two rounds of
clinical interviews. Computational errors led to incorrect solutions; geometric
54
misconceptions led to incorrect models. For example, in the transcript excerpt below, Jen
wrestled with whether air is the same thing as volume in the balloon problem.
Excerpt 1
1. Jen: Ok. So, I’m going to start by implicitly differentiating that ‘cause I
don’t know what else to do (laughs). And then, we want to know, radius decreases, I know dr/dt, and I want to do when r = 9. So, then I’m solving, but then I’m solving for dv/dt, right?
2. INT: Why do you think you’re solving for dv/dt?
3. Jen: At what rate must the air be removed? So, that would mean, that the volume, that the rate at which the volume is getting smaller?
4. INT: Ok.
5. Jen: Or, when the radius is 9 cm, yeah, because they gave us the rate at which the radius decreases. Hmm. (makes an unhappy face)
6. INT: What do you not like right now?
7. Jen: That I don’t know dv/dt. Like I don’t know what I’m solving for. Actually, like if I solve for dv/dt, then I’m going to need to do something else, I think. Because I want to know the rate at which air needs to be removed. And I’ll just know, like if I put in, the things I know, then I’m just going to get the rate at which the volume decreases, but how would I find the rate at which air needs to be taken out? Or are they the same thing and I’m just making this complicated?
8. INT: Do you think the air is the same thing as the volume?
9. Jen: Well, that’s what the balloon is filled with. That’s why I guess maybe it’s the same thing. Must the air be removed…
10. INT: Tell me more about this conflict that you have.
11. Jen: Well, I want to know the rate at which air needs to be removed, but I don’t know if that’s the same as the volume, the rate at which the volume is decreasing. Or does, no. It has to be right, because I don’t have anything else to…unless I’m totally off. I don’t have anything else to work with. Like no other numbers they give. Like I thought maybe it would be dr/dt, but they give me dr/dt so I can’t be solving for that. When the radius is 9 cm. I don’t know, cause they want to know, either the rate at which it’s decreasing or the rate at which air is being removed to keep the radius decreasing at a constant rate. I have to keep 15 and I have to keep 9, so I must be solving for dv/dt.
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It appears Jen was thinking about how the problem can be solved using her internalized
procedure, but there were cognitive conflicts in how she understood volume and her
experience with balloons. Jen did not appear to construct a mental model. She did draw a
circle on her paper and wrote down the formula for the volume of a sphere, but her next
step is to differentiate. Without a robust conceptual structure of the problem, followed by
the active engagement of her mind in covariational reasoning, she was only able to apply
a procedural approach to the problem.
Students appeared to focus on the three procedural steps that Martin outlined.
They generally drew a diagram and labeled the constants, chose a formula and
differentiated it, then substituted in values. This abbreviated procedure worked well on
standard problems, particularly ones that do not require an auxiliary problem to be solved
as in the balloon problem. The students did not appear to have much difficulty with the
plane and balloon problems; this may be the result of having seen similar problems
before.
The results also suggest that one of the critical components of solving related
rates problems is being able to visualize the dynamics of the situation. Students appeared
to be proficient at drawing an appropriate diagram. However, difficulties arose when
students began their labeling phase. Students labeled their diagrams with the constants
given in the problem; however, no attention was given to the quantities that are changing.
After drawing a diagram, students immediately looked for a geometric formula that
would fit the situation, differentiate it, and substitute in values. The focus was on the
procedural steps. What became apparent, particularly in the cup problem, was that
students did not typically account for the general relationships, for example between the
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radius and height of the cone. They did not appear to have a model that indicated these
two quantities were continually changing in relation to each other; revealing that students
were not actively engaged in covariational reasoning.
When the students did not know an explicit formula off the top of their heads,
they would ask the interviewer or want to consult their text. The students lacked
confidence about their ability to solve these problems during the interviews. As is
evidenced in the data from the cup problem in the interview, students frequently were
frustrated when they could not find a formula for that particular shape. When the book
did not provide an explicit formula, they frequently went to an alternate idea such as
viewing the cup as a trapezoid that had been spun around. After an alternate idea didn’t
seem to pan out, students were discouraged and some were ready to give up. In almost
every case, the interviewer needed to prompt the student to consider the entire cone
before any real progress could be made.
When a problem required the student to think beyond the classic examples from
class discussion and homework, the abbreviated procedure of choosing a formula,
differentiating it, and substituting in values did not work well. Difficulties arose for
students when a nonstandard question was posed. This is supported by the interview
excerpts below.
Excerpt 2
Betty: Ok, um, so then, how fast will the coffee level be rising when the coffee is
halfway up the cup? Um, so that would make the height 3 cm. So, first thing we have to do is take the derivative. Um, we want to find the change in the height. So, um, ‘cause we know the volume, the derivative of the volume. So, first we’ll solve for the height cause that’s what we’re looking for.
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Rather than attending to the dynamic nature of the problem (engaging in covariational
reasoning), Betty initiated her solution procedure by differentiating her chosen formula.
Betty demonstrated a manipulation focus; she actively sought out a formula on which she
could perform known operations and manipulate variables to find an answer. She did not
construct a mental model of the situation. Later in the interview, she appeared to
experience cognitive dissonance when her attempt to use the chain rule resulted in the
realization that one could not have two variables in the formula.
Excerpt 3
1. Betty: Then I … don’t need to solve for this separate derivative of r. ‘Cause I
think you still need to apply the chain rule, I think, don’t you?
2. INT: How are you applying the chain rule?
3. Betty: To that one, right? (pointing to right side of formula)
4. INT: Um hmm.
5. Betty: Well, we don’t want, this is hard. We don’t want two variables in our formula…
6. INT: No?
7. Betty: So we need to get rid of one of them. And I know I want the derivative of h, because how fast the height is changing. Then wait, I’m not, no I am cause how fast the level of coffee will be rising when the coffee is halfway…so I know the h is 3 here, but I want to be able to put that in.
Betty’s difficulty appears to be a result of having a static image of the problem. She was
able to select an appropriate formula for the situation, the formula for the volume of a
cone. However, she believed that she could not differentiate a formula with two variables.
In either case, she did not appear to link the chain rule to the concept of derivative and
rate of change. She may have recognized the need to use the chain rule or she may have
been ventriloquating terminology from class. She does not appear to have parameterized
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each variable as a function of time. Rather, she is focusing on the fact that she wants h =
3. Her diagram was not useful in helping her to engage in covariational reasoning
because she did not appear to be able to identify and relate the radius and height of the
cone as the relevant variables in the situation.
Another student had difficulty solving the cup problem. He did not draw diagrams
unless specifically asked about a diagram by the researcher as is evidenced in the
following transcript excerpt:
Excerpt 4
1. Mono: (Reads question 2 – cup problem) Truncated? What the heck is a
truncated cone?
2. INT: Ok. If you had a whole cone, you know what a whole cone looks like, it has a point and comes down in a circle, if you truncate it, you cut off the point.
3. Mono: Oh, ok, I gotcha. Ok. (Finishes reading problem) Ok, this is a volume problem.
4. INT: Ok.
5. Mono: Crap.
6. INT: It is a volume problem.
7. Mono: Wait a second, I know this. If I was doing similar triangles, that would be…no, I did this, this was in the homework. Um, what if, no that doesn’t help me. There’s some weird equation for a trapezoid or something. I’m going to say length, length one minus length two times probably height, yeah I’m going to say height, time width is its volume.
8. INT: Ok.
9. Mono: Whoa. Is there a real equation for this?
10. INT: Yes, there is.
11. Mono: Dang, is this part close?
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12. INT: Rather than trying to jump right to a formula, what do you usually do when you see one of these problems? What’s like the first thing you do?
13. Mono: You saying like the rate, like dv/dt equals 20 cubic cm?
14. INT: Ok, what do you do even before that?
15. Mono: Oh, well, I could draw a diagram.
16. INT: You could, couldn’t you?
17. Mono: I could. I shall.
18. INT: Do you think that might help you?
19. Mono: It may. I’ll try that. Cup whose inside is shaped like a truncated cone. Bad diagram. Ok, my truncated cone. And the upper radii is 4 cm (actually makes it a diameter), and the lower is 2 cm, and the height of the cup is 6 cm. Then…this is going to be a tough problem.
His solution procedure for every problem involved immediately searching for a formula
that fit the situation, differentiate it, and substitute in the given values. This suggests that
he is not constructing mental models for the problems. He appears to have an abbreviated
form of the book’s procedure in his mind that he uses to solve these problems. His
procedure worked for the plane and balloon problems, but created a major obstacle when
attempting to solve the cup problem because there is no formula for the truncated cone.
One student, Harry, did solve the cup problem successfully. He tried numerous
approaches before reaching his solution. First he tried to think of it in terms of the entire
cone with the bottom removed and then decided that was going to be hard; there had to
be a “trick.”
Excerpt 5
Harry: And then, um, ah…I know there’s something to do with like, similar
triangles in this we could say, but then we’re going to lop off this little, this thing doesn’t exist for the liquid’s going in there all automatically, so I
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suppose…yeah. It would make sense to find the dimensions of this cone here, maybe with the full volume of it, and then I could subtract this part (the little part at the bottom), and then that would give me the volume of just the cup itself.
It is interesting that he recognized that similar triangles should play a role in the solution.
He next tried a trapezoid approach. Eventually, he decided that viewing the cup as a
whole cone with an adjusted height would provide useful insights for solving the
problem.
Excerpt 6
1. Harry: Well, originally I thought I couldn’t pour coffee into a regular cone,
because that’s going to change the answer, but it won’t because it’s not going to matter because the coffee is always coming in at a constant rate, and even if I poured it into this imaginary cone, and it kept going in there, that once it got to this level here (bottom of cup), and that’s kinda where the problem starts, you know what I mean?
2. INT: Uh, huh.
3. Harry: And then it’s going to increase, so I just need to change the number, so instead of finding the height is 3, halfway up there, I could just find it as, um, 9, on the whole cone.
Here we see something of an “Ah Ha” moment for Harry. He appears to have engaged in
some sort of transformational/covariational reasoning as evidenced by the fact that he has
tried several approaches to the problem and seen what happened when he tried
implementing each approach. In the above excerpt, it is evident that he now has an
understanding of “how it works” and how the solution must play out. While Harry does
solve this problem successfully, he did not begin the process with
transformational/covariational reasoning and must try numerous approaches before
coming up with an appropriate solution.
As Martin found, when students were required to solve an auxiliary problem, they
struggled more. It is the solving of an auxiliary problem that requires the student to
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investigate and understand the general relationship between changing quantities. This
suggests that the students are not using transformational/covariational reasoning in
solving the problems; they do not have mental models that can be manipulated to play out
the situation at hand, and thus merely manipulate known symbols without regard for their
meaning. This lack of transformational/covariational reasoning appears to foster students’
dependence on the procedural steps, which in turn highlights students’ deficiencies in
basic algebraic and geometric knowledge. Thus, the research suggests that students’
transformational/covariational reasoning abilities and how they apply them, particularly
in the diagramming phase, may be critical to the successful completion of a related rates
problem.
Three major student difficulties emerged from the data: algebraic and/or
geometric deficiencies, student fixation on the procedural steps, and failure to recognize
and consider general relationships (Engelke, 2004). Further analysis of the data revealed
the students’ inability to develop a mental model of the situation. This hindered their
ability to understand “how it works.” It appeared they did not actively engage in
transformational/covariational reasoning at the beginning of the problem, as is evidenced
by their construction of static diagrams which appeared to result in their relying on
procedural steps. Their reliance on implementing procedures without a rational
foundation were also evident in what often seemed to me their random use of algebraic
techniques and misguided geometric associations.
4.3 Development of Classroom Activities
Building on the work presented in section 4.2, my pilot study took place in a
calculus seminar in the fall 2004 semester. I piloted classroom materials consisting of
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three activities that may allow teachers to more effectively teach related rates problems.
The first activity had the students working in Geometer’s Sketchpad (GSP) to reconstruct
their knowledge of similar triangles. The second activity presented related rates problems
in a manner that was designed to draw attention to quantities that change and quantities
that do not change. The third was a puzzle that would be solved by successfully
completing fourteen standard textbook type related rates problems. If the class
successfully solved the puzzle, I rewarded them with homemade cookies. Copies of these
activities can be found in appendix B.
The GSP activity took about 25-30 minutes for most groups (1-3 students per
group) to complete. While the similar triangles material was review, it also served to
introduce the students to GSP. In the activity on related rates where students had to solve
a similar triangles problem to simplify their formula, they quickly recognized that the
similar triangles relationship would be a valuable tool. I now turn my attention to how
students interacted with the related rates activities.
I examined the discussions of how the groups were solving the related rates
problems in two groups: Groups 1A (Laura, Joel, Jill, Jeff) and 2B (Keith, Amy, Jay,
Mark). There was more evidence that students were likely not engaging in
transformational/covariational reasoning. It was also found that, for most students, the
chain rule as applied in related rates problems was a rule that must be applied without any
regard for the reason why it must be applied.
When the activity requested that students draw a picture of the snowball at three
successive points in time, students drew three circles in descending order, not concentric.
This may suggest that the students were viewing the snowball at these points in time as
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static and not part of a dynamic situation. Similar difficulties arose in the trough problem.
I also found evidence that the students were not engaging in covariational reasoning in
the subsequent puzzle activity. In the following exchange, when Jeff asks Laura how her
answer is different from his, Laura explains to Jeff and the rest of her group the
difference between a changing quantity and a constant quantity:
Excerpt 7
Laura: You um, you didn’t do, you forgot to put in your variables like you’re
supposed to ratio 20 to 5 and then over h to r and then you’re supposed to find r with the variable h so that gives you what you have cause you’re getting ½ for your r but it’s supposed to be for any h… when you do that you’re making h a constant when it’s not or you’re making r a constant when it’s not
Thus, it would seem that Laura was engaging in covariational reasoning while the rest of
her group was not. Laura recognizes that h may take on any value and that she needs to
account for the general relationship between h and r, as evidenced by her constructing the
ratio of h to r and pointing out that it is not a constant. The rest of her group had
proceeded to substitute in a value for r rather than attend to the general dynamic nature of
the problem thus computing a numerical value for the volume of the trough. Even though
Laura appears in this passage to be engaging in covariational reasoning at some level, she
has difficulty with the concept of the chain rule.
The application of the chain rule in the related rates problems was troublesome
for many students. In examining a partial transcript of Group 1A’s discussion (see
below), we find evidence that Laura believes the chain rule is involved in the implicit
differentiation step of the problem; however, she has a difficult time expressing what the
chain rule is beyond computation. She would be at what Clark et al. (1997) describe as
the Inter stage of development for the chain rule. For her, it is a general derivative rule.
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She can provide numerous examples knowing that they are related by the chain rule, but
she has not developed a schema for the chain rule that involves function composition and
decomposition.
Excerpt 8: Group 1A
1. Laura: … ok implicit differentiation time
2. Joel: Alright
3. Laura: How does it work? I can’t remember
4. Joel: We know
5. Laura: Yeah that’s right
6. Joel: Wait we don’t dv dt is 1 = pi times 10 over 2 dd dt
7. Laura: For some reason I thought it had to be ½ dd dt
8. Joel: Is that right?
9. Laura: I don’t know, but don’t you want to use like the chain rule and you have to do the derivative of the inside so it would be ½ dd dt?
10. Joel: Yeah
11. Jeff: Just make it u or something like that, is that what you’re saying? do that, do it that way the u f u thing substitution I think it was called
12. Laura: Yeah but I don’t think that’s
13. Joel: What are you saying?
14. Laura: Like when we do the chain rule for d halves then like this is
15. Jeff: That’s what it’s called
16. Laura: Then don’t you have to multiply it by… this is really hard… because don’t you have to do the half? (pause) does that make sense?
17. Joel: I don’ know are we just saying d over 2 is like a variable x?
18. Jeff: Yeah
19. Laura: I always thought you had to do the chain rule for it
20. Jeff: Yeah like that? Is that what you’re talking about?
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21. Laura: Yeah that’s how I did it
22. Jeff: Cause d is the variable
23. Laura: Right, actually
Jeff has picked up on the chain rule discussion, but it is not clear if he is thinking about
the same thing as Laura. At this point, he throws out substitution as a possible name for
this approach to the problem. What does this indicate about his concept image of chain
rule? Is it conflated with u substitution? We cannot say anything either way on this as he
quickly says, “That’s what it’s called,” when Laura again says “chain rule” in response to
Joel’s request for clarification. There is no evidence to support that any of these students
have an understanding of the chain rule beyond the Inter stage because when pushed to
further explain what the chain rule is, they could not supply clarification beyond more
examples.
When pushed to further explain the chain rule and implicit differentiation, Laura
does not make significant progress and the rest of the group leaves it to her to provide the
answer to the questions:
Excerpt 9: Group 1A
1. INT: A moment or so ago I heard you say something about the chain rule.
Can you tell me what that was about?
2. Laura: Um we were implicit ly [sic] differ, we were doing this [struggling to say implicitly differentiating]
3. Jeff: Taking the derivative of that
4. Laura: Yeah
5. INT: Ok
6. Laura: I was like uhhhehh um we had to do d over 2 for r cause r is the formula we know and so we were differentiating we differentiated what were, like we differentiated all of this and then we had to differentiate the inside by taking a half times dd dt is that right?
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7. INT: Why?
8. Laura: Why? Why the chain rule?
9. INT: Why the chain rule?
10. Jeff: Because you have to
11. Laura: Because my calculus teacher told me to [laughs]
Group 2B had similar initial responses to the researcher’s request for further explanation
about the chain rule and implicit differentiation:
Excerpt 10: Group 2B
1. INT: So what does it mean to implicitly differentiate?
2. Jay: Find the derivative of both sides, that’s what differentiate is
3. Mark: Yes
4. INT: Ok
5. Keith: With respect to time
6. INT: With respect to time how do you know to do that?
7. Keith: Because Professor O said so
8. INT: What do you know about the chain rule?
9. Amy: Multiply everything by
10. Keith: Well it tells you how to get one of the derivatives but
11. Amy: Yeah
12. Keith: I don’t really see what it has to do with implicit differentiation
13. Amy: It’s like f prime times g of x plus g of prime times f of x
In this transcript excerpt from Group 2B, again the students say the chain rule is just
something that must be applied because that’s the way it was taught. Additionally, Amy
appears to have the chain rule confused with the product rule. It would seem that the
students understand that the chain rule, or at least that the addition of the symbol dd/dt is
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a necessary step to solving the problem, but it’s just “because my calculus teacher told
me to” (Laura in Group 1A) and “because Professor O said so” (Keith in Group 2B). As
the conversations continue, it becomes evident that the connection between the chain rule
and its role in related rates problems is weak at best for the students.
When trying to explain the chain rule, the students in the Group 1A transcript
provide additional examples beyond what is in the current problem, but they do not
articulate a generalized rule. Here, I also found there is reason to question what students
understand about the concept of variable. When writing out an example for the chain rule,
Laura uses an expression of the form: ( )534 ...4 ++ xx . Jeff interjects, saying that this is
not in one variable, but rather that 4x is one variable, 34x is one variable, and so on. Jeff
then proceeds to relate the d in the problem to distance before being reminded by another
student that it is diameter. It may be that his understanding of the variable d is that it must
always represent distance.
Group 1A’s conversation now examines more closely why the chain rule must be
used in these problems. They realize that they do this in physics problems as well. It
appears that this is merely a step in the problem for these students; they are simply
writing a symbol. This suggests that in attending to these problems, the students have a
manipulation focus.
Excerpt 11: Group 1A
1. INT: Right, you’ve said things about the chain rule and how when you
implicitly differentiate you just have to put it there
2. Jill: Because of the definition of
3. Jeff: Because we’re trying to find the change in what is it distance with respect to time
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4. Joel: Time
5. Jeff: So yeah change in diameter that’s what
6. Laura: Yeah we have to, you don’t get this in your formula then you can’t solve for it that’s what we’re trying to find, I don’t know why
7. Jeff: Well why
8. Laura: Why do we do what?
9. Jeff: I just always did it, I never knew why
10. Laura: Why what?
11. Jeff: The dd dt thing cause you do it in physics too all the time like on every single problem
12. Laura: I took like really easy physics an I was like
13. Joel: Um they never explain why we do it we just do it
14. INT: So it’s just something you just do?
15. Jeff: Well I’m sure there’s a reason, but
16. Joel: Yeah there’s a reason we just don’t know it
It would seem that when Laura says, “Yeah we have to, you don’t get this in your
formula then you can’t solve for it that’s what we’re trying to find, I don’t know why,”
she is indicating that it’s the only way to account for the time; she needs to get a dd/dt in
order to solve the problem. It would appear that for these students the symbol does not
represent that the d is in fact a function of time and that the rate of change we are
representing is changing with respect to time. While the students in Group 1A are
convinced there is a reason that they use the chain rule, they don’t know what that reason
is yet they are comfortable with writing down symbols that don’t necessarily carry, for
them, any mathematical meaning.
A final note of interest: In the first class period many students were heard saying
how much they disliked related rates problems while working on the structured activity.
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In the second class period when there was a puzzle to be solved (with a reward for
success - cookies), students were heard saying that the class had been fun. They did not
complain about solving related rates problems, but worked together to solve the problems
in the allotted amount of time. A copy of this puzzle may be found in Appendix B.
4.4 Limitations
All the students in this calculus seminar course had seen related rates problems in
their regular calculus class, taught in a traditional lecture style. The students were
questioning why they were being coached to solve the problem this way. They appeared
to perceive it as a new method rather than a means to further investigate the properties of
this type of problem. Thus, it was difficult to determine exactly how effective this new
method was at this time.
4.5 Summary
The results presented in the preceding sections informed my research in many
ways. In the previous research on related rates, the role of transformational and
covariational reasoning appears to be notably absent. It seems likely that these reasoning
abilities are necessary to construct an accurate mental model of the situation. It would
also appear that the students’ understanding of the concepts of variable and function may
be more critical than previously believed. The implicit variable of time in these problems
appeared to be a source of students’ difficulty.
These observations led to revisions for the in-class activities designed to introduce
related rates problems (see Appendix C). The most notable revision is the inclusion of a
custom computer program which models related rates problem situations. It is expected
that the computer program will facilitate the students’: 1) visualization of related rates
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problem situations; 2) identification of and attention to which quantities are changing
versus those that remain constant; 3) identification of variables in the problem situation;
and 4) observation of time as an integral part of the problem. These revised materials
were used in a teaching experiment which is described in Chapter 5.
Chapter 5: Methods
In understanding how students solve related rates problems, we are interested in
their problem solving processes, cognitive constructions of mental models, and formal
mathematical knowledge. Thompson (1982) remarked that statistical methods are
legitimately applied when subjects are passive, but are inappropriate when subjects are
active, flexible, adaptive processors of information as in the case of education. Thus, I
have qualitative data to augment the quantitative data from exams to construct a theory of
how students are thinking about related rates problems. I am not merely interested in
knowing if students are able to learn the procedures for solving some standard related
rates problems. Rather, I am interested in whether or not they are able to engage in
effective problem solving strategies that allow them to solve related rates problems that
are unfamiliar. In examining video from the classroom or individual interviews, we are
more likely to see students verbalizing why they are writing things down and how they
are thinking about the mathematical problem solving process, especially when prompted
by their peers, instructor, or interviewer.
Dubinsky’s (1991) suggestion for fostering conceptual thinking in mathematics
has the following four steps: observe students, analyze the data, design instructional
materials, and repeat the process until stabilization occurs. This idea was further
developed by Steffe and Thompson (2000) as the teaching experiment and by Cobb,
Confrey, diSessa, Lehrer, and Schauble (2003) as the design experiment. The design
experiment seeks to develop theories about the ways students learn and what types of
activities support this learning. The researcher/teacher designs instructional materials,
building and capitalizing on existing literature when possible in addition to employing a
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hypothetical learning trajectory. These materials are used with students and analyzed
after each session. The ongoing analysis may lead to immediate changes in the materials
for future use and may require subsequent activities to be modified to reflect what has
already occurred. After the completion of the study, a retrospective analysis is usually
conducted to construct a framework and further revise the materials. Thus, a subsequent
cycle of the experiment is generated. This process may be iterated several times before
stabilization occurs.
5.1 Data Collection
I taught a section of first semester calculus in the fall 2005 semester. The course
was taught utilizing supplemental materials that Carlson and Oehrtman had previously
developed to aid students’ conceptual development of key calculus concepts. These
materials covered concepts such as function, derivative, the Mean Value Theorem,
accumulation, and the Fundamental Theorem of Calculus. I also administered the
Precalculus Concept Assessment Instrument to all students on the second day of class.
This data provided insights about the students’ understanding of the concept of function
at the beginning of the course.
5.1.1 Subjects
I recruited three students to participate in a series of six teaching episodes outside
of their regular class. The three students that participated in this study were chosen from a
group of volunteers based on their schedule of availability. This group of three students
did not attend the regular class sessions while related rates problems were being taught to
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the complement of the class. Amy, Ali, and Ben (pseudonyms used to protect students’
identity) were all freshmen and were able to meet every afternoon for a week and a half.
Amy is majoring in microbiology and marketing. She had a calculus course in
high school and admits to having trouble manipulating numbers though she still enjoys
math. On the PCA, Amy scored 13 out of 25 and was successful in completing the
problems which used algebraic representations for function composition and the bottle
problem which required her to engage in covariational reasoning. She had difficulty with
the problems that involved graphical, tabular, and contextual representations for function
composition and average rate of change. Amy had a semester exam average of 81.8% and
a homework average of 76%, earning her and overall average of 78% and a grade of C+
for the course.
Ali is majoring in secondary education and planning to teach mathematics. She
also had calculus in high school and stated that she likes to have steps to solve problems.
Ali scored 16 out of 25 on the PCA and successfully completed problems that used
algebraic, graphical, tabular and contextual representations for composition and average
velocity. She had difficulty with the bottle problem. Ali had a semester exam average of
83.6% and a homework average of 89%, earning her and overall average of 79% and a
grade of C+ for the course.
Ben is majoring in aerospace engineering, and he wants to design space shuttles
for NASA. He took advanced placement calculus in high school and stated that math has
always come easy for him. Ben scored 12 out of 25 on the PCA. He successfully
completed the problems that used algebraic, graphical, tabular and contextual
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representations for composition and the bottle problem. Ben had trouble with the average
rate of change problem. Ben had a semester exam average of 83.3% and a homework
average of 97%, earning her and overall average of 84% and a grade of B for the course.
Ben did not complete the study as he sustained a sports injury over the weekend and
returned home for surgery.
5.1.2 Individual Interviews
Individual interviews were conducted with each of the three subjects on the day
before the teaching experiment began. The purpose of the individual interview was to
ascertain the students’ understandings of the concepts of variable and function and their
ability to engage in covariational reasoning at the beginning of the study.
In an attempt to describe how students understand the concept of variable, the
students were asked to choose items that describe the role of the letter x in a mathematical
expression from a given list. Students were also asked to name the independent and
dependent variables for given expressions. These items were taken from Jacobs (2002)
study of how calculus students understand the concept of variable.
With regard to the concept of function, it was important to describe not only what
they understood about the nature of functions, but how they interpreted and solved
problems that utilize function composition. The literature suggests that students’
difficulties with the chain rule and its applications are closely related to their
understanding of function composition (Clark et al., 1997; Cottrill, 1999). The students
were asked to describe the relationships expressed in an algebraic formula and to solve
two contextual problems. One of the contextual problems they had seen previously on the
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Precalculus Concept Assessment Instrument (PCA) administered on the first day of class;
the other contextual problem had not been presented to them prior to the interview.
Finally, to describe the extent to which the students engaged in covariational
reasoning, the students were asked to imagine a given bottle filling with water and graph
the height as a function of volume. A question similar to this had also appeared on the
PCA at the beginning of the semester and was also used in Carlson, et al.’s study of
covariational reasoning (2002). See appendix C for the complete individual interview
protocol.
I also observed and videotaped three mathematicians solving the trough problem,
the plane problem, and the coffee cup problem in a think aloud problem solving session.
When observing the mathematicians solve related rates problems, it was my goal to be
able to characterize their understandings of the concepts of variable and function and the
role of those understandings in completing related rates problems. I also wanted to
observe their thought processes and problem solving behaviors.
5.1.3 The Teaching Experiment
From the literature review and the results of my pilot studies, it would appear that
most students have an impoverished view of the concepts of variable and function when
they begin calculus. Students may not engage in covariational reasoning to construct a
mental model and may not draw diagrams as part of their problem solving process. While
students may be adept at calculating derivatives, they are usually just carrying out a set of
known rules. Ideally, students would have a process view of function that allows them to
think about every quantity that is changing as a function of time, that is reconceptualizing
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the variables in a geometric formula as functions of time. The students would be able to
actively engage in covariational reasoning that allows them to imagine the system
changing and attend to how these quantities are changing in relation to each other. The
student would then be able to identify the relationship between these functions as either
composition or parametric. Instead of the chain rule being a series of steps used to
compute a derivative, we would like to see students understand the role of function
composition in the chain rule; specifically, we would like them to understand how the
variable t comes to exist in our solution of a related rates problem such as those
previously discussed.
To investigate my research questions, I conducted a small-scale teaching
experiment. A teaching experiment consists of a sequence of teaching episodes each of
which involves a teacher/researcher, one or more students, a witness, and a method of
recording what transpires in the episode (Cobb, 2000; Steffe & Thompson, 2000). The
content and sequencing of the instructional intervention draws heavily on the knowledge
gained during my pilot studies. I videotaped each of these class sessions because the
videotapes allowed me to observe how students discuss concepts among themselves,
including the gestures (which may correspond to and provide evidence for covariational
reasoning) and the terminology they use. This also provided me the opportunity to
describe their understandings and answer questions such as: How do they describe the
changing quantities? What is the role of covariational reasoning in the students’ solution
process? What, if any, hand gestures do they use to illustrate motion/rates of change?
What understandings of the chain rule are necessary to complete the problem?
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In conjunction with my pilot studies and the development of my activities, I
developed a hypothetical learning trajectory (HLT). Simon (Simon, 1995) described the
HLT as having three components: 1) learning goals for the students, 2) planned
instructional activities, and 3) a hypothetical learning process in which the teacher
anticipates how students’ thinking and understanding might evolve in the context of the
instructional activities. In table 8 below, I have briefly outlined the instructional sequence
along with the anticipated hypothetical learning trajectory. In the first session, the
students interact with a computer program designed to let them manipulate the problem
situation, thus encouraging them to construct dynamic mental models. Another important
feature of the program is that it should generate time as a necessary variable and not
something that “magically” appears. Thus, even though we cannot label time in a static
diagram on paper or in the dynamic computer program, time is still an integral part of the
problem. In session 2, the students explore the multiplicative nature of the chain rule. The
students begin exploring related rates type problems in sessions 3 and 4; however, these
problems are stripped of most, if not all, numerical data. This should encourage the
students to think about general relationships rather than solving for a specific value. In
the final sessions, the students are presented with novel problems stated in a more
traditional manner. This will afford the students an opportunity to apply what they have
learned. The complete lesson logic for the entire instructional sequence and the students’
activity sheets may be found in Appendix C.
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Table 10: Hypothetical Learning Trajectory
Instructional Sequence Learning Goals Hypothetical Learning
Process
Session 1: Average Rate of Change and instantaneous Rate of Change
• Clarify average rate of change.
• Get the students to think about the rate of change of one quantity, u, with respect to changes in another quantity, v, when neither u nor v vary constantly with respect to time through an exploration of the plane problem.
• Be able to solve for a particular quantity when given partial information.
• Introduce the variable of time and ask the same questions as in the previous session.
• What happens as the intervals get very small?
• Formalize: instantaneous rate of change.
• Students will better understand average rate of change.
• Students will be able to coordinate changes in one quantity with changes in another quantity.
• Students will formalize the idea of instantaneous rate of change by using the concept of limit and “very small” intervals.
• Students interact with the computer program to measure changes over various intervals.
• Students construct ratios to represent the rate of change in one variable with respect to another variable.
• Time becomes a variable that each geometric variable can be related to independently.
• Students now have the rate of change of each variable with respect to time.
• Construct the ratio of the rates.
• Smaller and smaller intervals of time mean that the average rate of change is getting closer and closer to the instantaneous rate of change, i.e. the derivative.
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Instructional Sequence Learning Goals Hypothetical Learning
Process
Session 2: The Chain Rule
• Revisit the chain rule with application problems similar to what may be found in a related rates problem.
• Investigate the multiplicative nature of the chain rule.
• Students will understand that the chain rule is the overall rate of change for a function created by composing functions and that this is obtained by considering the rate of change of the individual functions that went into the composition.
• Students will naturally want to multiply the rates given in the problem.
• Students will express the rates with the delta notation adopted in session 1.
• Students will discuss why the multiplication provides the desired rate.
Sessions 3-4: Related Rates
• Introduce a related rates context – plane problem
• See static relationship: 222 zyx =+
• Make the static relationship dynamic:
( )[ ] ( )[ ] ( )[ ]222tztytx =+
• Make a substitution to create an equation in t.
• Solve for the variable of interest.
• Find the rate of change with respect to time, differentiate with respect to time.
• Reflect on substitution, we didn’t need to do it!
• A second related rates context – trough problem.
• Can be solved using only chain rule.
• Students will develop an approach to solving related rates problems.
• Students will begin to develop the concept of parametric functions.
• Students will develop a deeper understanding of the chain rule and its applications.
• Students draw their own diagrams (extension of sessions 1-2 when diagrams were provided).
• Students analyze the situation by investigating a series of questions to explore the existing relationships of the problem situation.
• Students differentiate with respect to time (building from session 2) to get a general form for the desired rate.
• Substitute in given values to find a particular solution.
• Expect students to have difficulty determining when to use parametric vs. compositional relationship.
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Instructional Sequence Learning Goals Hypothetical Learning
Process
Sessions 5-6: Solving Novel Related Rates Problems
• Provide additional related rates problems – including the coffee cup problem.
• Allow the students to apply what they have learned. How do students approach these problems?
• More practice, applying new knowledge.
• Build confidence on problem solving abilities.
• Develop adaptability of the solution procedure they have constructed.
• Cup problem will require them to think of an innovative way to use the conventional geometric formulas.
After each teaching episode, I evaluated the session for clarity of the activity, the
questioning that occurred, mathematical constructions that were made, and the
effectiveness of the activity. This ongoing analysis allowed me to modify the activities in
minor ways throughout the teaching experiment so the students were continually building
on the knowledge and experiences of the previous sessions. At the completion of the
teaching experiment, I conducted a retrospective analysis of the data collected.
My learning goals for the students included: 1) understanding the relationships
between changing quantities, 2) an improved understanding of the concepts of variable,
function, and derivative (specifically how it applies to related rates problems), 3) the
ability to engage in covariational reasoning as part of the problem solving process, 4) the
ability to reconceptualize variables in a geometric formula as functions of time, and 5)
the ability to recognize if the relationship between the variables is parametric or
compositional in nature.
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It is my hypothesis that this sequence of activities may aid students in
constructing mental models and developing the problem solving abilities necessary for
solving related rates problems. While these activities may take more time, it may be that
students are better able to understand two fundamental calculus ideas: the chain rule and
parametric functions.
5.2 Data Analysis
The videotapes of the individual interviews and each class session were
transcribed to facilitate the open coding techniques of Strauss and Corbin (1990) as
described in chapter 4. Following Dubinsky’s (1991) suggestion, I have observed
students learning and solving related rates problems; I have developed instructional
activities; I have tested and revised the instructional activities; and I have now tested a
revised sequence of instructional activities.
The originally proposed sequence of activities was modified in small ways as a
result of ongoing analysis that occurred during the teaching experiment. These small
modifications were generally of the nature: change the order of the problems, start by
asking this question as I believe it will encourage them to build off something that was
said in today’s session, etc. Through this ongoing analysis, I intended to make each
teaching session relevant to what had been discovered and discussed in the previous
sessions.
At the completion of the teaching experiment, I began the retrospective analysis. I
watched all the videos in succession and made notes of my first impressions. I proceeded
to transcribe each video in order and make further notes of what I observed as I
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transcribed. Finally, I analyzed the transcripts using discourse analysis and reviewed the
original video data as needed.
Discourse analysis examines how one uses language to communicate. For my
analysis, I used the pragmatics approach described by Deborah Cameron (2001).
Pragmatics “deals with how language can be used to do things and mean things in real-
world situations.” (Cameron, 2001, p. 68) This may be characterized as studying what is
meant by the speaker and what the listener interprets from an utterance. However,
Cameron quoted Thomas (1995): “Making meaning is a dynamic process, involving the
negotiation of meaning between speaker and hearer, the context of utterance (physical,
social, and linguistic) and the meaning potential of an utterance.” (Thomas, 1995, p. 68)
The context of the utterances in this study is the teaching episode or individual interview.
In summary, I have multiple data that will be interpreted in different ways. The
following is a summary of the data I have collected and how it was analyzed.
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Table 11:
Data Sources
Data Source Purpose Method of Analysis
PCA To assess students’ understanding of the concept of function
Pretest, standard statistical methods
Individual Interviews To provide a baseline of the students’ understandings of the concepts of variable and function and the extent to which they may engage in covariational reasoning
Discourse Analysis
Videotaped Instructional Sessions
To investigate how students interact with the related rates activities and how their reasoning about related rates problems develops
Discourse Analysis
Exams To assess individual understandings of calculus concepts including how to solve related rates problems
Grading rubric
Mathematicians Solving Related Rates Problems
To observe how professional mathematicians think about and solve related rates problems
Discourse Analysis
The Precalculus Concept Assessment was administered to all students in my
calculus class and has provided data about the students’ understanding of the concept of
function at the beginning of the course. The videotapes of the class sessions allowed me
to observe how students discuss related rates among themselves, including what
terminology they use. Exam data has been analyzed to determine students’ individual
understandings and ability with regard to related rates problems, as well as other related
topics in calculus (e.g., function, derivative, the chain rule). This study also allowed me
to address whether the use of a computer simulation promotes students’ formation of a
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mental model of the problem situation and an ability to imagine the how the situation is
transformed by changing the variables of the model.
Chapter 6: An Emergent Framework
In this chapter, I formulate a framework that characterizes what solving a related
rates problem entails. This framework will also be used for analyzing my teaching
experiment data in Chapter 7. While this framework was informed by my pilot study, it
mainly grew out of analysis of interviews with problem solving experts.
In the data analysis, I will use the terms mental image and mental model. A
mental image will mean a static representation of some aspect of the problem situation in
the problem solver’s mind. A mental model will mean a representation of the problem
situation which has additional (usually conceptual) knowledge associated with it in the
problem solver’s mind. One may use a mental model to think about the problem situation
changing and anticipate how different actions impact the system.
I observed and interviewed three mathematicians as they solved three related rates
problems discussed earlier (See Chapter 1, Section 1 [trough problem] and Chapter 4,
Section 2 [plane problem and coffee cup problem]). As the mathematicians solved each
problem, they drew diagrams and wrote functions on their paper or whiteboard. These
physical pieces of information represented the result of mental activities and became part
of their solution. I use the term artifact to refer to these physical pieces of information.
The solution to the problem is the collection of artifacts.
As the mathematicians attempted each problem, they employed many general
problem solving strategies, which included making sense of the problem statement,
constructing a diagram, and labeling known values and unknown values. They also
consistently drew on a rich source of content knowledge as they attempted to construct a
functional relationship and relate the rates. Throughout their problem solving process,
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they were able to imagine the situation changing and attend to how various attributes of
their model were transformed. It is the purpose of this chapter to describe how one’s
mental model and content knowledge inform the construction of the solution artifacts that
constitute the solution to the problem.
6.1 Mathematicians Solve Related Rates Problems
In this section, I present the results of the three mathematicians solving related
rates problems. They were asked to solve the trough problem, the plane problem, and the
coffee cup problem as stated in Table 1 in a think aloud interview. After they completed
each problem, the interviewer asked questions she believed were needed to clarify their
thinking. These interviews were videotaped and transcribed for analysis. Each interview
was analyzed using the pragmatics approach from discourse analysis described in
Chapter 5.
6.1.1 The Trough Problem
All three of the mathematicians successfully solved the trough problem:
A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3
feet across at the top and have a height of 1 foot. If the trough is filled with water
at a rate of 12 ft3/min, how fast does the water level rise when the water is 6 in.
deep?
Consistent with what was reported in the literature review, the mathematicians
were inclined to draw a diagram and label it with constants and variables. They appeared
to have a strong image of variable as a varying magnitude and proceeded to label their
unknowns with letters that helped them keep track of what was being represented – for
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example, h represented height. At some later point in the problem solving process, they
used a function to represent the relationship between the variables - for example the
volume of the trough was represented by bhlV2
1= . The mathematicians then used
function composition to express the volume in terms of height. This function was
differentiated with respect to time. Known values were then substituted into the
differentiated function and a solution was computed.
Notice that in the statement of the problem, time is stated only as part of the rate.
Time was not labeled in any of the diagrams that were drawn, and the variable t was not
represented in any of the algebraic representations. Time was usually only indicated
verbally as part of the rate or in conjunction with the phrase “with respect to.”
The literature does not address how building a mental model and accessing one’s
content knowledge facilitates one’s solution process. It also does not address how one
deals with time as a variable. In the following sections, I describe what I observed in the
mathematicians’ solutions relative to these points.
Dan’s solution. Dan began reading the question and stated:
Excerpt 12
Dan: (1) (reads question) So, as I’m reading this, I guess I’m making a, I’m
picturing what this is describing (2) Um, if the trough is filled with water at a rate of 12, ok, so I see that this is
some kind of related rates problem (3) Um, ok and, uh, I mean, just because I haven’t even really read the last
sentence very completely, but it’s an isosceles triangle. So I’m expecting some kind of um similar triangles sort of issue to generate relationships.
(4) Um, but ok, let’s go ahead and see what I’m looking for. The trough is filled with water at a rate of 12 cubic feet per minute. How fast does the
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water level rise when the water is 6 inches deep? … Ok, so how fast does the water level rise? So we’re looking for rate of change of height with respect to time.
(5) And um, it gives me a flow rate of water is coming in at 12 cubic feet per minute. So I’m given the rate of change of volume with respect to time. So I need a relationship between volume and height, and that’ll let me connect the rates, ok.
(6) Um, also since this a a triangular prism, I don’t have to really draw a completely 3 dimensional picture, but I can just draw the end [artifact – 2-d diagram]. So I can draw that, even though what I was thinking was, you know, a picture that looked, you know kind of more like that [artifact – 3-d drawing] So, I was picturing that, but um, I don’t necessarily need to show this part because the relationships, I can capture just here.
(7) Um, let’s see, ok. So, this is, uh, 10 feet long. So, I’ll go ahead and label that over there since I’ve got it up there, and three feet across, three feet across the top, and a height of 1 foot, ok. So, this is not at all to scale. That’s one [artifact – labeled diagram]
Dan commented that this is “some kind of related rates problem.” He appears to
have recognized that this problem fits into a class of problems known to him, related rates
problems, and thus recalled some heuristics that have worked for him when previously
solving related rates problems. One heuristic appears to be “draw a diagram” because that
is the first thing he actually does as part of his solution. Dan keyed in on the words
“isosceles triangle” in the problem statement and mused that he expected a “similar
triangles issue to generate relationships.” (Line 3) This seems to be another strategy he
identified for possible use. He also stated, “I need a relationship between volume and
height and that’ll let me connect the rates.” Yet another heuristic is suggested by the need
to “connect the rates” using the relationship. (Line 5) This suggests that Dan was
planning out a series of actions that he would be taking to complete the problem. It also
seemed that he had begun the process of building a mental model.
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After Dan finished reading the problem statement, he recognized that the problem
is asking “how fast does the water level rise” and restated it in his own words; he stated
that he wanted to find “the rate of change of height with respect to time.” (Line 4) It
would appear that Dan identified the water level as the height in his mental image. The
phrase “with respect to” appears to be part of his content knowledge of derivatives. By
using the phrase “with respect to” to restate the problem, he identified time as the
independent variable in the problem. He then conveyed that “the rate of change of
volume with respect to time” and “the rate of change of height with respect to time” were
the rates he wanted to relate. (Line 5) Dan commented that he wanted to construct a
relationship between volume and height because that would allow him to “connect the
rates.” His clarity regarding the variables and the rates he needed to relate to solve the
problem suggest that Dan constructed a meaningful mental model of the problem
situation. Dan’s mental model of the problem situation supported his reasoning about the
varying quantities and the constant quantities.
After Dan constructed his mental model and outlined his solution process (Lines
1-5), he began drawing diagrams. Dan considered at least two perspectives of the
problem situation before choosing the perspective that would be most helpful to him.
(Line 6) While he noted that in his mind he was picturing the 3-dimensional trough, he
initially drew a diagram that represented the cross section, claiming that this would
provide him all the information he needed about the relationships in the problem. This
suggests that Dan’s mental model of the problem situation contained information about
relationships that exist between the changing quantities. Specifically, his mental model
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afforded him the opportunity to consider an amount of water in the trough and the
relevant relationships between the water and the whole trough. The cross-sectional view
also allowed Dan to easily identify the similar triangles he thought would generate
relationships. Dan then constructed a 3-dimensional diagram as he explained, “I don’t
necessarily need to show this part because the relationships I can capture just here”
(indicating the cross-sectional diagram). Thus, Dan constructed two diagrams and labeled
the 3-dimensional one with constant values. (Lines 6-7)
After drawing his diagrams, Dan continued with:
Excerpt 13
Dan: (1) Ok, so, um, so, what I want to look at is at any particular kind of uh
configuration for this thing as it’s filling up with water, and it’s going to be filled up to some point, [artifact-modified diagram by drawing a line to indicate the water level]
(2) and so I want to know that height, [artifact-modified diagram by drawing a line to indicate height and labeled it with an h] because that’s going to be what’s related to the uh volume, and that’s what I want to know. How it’s changing.
(3) Extend that axis up, and so, the similar triangles are going to come from, um, the big isosceles triangle to the small isosceles triangle. So, that’s 3 all the way across, that’s 1 all the way top to the bottom, there to there, and h is um the corresponding side to the one, and that width is the corresponding side to the 3. So, I’ll put w. [artifact –labeled diagram]
Dan noted that he was interested in a particular instant in time – when the trough
is “filled up to some point.” (Line 1) He also anticipated how he might represent the
volume with a function. (Line 2) Dan’s mental model allowed him to consider what the
volume of water in the trough would be at any given instant, and he indicated this by
modifying his diagram to include a horizontal line which represented the water level. He
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subsequently drew in a vertical line to represent the height of the water because he
wanted “to know how it’s changing.” Thus, Dan’s mental model incorporated the height
and width of the water in the trough as relevant changing quantities.
Dan’s next goal was to determine the amount of water in the trough at any point
in time. He modified his diagram by extending “that axis” of the triangle that represented
the height of the water and observed where the similar triangles were. (Line 3) As Dan
labeled his variables for the height and width of the trough he stated “h is the
corresponding side to one” and “width is the corresponding side to the 3 so I’ll put w.”
(Line 3) The labels that Dan chose for his variables were the initial letters of the nouns he
used to name the properties: h represents the height of the water and w represents the
width of the water. Dan’s mental model now included a similar triangles relationship and
variable names for the changing quantities, namely the height and width of the water in
the trough. Thus, Dan appeared to have completed the first part of solving a related rates
problem, draw a diagram and label it.
After he modified his diagram by labeling the height and width of the volume of
the water in the trough, Dan continued and constructed a relationship between height and
volume.
Excerpt 14
Dan: (1) And so, what we need then is that h to w is one to three [artifact –
algebraic relationship] (2) And I know that I’m eventually going to want a relationship between
volume and height. So, I’m going to need to get rid of the w, so that the relationship that I’m going to need is to have w as a function of h. So, I want, uh, w is, um, 1 over 3 h. [artifact – solved equation]
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(3) Ok, let’s see, ok, so volume then is, um, the area of the base of this triangle, and then times the length which is 10. So, that’s one half w times h for the area of the triangle times 10. So, that’s the volume of water. [artifact-volume relationship] Um, so, then putting that in for w I get, um, let’s see. So, that’s five thirds, let’s see. [artifact-solved equation]
(4) Ah, this is interesting cause the h’s are going to cancel. So that is wrong. Uh, ok, so let’s see. Ah, it’s because I can’t solve algebraic expressions. So w is h over 3. Let’s see, let’s make sure I’m doing this, no, come on, w is 3 h. Ok, I can do this. So what I want is,
(5) Um, see what I’m trying to do is intentionally put in some metacognitive checking
(6) Ok, um, ok, all right w, w, so I want then, so that’s 5, 3 h, and h. So, I get 15 h squared. [artifact –solved equation] ok
(7) And, um, that makes sense that I should get something squared because I’ve got, of the two dimensions that I have, two of them are linearly dependent on h, and one of them is constant. So, that makes me happy. Um, as opposed to a constant, which didn’t make me happy because I was pretty sure that as it filled up the volume changed. [check]
It would appear that Dan accessed his content knowledge of geometry to set up a
proportional relationship between the height and the width. (Line 1) Dan reiterated that
he wanted to construct a relationship between the volume and the height of the trough
and continued, “I’m going to need to get rid of the w.” Subsequently he stated, “the
relationship that I’m going to need is to have w as a function of h.” (Line 2) Thus, it
would appear that Dan wanted to construct a relationship in which h is the independent
variable and w is the dependent variable as is indicated by his use of the phrase “as a
function of.” This allowed him to use function composition to express volume as a
function of height. (Line 3) He appears to have accessed his content knowledge of
functions to set up the relationship between the volume and the height. However, Dan
appeared to mentally solve for w and wrote downh
w3
1= , an incorrect statement.
However, Dan’s subsequent statements provide evidence that he checked that the
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relationship he constructed between the volume and height was reasonable. He
recognized that something was wrong with his relationship; he stated, “this is interesting
cause the h’s are going to cancel so that is wrong,” and proceeded to check his
calculations. (Line 4) He noted that it was reasonable to expect something squared
because two of the dimensions are linearly dependent on h and one is constant. Dan
articulated that it was not logical to get that the volume was constant as he did in his first
attempt at composing the functions “because I was pretty sure that as it filled up the
volume changed.” (Lines 6-7) It would appear that Dan imagined the trough filling with
water and coordinated how the height and volume would change in relation to each other.
He reasoned that the volume would change. While he did not state it explicitly, it would
seem that he would expect both the height and the volume to increase as the trough filled
with water. When his answer did not correspond with how his mental model played out,
he knew something was wrong and checked his first calculation. This suggests that Dan
used his mental model as part of the checking phase at this point in his problem solving
process. After he was satisfied that his relationship was acceptable, Dan was ready to
move to the next step in solving this problem which was to “connect the rates.”
Note that Dan had still not labeled time in his diagram or used it in his
representations of the relationships that exist; however, he proceeded to compute the
derivative with respect to time as is seen in the following excerpt.
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Excerpt 15
Dan: (1) Ok, all right. So, now, um, like I said before I want, um, uh, information
about the rates of change. So, I want to differentiate both of these with respect to time.
(2) So, that’s easy on the left. It’s just dv dt . (3) But on the right, I need the chain rule. So, this is 30 h times derivative of h
with respect to time. [artifact-differentiated equation] (4) And now, I think I have everything I need. I can just plug in, see it says it
wants to know when the water is 6 inches deep. So, um h is going to be 6, and um the flow rate dv dt is a constant of 12. [artifact] This a funny way to write it; h equals 6 inches, and oh, great one half foot. [artifact] All right, cubic feet per minute. All right, um, ok. So, now I should be able to plug these in, and so I take 12 is 30 times one half times dh dt [artifact-substituted in values] and so dh dt is 12 fifteenths, and this should be um feet per minute. [artifact-solution]
(5) Just thinking about the context in terms of the units, but let’s see if that makes sense. Cause this is cubic feet per minute, 30 is uh… Oh, you know I wasn’t really keeping track of units back there. Let’s see, 30 has units of feet I think because it came from the 3 times the 10. One half and then dh
dt. What’s the one half? The one half has units of feet, that has units of feet, that has units of feet per minute, cubic feet per minute. So, it matches up. [check] Um, so, barring another algebra or um arithmetic mistake that’s my answer.
Dan applied the chain rule to compute the derivative with respect to time. (Lines 1-3)
Even though Dan had not labeled time in his diagram, it would appear that he was able to
imagine each variable in his mental model changing as a function of time. Thus, he could
differentiate his function “with respect to” time. He substituted known values into his
differentiated equation and used algebraic manipulations to compute an answer which he
checked by conducting a unit analysis. (Lines 4-5) It was not a problem for Dan to
operate on 2h as a composite function of time to be able to express the derivative
asdt
dhh
dt
dV30= . When asked why he needed the chain rule, he responded:
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Excerpt 16
1. Dan:
a. What tells me to do that? Um, well to take the derivative with respect to time, um, I have to think of these, both sides as functions of time.
b. Um, so, here that’s no problem. I don’t think I even thought of the left hand side in terms of a chain rule I just said derivative with respect to time.
c. But here I’m saying, well, ok this a, I want to think of this as a function of time. Well, to do that I think of h as a function of time, and that’s what we’re af, that’s what we’re after anyway is dh dt, the rate of change of height with respect to time.
d. So, um, I want to think of h as a function of time, but then the expression on the right hand side is a function of h or an expression written as a function of h. So, um, as function h which is itself a function of t. So, I need to think about that as a composition to use the chain rule, to take the derivative. Does that answer your question?
2. NE: So, the composition that’s in there
3. Dan: The 15 h squared
4. NE: Yeah
5. Dan: Do you want me to write that out like in function notation?
6. NE: Yeah, how would you?
7. Dan: [writes ( )[ ]215 th ] Um, I don’t know that I necessarily was mentally
picturing this notation or not, but I definitely was picturing h being a function of t, if that makes sense
8. NE: Ok
9. Dan: I was picturing h as meaning some function h is some function of time, but not necessarily written out with function notation. Although, I mean that’s an easy um step.
What is evident in the transcript excerpt above is that Dan thinks of each variable
as a function of time. (Line 1) In Line 7 above, he demonstrated that he could write out
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the functional relationship with time represented as the independent variable when
prompted. However, he did not express time as part of his algebraic representation of the
function nor did he believe that he consciously thought of time as the independent
variable as is evidenced in his statement in Line 9 above. It would appear that Dan was
able to imagine the objects in his mental model change with time, and thus he was able to
think about each variable as a function of time. Dan appeared to access relevant content
knowledge of function and derivative to apply the chain rule and complete his “connect
the rates” step.
Adam’s solution. Adam also obtained a correct solution to the trough problem.
His steps are similar to those performed by Dan and are detailed here. Adam began the
problem by stating:
Excerpt 17
Adam: (1) Ok, a trough. Ok, so do you mean like a ditch? (2) Is 10 feet long and ends have the shape of isosceles triangles. I have to
remember what isosceles is. That’s equal on two sides, right? It’s not equilateral.
(3) Ok, well, so the thing I would always do on this, I think, is I always tell the students to draw a picture because otherwise I have no idea what the problem is talking about.
(4) Um 10 feet long, and it’s ends have the shape… I’m trying to understand, I’m trying to get the picture here. So, oh, I see so the cross section is an isosceles triangle, yeah. [artifact – diagram] So, I’m a terrible artist. So, something like that.
(5) So, triangular trough, I think. All right, across at the top. So this is three feet this way, and it’s 10 feet this way, and it’s a foot high. So, so I think that’s what we’re talking about.
(6) And I am going to pour water into this thing at the rate of 12 cubic feet per minute, and how fast does the water rise when the water is 6 inches deep. So, presumably 6 inches deep means, um, so one half [artifact-labeled diagram] Ok, I get it. I haven’t done these in a long time.
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While Adam’s first mental image was that of a ditch, he proceeded “its ends have
the shape, I’m trying to understand, I’m trying to get the picture here, so oh I see the
cross section is an isosceles triangle” and subsequently constructed an appropriate 3-
dimensional diagram. (Lines 1, 4) He appeared to be comfortable with triangle, but
needed to recall what it means to be isosceles and accessed his content knowledge of
geometry. (Line 2) It would appear that Adam manipulated his mental image of the
problem situation in some way (we do not know exactly how from what he said) which
allowed him to recognize that the cross section may be represented by an isosceles
triangle. He then drew a diagram on his paper. (Line 4) After constructing his diagram,
Adam labeled the dimensions of the trough with constants. (Line 6) This suggests that
Adam’s mental image of the trough is still static. Thus, Adam carried out the first part of
solving the problem, draw a diagram, which he stated was necessary, “because otherwise
I have no idea what the problem is talking about.” (Line 3)
This suggests that the mind may conjure multiple images, some of which may be
incorrect, in the process of reading, interpreting, and trying to understand the problem
statement. These mental images appear to then be refined until a mental model has been
created which is in accordance with the problem solvers’ interpretation of the problem
statement and his content knowledge of specific mathematical domains (in this case,
geometry).
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Adam continued:
Excerpt 18
Adam: (1) All right. So, let’s see. What’s the volume of this thing? So, the volume uh
is the area of the triangle times the length. So, it’s 10 feet long times the area of the triangle, which is one half the base times the height, and the base is 3, cause it’s 3 feet across, and it’s one foot high. So this is, let’s see 15. [artifact-volume of trough] Right, ok. So, we have that the volume then is 15 cubic feet.
(2) Um, now the other thing the expert always does of course is assign letters to everything so that we remember what the calculations are. Ok
(3) Um, all right. So now the other thing about this is the length, which is constant. So, we can call it 10, and um, let’s see. So, now if, to do the related rates, you have to ask how much water is in the thing if it is at a height of depth h
(4) Well, that’ll be 10 feet long times one half times 3, cause the top of this trough is constant. It’s the water that’s rising. So, again this comes out to be um so 15 h, and h has to be in feet. [artifact-volume equation
( ) hhV 15= ]
While Adam proceeded to compute the volume of the whole trough, he never
used it as part of his solution. (Line 1) It may be that this afforded him the opportunity to
access his content knowledge of geometry, to recall what the volume of a prism is, and to
orient himself to the space being described in the problem. It would seem that by
computing the volume of the trough, Adam thought about the dimensions of the trough as
potential variables. Adam assigned letters to each dimension of the trough “so that we
remember what the calculations are.” (Line 2) Adam’s mental model now seems to
account for a distinction between varying quantities and constant quantities because he
identified the length as a constant quantity. Furthermore, he asked, “how much water is in
the thing if it is at a height of depth h?” This suggests that he identified height as a
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changing quantity. In Line 4, Adam identified the width of the trough as a constant. Thus,
he modified his diagram by labeling the constants and variables. (Line 3)
Adam also noted, “let’s see so now if to do the related rates you have to ask how
much water is in the thing if it is at a height of depth h” and proceeded to construct a
function for the volume as a function of height. (Lines 3-4) This statement suggests that
Adam considered what the volume of the water in the trough is at a particular instant in
time. Thus, write an equation to represent the amount of water in the trough seems to be
the next step in Adam’s solution process. To create this equation, Adam appeared to
access his content knowledge of geometry to determine a general formula, but then
accessed his content knowledge of function to express the volume in terms of height.
(Line 4) It is interesting to note that Adam’s algebraic representation for the volume of
water in the trough was ( ) hhV 15= . This function is not the correct result and is due to a
flaw in his mental model. Adam identified the width of the trough as a constant rather
that considering the width of the water in the trough as a variable. However, Adam still
obtained the correct answer because of the values given in the problem.
Adam continued solving the problem:
Excerpt 19
Adam: (1) Um, ok, so from some expression like this you could ask, uh, if h is
changing at a certain rate, then um how much is the volume changing? (2) And so this means you need dv dh. Hmm, that doesn’t sound quite right,
but that doesn’t… dv dh is constant. This is embarrassing to have this all on video tape.
(3) Let’s see, all right. So, we are given that this thing is filling up at the rate of 12 cubic feet per minute. So that means that dv dh, well no, dv dt is 12
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cubic feet. [artifact – expressing the known and unknown rates with Leibniz notation]
(4) And so our units are consistent, right? This is 15 square feet, and h is feet. So everything is cubic feet, all right. [check]
(5) So dv dt is 12, and we want to find out what is dh dt , that’s unknown (6) So, let’s see. So, then the chain rule will give us that dv dt is dv dh times
dh dt [artifact-chain rule equation dt
dh
dh
dV
dt
dV⋅= ]
Adam initially considered the rate of change of volume with respect to height. It
is possible that was a natural thing to consider based on visual clues from his functional
relationship which related volume and height. (Lines 1-2) This is suggested by his
statement, “so from some expression like this you could ask uh if h is changing at a
certain rate then um how much is the volume changing so you need dv dh.” However, he
then appeared to think about the rate that was given and subsequently identified time as
the independent variable. (Line 3) This was done through the symbolization of the rate
using Leibniz notation as Adam noted, “so we are given that this thing is filling up at the
rate of 12 cubic feet per minute so that means that dv dh well no dv dt.” (Line 5) It seems
that the units of feet per minute associated with the rate of change of the volume indicate
to Adam that the derivatives should be taken “with respect to time.” These
understandings are likely part of his content knowledge of derivative. He next identified
the need to apply the chain rule as had Dan. (Line 6) However, Adam did not apply the
chain rule to his functional relationship between volume and height the same way that
Dan did. Rather, Adam wrote out the relationship dt
dh
dh
dV
dt
dV⋅= and proceeded to
identify what each symbol in this representation was. Thus, Adam had completed the
next part of the problem which was to relate the rates.
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Adam finished the problem by identifying the pieces of his chain rule equation as
is seen in the following excerpt:
Excerpt 20
Adam: (1) So, dv dt is 12 cubic feet per minute, um, dv dh is 15 [artifact- substituted
in values] (2) Um, what about units here? Let’s see, and we want to know dh dt, which
is our unknown. So, this in cubic feet per minute, and dv dh… let’s see. So, this is feet squared, and dh dt is feet per minute. So, we have cubic feet per minute equals cubic feet per minute. So, that’s fine [check]
(3) So, I would say then that um dh dt is 12 over 15, and so, that’s whatever it is. That’s four fifths [artifact – solution]
Adam identified that dt
dV was 12, that 15=
dh
dV, and that
dt
dhwas what he wanted
to find. (Lines 1-2) He substituted those known values into his differentiated function and
performed appropriate algebraic manipulations to find an answer. He checked his answer
by performing a unit analysis which relies on his content knowledge of units and algebra.
(Line 2)
Bob’s solution. Bob’s solution to the trough problem is slightly different from
those of Dan and Adam. Bob began:
Excerpt 21
Bob: (1) Ok, a trough is 10 feet long and its ends have the shape of isosceles
triangles, isosceles triangles [artifact - draws diagram] that are 3 feet across. Oh, ok, all right. Minor, my diagram is upside down, but that won’t change anything. Ok, 3 feet across, one foot high.
(2) Oh, it will make a difference. Sorry, because you’re asking about rate of change. [artifact - redraws diagram] One foot high, 3 feet across the top, 10 feet long. Ok, um, I was just thinking, all right, I was supposed to think out loud
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(3) I was just thinking, let’s see, will the depth matter? Um, yeah the depth will matter because the water’s coming in at a constant rate. So the longer it is, the slower it will fill.
Bob’s initial diagram was of the 3-dimensional trough oriented upside down with
the 3 and 1 labeled, and he thought this would not matter. (Line 1) Thus, Bob’s first
mental image of the trough is incorrect. However, it appears that as he continued to read
the question, he imagined what would happen to the depth of the water as the water was
being poured into the trough. (Line 2) This caused him to change his mind. He stated, “I
was just thinking let’s see will the depth matter um yeah the depth will matter because the
water’s coming in at a constant rate so the longer it is the slower it will fill.” (Line 3) It
would appear that Bob considered the dynamic process of the trough filling with water
and the relationship generated between the volume of water in the trough and the height
of the water in the trough as time passes. He observed that as the water was poured into
the trough, the volume increases at a constant rate and the height increases as a
decreasing rate. Thus, it would appear that initially reading the problem statement caused
Bob to apply the step of draw a diagram, possibly based on key words. However, as Bob
further read and analyzed the problem statement he built a mental model which
accurately characterized the problem situation. Bob decided that it was necessary to draw
a second diagram to correctly reflect the problem situation and proceeded to redraw his 3-
dimensional diagram with the proper orientation. He then labeled the dimensions of the
trough (3, 1, and 10) on his diagram, but he never labeled any variables on his diagram. It
is possible that he assigned them to his mental model as he did talk about height and base,
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but he did not indicate these variables on his diagram. This may have caused some of the
confusion Bob experienced later in the problem solving process.
Bob continued:
Excerpt 22
Bob: (1) Ok, if the trough is filled with water at 12 feet, 12 cubic feet per minute,
how fast is the water level rise when the water is 6 inches deep? Um, how fast will it, so you want um the rate of change, you want, you want the rate of change of the height of the water, I guess with respect to time. At least time is what makes sense. Well, I suppose you could do it with respect to volume, ok, but I’ll do it with respect to time.
(2) So, water’s coming in at 12 cubic feet per minute. [artifact-rate] How fast is and you want height per minute, ok, so, [artifact -writes dh/dt]
(3) Um, so, the volume of this thing, it will be one half the base times the height. [that’s right no length] [artifact – volume equation] Yeah, and so the base, the base is equal to, the base is a function of height. So, um,
(4) Ok, so, um, let’s see. So, 3 over 1 will be equal to um b over h. So, b is equal to [artifact-equation b=3h][long pause] 3 h so,
(5) so you want one half times 3 h times h [artifact-equation ( ) hhV ⋅= 32
1]
Note that it is again the rate that appears to indicate that time is the independent
variable. Bob stated, “the rate of change you want you want the rate of change of the
height of the water I guess with respect to time at least time is what makes sense well I
suppose you could do it with respect to volume” which suggests that he is not sure if he
wants time or volume to be the independent variable. (Line 1) This suggests that Bob’s
mental model allowed him to observe that the height was changing as a function of time,
but he was not sure if he should consider the rate of change “with respect to” time.
However, as he reread the problem statement, he noted, “so water’s coming in at 12 cubic
feet per minute how fast is and you want height per minute ok” which suggests that
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reading the rate for the volume of water being poured into the trough with “per minute”
as part of the rate indicates that he should also be looking for the rate of change of the
height with respect to minutes (time). (Line 2) Thus, the statement of the given rate in the
problem appears to cause him to draw on his content knowledge of derivatives.
When constructing his functional relationships, Bob recognized that he wanted
volume as a function of height, but he left out the dimension of length. (Line 3) The
relationship he constructed was given by bhV2
1= . He constructed an appropriate
proportional relationship between the height and base which he correctly solved. (Line 4)
This suggests that Bob’s mental model has variables assigned to the relevant changing
quantities and contains the similar triangles relationship. He then substituted h3 for b in
his volume equation to get the relationship ( )hhV 32
1= . (Line 5) Thus, Bob appears to
have accessed his content knowledge of geometry to write down an equation for the
volume of water in the trough and then access his content knowledge of function to
perform the composition. However, because his first volume formula was incorrect, his
subsequent relationship was incorrect. This created some problems when he attempted to
relate the rates.
Bob proceeded:
Excerpt 23
Bob: (1) and you know that um dv dt is 12 cubic feet per minute. So, 12 cubic feet
per minute is equal to, going to do this [artifact - derivative on equation] 3 h.
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(2) No, wait a second, dv dt is equal to 3 times the height. I’m pausing because um um I just got that um the rate of change of volume with respect to height is 3 times the height, which is also equal to the base, and I’m saying, now is that a coincidence? All right, well I’ll put that aside for now. [artifact-checking-makes notation on paper]
(3) So, no, that, this can’t be right. What’s going on? 3 h, sorry, duh 3 h dh dt [artifact-differentiated equation]duh
As Bob began to relate the rates, he noticed that something might be amiss when
he said, “I just got that um the rate of change of volume with respect to height is 3 times
the height which is also equal to the base and I’m saying now is that a coincidence?”
(Line 3) Bob had written down that hftdt
dV3min/12 3 == . He decided to continue
working out the problem and proceeded to recognize that he had neglected to write down
dt
dhas part of his derivative and then computed an answer. It is not clear what caused Bob
to recognize that he had forgotten thedt
dh.
Bob continued:
Excerpt 24
1. Bob: a. So, you’re asking this, when um the height is equal to what? Oh, this is
12 cubic feet, duh. So, [long pause] 1 foot, 6 inches, so point five. So h is point 5. So it’s 3 times point 5 dh dt. So whatever 12 divided by 1 point 5 is [artifact -solution]
b. Um, and let me look back at this. Does that make sense? 3 halves, two thirds 8, so 8 um feet per minute. [long pause] Does that make sense? [check] Ok, what I would like to do is to graph this
2. NE: Ok 3. Bob: But I don’t have, I don’t have the tools that I would use I 4. NE: What tools would you use? 5. Bob: I’d use, um my you know graphing calculator, not the hand held, but
you know, the program. Um, does that 8 feet per minute, does that make sense?
6. NE: What would you graph?
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7. Bob: a. Um, I would graph, um let’s see. Uh, I would graph height as a
function of time, and to get that b. Ok, let me go back here just a second. [long pause – looks at previous
work] So, if this were at half a foot, then the volume of this is a half a foot. So, the height would be one and a half feet. The um, the height would be one and a half feet. So, so, that’s three eighths feet squared
[artifact- 8
3
2
1
2
3
2
1=⋅⋅ ], thirty eighths feet cubed, and then at uh, at one
foot it would be, um. So, the volume at half a foot is thirty eighths cubic feet, um at one foot it is one half, doo doo doo. So, half base time height. So three halves square feet. So thirty halves cubic feet at
one foot. [artifact- 3
8
30ft at .5 ft] So, it would be, um, see we
have…So it would be uh twenty two point five halves feet cubed in one half. [artifact-writes average rate of change]
c. No, wait. Ok, I’m doing this volume with respect to height, ok. So, so, it changed twenty two and a half cubic feet. I’m sorry, eleven point two five cubic feet, [artifact] and how long did it take to do that? Uh, it took about a minute, um a little less than a minute. So, the average rate of change over that half of an interval… [shakes head] Shit. I’m getting myself all confused. [scribbles out previous calculation]
d. Ok, because I want, you’re asking for dh dt. So, the question is, how long did it take? how long does it take to go up? So how long does it take to go up one half foot from point five to one foot? [artifact-writes out question]
e. Ok, it takes, ok so, it’s eleven point two five cubic feet. So, it takes about a minute. [artifact-writes that out] So, the average rate of change from point 5 to 1 is about a half a foot per minute. So, 8 feet per minute doesn’t look right. [check]
f. Unless up here I made an arithmetic error. 3 eighths cubic feet, one half foot times. So, three eighths feet squared. So at thirty halves at one, ok. So three halves squared feet times ten is thirty halves cubic feet at one foot. So, that’s seven point twenty two point five halves, eleven point two, five cubic feet which takes about a minute, and so it looks like it goes up at about half a foot per minute. So, 8 feet per minute can’t be right.
g. Duh, I left off the times 10. As Bob began to check his answer to the problem, he wanted to employ a
graphical representation of his function. (Line 1b) He commented that he would like to
use the “My Graphing Calculator” program to graph his functions. When pressed about
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what he would graph, he stated that he would graph height as a function of time but
quickly went back to the problem. (Lines 5, 7a) He never did elaborate on how he
thought he could graph height as a function of time and how it may help him solve the
problem. To help him understand where the error might be, Bob computed the average
rate of change and compared that to his answer. (Line 7b) Bob also restated what the
question is asking for, “because I want, you’re asking for dh dt so the question is, how
long did it take, how long does it take to go up, so how long does it take to go up one half
foot from point five to one foot.” (Line 7d) Once Bob realized that he had forgotten to
multiply by the length of the trough, he quickly computed the correct answer.
After he computed his answer, Bob appeared to consider how the rate he found
matched up to what he had imagined with his mental model. It appears that when Bob
used his mental model to imagine the trough filling with water, he did not believe that the
trough could possibly fill up in less than a minute because he knew the volume of the
whole trough was 30 cubic feet and the rate at which the water was coming in was 12
cubic feet per minute. He reasoned that in one minute, the trough was only going to have
12 cubic feet of water in it, which would certainly not fill the trough. Thus, he knew he
had an error somewhere and decided to compare his answer to the average rate of change
to gain a sense of what his answer should be. This suggests that Bob has a strong
understanding of the relationship between the average rate of change and the
instantaneous rate of change.
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Summary. As the mathematicians’ solved the trough problem, it appears that they
successfully created a good mental model and employed a series of problem solving
strategies to guide their solution process. These strategies appear to be: draw a diagram
and label it, construct a functional relationship, relate the rates, solve for the unknown
rate, and check the answer for reasonability. Each of these strategies generates one or
more solution artifacts, a written piece of information that documents one’s thought
process. It is also evident that the mathematicians frequently related the artifacts they
produced back to their mental models. In the case of Dan and Bob, this allowed them to
discover an error in their artifacts and to correct it before much backtracking was needed.
It would seem that each of the mathematicians constructed a meaningful mental
model of the problem situation which was reflected in the diagrams they drew. The
mental model and diagram were then accessed and referenced at various points in the
problem solving process. After constructing the diagram, it was necessary to relate the
variables in the diagram with a functional relationship. To express the relationship
between the variables with a function, the mathematicians appeared to access their
content knowledge of geometry and functions. This function was then used to relate the
rates by differentiating it with respect to time which required them to access their content
knowledge of derivative and function. When it came to relating the rates, the
mathematicians seemed to identify time as the independent variable possibly as a result
of imagining the problem situation and attending to the nature of the quantities that
change or as a result of interpreting the given rate. While none of the mathematicians
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explicitly labeled time in his diagram, throughout the problem it appeared each was
thinking about each variable as a function of time because they each talked about time as
part of the rate. Note that this may be subconscious. When constructing his relationships,
Dan wrote a function using the variables from his diagram. However, it is likely that he
was still thinking about them as functions of time subconsciously; each variable could be
expressed in the form ( )th if need be. To relate the rates, Dan consciously thought about
his function as a function of time, even though he said was likely not mentally visualizing
each variable as an expression of the form ( )th , and computed the derivative with respect
to time. Being able to recognize that the function that has been written down is a
composition of functions with time as the independent variable appears to be critical, that
is thinking about the relationship as ( ) ( )[ ]215 thtV = (either consciously or unconsciously).
Finally, known values were substituted into the differentiated equation and the desired
rate was calculated. Each mathematician also checked that his answer was reasonable.
6.1.2 The Plane Problem
During the problem solving session, each of the mathematicians successfully
solved the plane problem:
A plane flying horizontally at an altitude of 3 miles and a speed of 600 mi/hr
passes directly over a radar station. When the plane is 5 miles away from the
station, at what rate is the distance from the plane to the station increasing?
The same related rates problem solving strategies of draw a diagram and label it,
construct a functional relationship, relate the rates, solve for the unknown rate, and check
the answer for reasonability were also seen in the mathematicians’ solutions to this
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problem. For brevity, I will focus on Bob’s solution as all three mathematicians produced
very similar solutions.
Bob’s solution to the plane problem began:
Excerpt 25
Bob: (1) A plane flying horizontally at an altitude of three miles and a speed of 600
miles per hour passes directly over a radar station. When the plane is 5 miles from the station, at what rate is the distance between the plane and the tower increasing?
(2) It’s traveling horizontal. It’s 3 miles up. There will be a straight line distance [artifact -diagram]
(3) I’ll call this, I’ll call this d. I’ll call it r, plane is 5 miles from the station. So, it’s going away. I’m calling this d. [artifact – labeled diagram]
Bob quickly drew a diagram and labeled it with variables and constants. Bob’s
diagram was of a plane traveling to the right over a tower and formed a right triangle.
(Line 2) This implies that he accurately interpreted the problem statement. Bob assigned
letters to the changing quantities which suggests that he accessed his content knowledge
of variable. (Line 3) Thus, it would appear that Bob had completed the first step in his
problem solving process, draw a diagram. While the plane could be traveling to the left,
none of the three mathematicians drew the diagram with that orientation.
Bob continued:
Excerpt 26
Bob: (1) You’re asking for a rate. So, I imagine you’re asking for the rate of change
of this distance with respect to time. (2) And we’re working in miles and hours, ok. So, d is a function of time, r is
a function of time, ok. So, dr dt is 300. Is that right? Or 600 miles per hour [artifact-rate next to diagram]
111
(3) Call this cap d, ok. So, I can relate these by d squared is equal to uh 3 squared plus r squared [artifact-equation]
Bob reread part of the problem to remind himself of the rate he wants to find. It
appears that he took a cue from the given rate and thought about each of the variables he
labeled in his diagram as a function of time. (Line 2) This is evidenced in his statement,
“we’re working in miles and hours ok so d is a function of time r is a function of time.”
He proceeded to write down the relationship 222 3 rD += which does not explicitly
represent time as the independent variable even though that appears to be how he is
thinking about D and r. (Line 3) Bob does not explicitly state that he is using the
Pythagorean Theorem to relate the changing quantities in the problem situation yet this
appears to be part of his content knowledge of geometry. Thus, Bob completed the
second step in his solution process, construct an algebraic functional relationship between
the variables in the diagram.
Bob proceeded:
Excerpt 27
1. Bob: So, using implicit differentiation. I’m using implicit differentiation because I’m imagining that each of these is a function of t, ok.
2. NE: Ok 3. Bob: And so, since I’m pretending that each of these is a function of t, then
I’ll just look at its rate of change with respect to t. So, 2 d dd dt is uh 2 r dr dt. [artifact-differentiated equation]
Bob appears to have accessed his content knowledge of derivative to state that he
will be using implicit differentiation to compute the derivative of his functional
relationship with respect to time. He noted that he was “imagining” and “pretending” that
each variable is a function of time so you look at the “rate of change with respect to t.”
112
(Lines 1, 3) Thus, it would appear that the phrase “with respect to” conveys information
about a relationship between the independent and dependent variable in the functional
relationship. As a result, Bob had completed the third step in his solution process, relate
the rates.
Bob summed up the key to being able to relate the rates for both the trough
problem and the plane problem in the following excerpt:
Excerpt 28
1. NE: How do you know to think about implicit differentiation?
2. Bob: Oh, because what I was saying was that I’m pretending, ok, you’re asking about the rate of change of this distance with respect to time
3. NE: Um hm
4. Bob: Right, and um, so I don’t see anyway to look at this as a composition of functions. Um and so, instead what I’ll do is I’ll just imagine that each of these is a function of time.
5. NE: Ok
6. Bob: So, they’re all functions of time, and that way then, uh then I can relate, you know, the the rate of change this way, with the rate of change this way, because both functions now both functions are functions of time.
7. NE: Ok
8. Bob: And it isn’t, ok, it isn’t that I’m… to me the key isn’t that I decide to use implicit differentiation
9. NE: Ok
10. Bob: The key to me is that I decided to think of all of these as a function of time.
11. NE: Ok. So, how is this different from the first problem?
12. Bob: Um [looks back at first problem] in the first problem, they’re all a function of time, but they’re separate functions of time. That volume is a composite function.
113
13. NE: Ok
14. Bob: So, then I suppose that’s where the chain rule comes in, you know. The emphasis on the chain rule because I was thinking, see I ended up thinking of the base as a function of height. So, it then, it became, the volume became a composite function where b is actually b of h and h is h of t, you see?
Notice that in Line 2, Bob stated that he is pretending each variable is a function
of time, and in Line 4, Bob stated that he imagined each variable is a function of time.
Then, in Line 6, he noted that everything is a function of time which allows him to relate
the rates. He concluded that the key is not that he decided to use implicit differentiation,
but rather that he decided to think of each variable as a function of time. (Lines 8-14)
Bob finished the problem:
Excerpt 29
1. Bob: And you’re looking at the distance is 5 miles away. So, 10 times the rate of change of the distance with respect to time is 2 r. [artifact-partial substitution into differentiated equation] Hmm, ok, could I ask a question?
2. NE: Um hm 3. Bob: Do you mean 5 miles away horizontally or do you mean 5 miles away
this way? 4. NE: I would say the 5 miles this way, not horizontal 5. Bob: This way 6. NE: Right 7. Bob:
a. Ok, ok. So, I need to find out what r is when it’s straight line 5 miles away. So, 3, 5, oh it’s 4. [artifact- draws triangle] How about that? So, 2 times 4 times dr dt, and we know the rate of change of r with respect to t is 600 miles per hour. [artifact-completes substitution] So, the distance is changing at uh 8 times 60, 480 miles per hour. [artifact-solution]
b. Does that make sense? [check] So, if it goes out that much [motions with hands then artifact- draws triangle on paper] Ok, so, what I’m wondering is, does it make sense that this hypotenuse distance is changing at a slower rate than the horizontal distance? [motions with hands]
114
c. So, if it goes out this much, then if we come down here, this would hmm. [artifact-draws another triangle] Yeah, it looks like, yeah I guess that makes sense because if you make a triangle longer this way, say you go out to infinity, then the two, the hypotenuse and this side are going to get closer and closer to the same length in ratio. [check]
8. NE: Um hm 9. Bob: So, this is going to become a larger percent of the diameter, and so that
means that as you make that horizontal length longer, you have to make it longer. Well, if you make it a little bit longer, then you’re making the hypotenuse longer. So, that they can approach each other in ratio. Ok, so I guess it sort of makes sense.
10. NE: Ok 11. Bob: All right. At least it being less than 600 miles per hour makes sense
[long pause –goes over calculation again] ok After computing an answer, Bob considered the possibility of the plane going out
to infinity and what would happen to corresponding lengths. (Line 7b, 7c) Bob also drew
additional diagrams to help him attend to the how the lengths will change in relation to
each other. This was evidenced in his statement, “I guess that makes sense, because if
you make a triangle longer this way, say you go out to infinity, then the two, the
hypotenuse and this side are going to get closer and closer to the same length in ratio.”
(Line 9) Dan made a similar statement:
Excerpt 30
Dan: That should be, uh faster than what this is changing at, which was, it
should be faster than, shouldn’t it be faster, it should be… [long pause] No, it’ll be slower because when it’s directly overhead the rate of change of d will be zero, and when, as we go to infinity, the rate of change of d approaches the rate of change of p. So, it should get closer to 600. So, that should be slower. So, that seems like a reasonable answer.
While Dan initially thought the rate between the plane and the tower should be faster than
the plane, he considered what happens as the plane goes to infinity and reached the same
conclusion that Bob did.
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All three mathematicians began the plane problem by reading the problem
statement and drawing a diagram which was labeled with constants and variables, a
solution artifact. The next solution artifact that was generated was an algebraic
relationship between the variables based on their content knowledge of geometry,
specifically the Pythagorean Theorem. This algebraic relationship was differentiated with
respect to time to relate the rates producing another solution artifact. Finally, the known
values were substituted into the differentiated equation and algebraic manipulations were
performed to obtain an answer which was then checked for reasonability. These steps are
similar to those that were used when solving the trough problem and generated similar
solution artifacts.
In the plane problem, the key to successfully relating the rates appeared to be the
ability to think about each variable as a function of time. While Bob expressed the
algebraic relationship between the variables as 222 3 rD += , he was actually thinking
about it as ( )[ ] ( )[ ]2223 trtD += , but likely subconsciously. Thus, Bob expressed the
derivative as dt
drr
dt
dDD 22 = while a more formal notation might be ( ) ( )
dt
drtr
dt
dDtD 22 = .
Even though one does not know the value of t or an explicit function for D(t) or r(t), the
problem provides enough information to obtain values for D(t) and r(t) at the instant we
are interested in.
6.1.3 The Coffee Cup Problem
Dan and Adam successfully solved the coffee cup problem:
Coffee is poured at a uniform rate of 20 cm3/sec into a cup whose inside is shaped
like a truncated cone. If the upper and lower radii of the cup are 4 cm and 2 cm,
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respectively, and the height of the cup is 6 cm, how fast will the coffee level be
rising when the coffee is halfway up the cup?
Bob did not complete this problem due to time constraints. Hence, the data presented in
this section is limited to the solutions provided by Dan and Adam.
For brevity, only Adam’s solution is presented as it highlights some interesting
thinking. Adam began:
Excerpt 31
1. Adam: All right. Well, this is my kind of problem. Coffee is poured at a uniform rate of 20 cubic cm per second into a cup whose inside is shaped like a truncated cone. Oh, this must be one of those horrible, uh um, hotel things, right? They have the little plastic deals with the little conical cups, and that always signals bad coffee. So, cause they always make it much too weak, you know. So, there you are in this windowless ballroom, drinking out of this flimsy. [hand motion] Yeah, I know what you’re talking about. All right, a truncated cone. All right, upper and lower radii of the cup are 4 and 2, um this makes a difference. I don’t know what the upper and lower radii means. Do you mean, is the cup 4 inches wide, so it’s a shallow cup, or a tall narrow cup?
2. NE: The cone has actually been truncated, so the bottom doesn’t actually exist.
3. Adam: Oh, oh, oh, I see. I’m still thinking about those horrible hotel cones. All right. So, it’s truncated like this. All right, gotcha, like that. [artifact-diagram]
4. NE: Right. 5. Adam: Ok, ok. When the coffee is half way. All right. All right. So, we have 4
cm. It’s a radius, not a diameter. So, it’s 4 cm across the top, 2 across the bottom. I see, all right, and the height of the cone is 6. [artifact- labeled diagram]
When Adam read the first sentence of the coffee cup problem, he interpreted the
words to construct a mental model of a conical coffee cup with which he was familiar,
those at conferences in hotels. However, as he continued to read the problem, the
language of upper and lower radii created a problem with his mental model. The way he
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interpreted those words did not support his understanding of a conical cup. It would
appear that he missed the term “truncated” in the first sentence or possibly did not attach
any relevance to the term. After it had been pointed out that the bottom of the cone does
not actually exist, he seemed to modify his mental model to accurately reflect the
problem situation and constructed an appropriate diagram. He also appeared to access his
content knowledge of geometry to make the distinction between a diameter and a radius
which allowed him to accurately draw and label the radii in his diagram. As we have seen
in the previous problems, the first step of draw a diagram had been completed.
Adam continued:
Excerpt 32
Adam: (1) I hope I’m getting these answers right. It’s kind of embarrassing; you
show this at my post tenure review, and I’ll get fired. (2) Let’s see. Ok, so now, we’re going to pour coffee into this thing, and we
want to know how fast the coffee is rising when the coffee is halfway up. So, this means we have to figure out first of all what the volume of this thing is. So what’s the volume? The volume is, um, is like one half pi r squared h, or something like that [artifact-equation]
(3) cause this is where you could give them the whole deal. There’s a factor of 2 pi about the, you know. Oh, there’s a factor of 2 pi here cause you do the little shells, and you rotate it around, right? Yeah. [hand motion] I don’t remember it, Nicole. pi r squared h. Does that sound right? I will go on that assumption here. You know, I’m terrible at memorizing formulas. Let’s see. Is one half the base times the height, and you rotate it around. No, that’s not right. That would be like a cylinder, a cylinder. [artifact-diagram]
Adam restated that he wanted to find how fast the coffee is rising when the coffee
is half way up the cup which caused him to identify that he needed to know what the
volume is. (Line 2) Thus, it would appear that the next step for Adam is to determine a
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formula for the volume of a cone. The volume of a cone formula is not readily accessible
as part of Adam’s content knowledge. In an attempt to reconstruct the volume of a cone
formula, Adam appears to have accessed his knowledge of integral calculus and the shell
method. (Line 3)
Adam spent a considerable amount of time trying to reconstruct the volume of a
cone equation as is evidenced in the following excerpt:
Excerpt 33
Adam: (1) Ok, so let’s see. So, what are we going to get? A cone. So, a cone is
something like this. [artifact-modified diagram] Um, so what happens to it? So, we have unit one half here, and oh, I think I’m missing a factor of a third or something like this hmm. I’m too lazy, let’s see, I did it this way [artifact-modified diagram]
(2) I’m trying to look at the cross sections, and remember how this goes. Um, so, pi r squared which is the volume of the cylinder, and a cone. Darn, I can’t remember. This is a problem when you haven’t taught the course in a while.
(3) All right, so we can take, let’s see. We can take this, this has height h, and radius r , and this area is one half uh h, r, and then we have to rotate this around 2 pi [artifact-another diagram-a triangle]
(4) Um, so, the volume here is dv θd . Would be, let’s see, one half h, r, um ok. So if we slice this a little bit, um we can call this delta theta, [artifact-modified diagram-a labeled triangle] and so the volume of this little wedge is going to be the area, which is going to be one half the base times the height times a delta theta, [artifact-equation]
(5) Uh, all right. So, that says that dv is one half h, r, θd . Um, and so that would be dv θd is one half h, r, and theta is going to go from 0 to 2 pi. So that means that the volume is one half h times r times 2 pi. [artifact-series of manipulations to get equation] Um, so the one for the cylinder is pi, r,
h. I’m assuming I’m doing this right. So, let’s see. So, theta has, oh units of, oh I see. So, this length here is θd ok. [erases previous work] So that means that the, that means that the little circumference here, the circumference is θ∆ times, I don’t know, r cause we had a circle of radius r, and all the way around is 2 pi r. All right, so that means that the volume of this wedge is one half h r squared delta theta. [artifact – equation] So that says the dv θd is one half h r squared, [artifact – equation] all right.
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So v is one half h r squared times 2 pi. [artifact – equation] So, so we have a factor of one half. So the volume of a cone is half that. [artifact – equation] It looks right cause I never remember what the formula is. Ok, let’s see, all right. So now this is taking forever. I’m sorry.
NE: That’s ok. Adam constructed a series of diagrams in an attempt to help him develop the
volume of a cone formula. (Lines 1, 3, 4) He drew a triangle to represent the elements
that are used in the shell method and labeled it with the radius, the height, and θ∆ . From
this diagram, he constructed a function that he could integrate. (Line 5) While this
approach did not produce the correct formula for the volume of a cone, it did suggest that
he had knowledge of how one might reconstruct it from first principles.
After Adam decided on the formula for the volume of a cone, he proceeded:
Excerpt 34
Adam: (1) I’m glad it’s not a timed test. All right. So, now we’re going to slice this
off. So, we have a cone. [artifact-modified diagram] Um, all right. So, now we have to figure out how much this is, and so we have to figure out the volume of the missing piece. So, what I’m going to do here is find the volume of the whole thing minus the volume of the missing piece. If I took this thing down to the tip and then that will give me the volume of the remainder.
(2) Maybe there’s an easier way to do this problem. That’s why we always have example 9, cause somebody else has actually thought about it, and so we can follow the pattern.
(3) All right, let’s see, uh is this going to get me anywhere? [check] As the radius decreases linearly, it goes from 4 cm to 2 cm in a distance of 6. Um, so that radius is going to be decreasing uniformly with height. So, um, if we went down another 6 cm, we’d have zero radius. So that would be the tip. So the whole the missing piece here would be a cone of height of 12 cm. So the volume of the missing, of the bottom is one half times pi times r squared, which is 2 squared times the h, which is 6. Um, all right. So this is going to be 3 times 4 times pi, I think, right. [artifact-volume of missing bottom]
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(4) And then the volume of the, the volume of the whole thing, that’s the volume that’s in the bottom. The volume of the complete cone is one half times pi times uh 4 squared times 12. Um, so this is 6 times 16 is 96 pi. [artifact-volume of complete cone]
(5) All right, the volume of the cup is the difference between the two. So we are 86, 84 pi [artifact-volume of cup] that’s a pain, let’s see [25:31]
Adam found the volume of the actual coffee cup by computing the difference
between the volume of the entire cone and the volume of the piece of the cone that does
not really exist. As in the trough problem, these calculations appear to allow Adam to
better understand what is happening in the problem situation.
After completing these calculations, Adam reconsidered how he should be
thinking about the problem situation.
Excerpt 35
Adam: (1) All right now, ok. So now actually what I need to think about here is, um,
should I think about height from the, um, the imaginary tip here? Or should I, should I start thinking about height from here on up? And I think about height from the bottom of the imaginary tip and hope that I made the correct adjustment when I get over here
(2) Um, cause all you want to know is the rate of the change in the coffee. So whether I measure from here or from down here, um, that doesn’t make a whole lot of difference. So, that’s important, ok.
(3) Let’s see. Um, yep, ok. Now in this case, since we have a cone, the volume depends upon the height. So in this particular case then, the radius is, um, so the radius… We’re going down here what, 2 cm in the space of 6 cm. So it’s changing at the rate of um one third cm per cm, so the slope is a third. So that means that this is one half times pi times one third h [artifact-volume equation]
(4) Ahh, I’m looking back to worry about whether to measure from my imaginary tip, or whether I’m going to start measuring from um the actual bottom. Um, what do I want to do here? This is much, boy this is, there must be a slicker way of doing this in the book cause it would be too hard for the students to be talking about, well, measuring from the tip or here or whatever. Um, so what should I think about? I’m going to go with the
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imaginary tip. So this is one third h, zero is here, 6 is here, and 12 is here. [artifact-modified diagram]
Adam was concerned about how he should measure the height of the cone. (Line
1) This lead him to the revelation, “cause all you want to know is the rate of the change in
the coffee so whether I measure from here or from down here um that doesn’t make a
whole lot of difference so that’s important,” and decided that he would measure from the
imaginary tip of the cone. (Lines 2, 4)
In the middle of deciding how he should measure the height, Adam also noted
that the volume depends on the height. (Line 3) To relate the volume and height, he
observed that, “so the radius we’re going down here what 2 cm in the space of 6 cm so
it’s changing at the rate of um one third cm per cm so the slope is a third so that means
that this is one half times pi times one third h” and modified his volume equation so that
it was in terms of height. He constructed a linear relationship between the radius and the
height of the cone and used composition to express the volume in terms of height. Thus,
it would appear that his content knowledge of functions was accessed to allow him to
relate the volume and the height of the cone. Adam also commented that an example in
the book would likely be helpful and provide a “slicker” way to solve the problem. (Line
4) However, this resource was not available to him in this session.
He continued:
Excerpt 36
Adam: (1) So this is what I would do in my office, and then I would think of a slicker
way of doing it. Then it would all look nice for the students. That’s why we always look like geniuses in front of the black board, cause ok.
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(2) Um, so do I have to worry about my offset here? Maybe not, ahh. I’m worrying about what this offset means here. I’d have to offset the bottom, ehh. I’d have to worry about, see what I’d have to worry about is this, uhh, maybe. Let’s see. I have to remember to subtract 12 pi from everything. I don’t know. What would I do?
(3) If I were to think out loud about this, I would say, ok my guess is it doesn’t matter. See, because if you had a, if you had a really truly conical coffee cup, not one that was lopped off here, you could just as well imagine. So instead of worrying about, so what I would tell the students is, I would say, well instead of worrying about this kind of cup, um, think about this kind of cup. Just imagine that you already filled it up with coffee to here, um, because who cares about the total volume. All you want to know is how fast the coffee is rising when you’re at this volume. So you’re, it doesn’t really [hand motion] So it’s the same answer. Getting the students to see that is perhaps one of the things that distinguish the expert like me from them.
(4) Of course, I could be getting all the answers wrong, and you’ll be laughing at me later on in the cutting room. Look at this professor who can’t even do the problem. Ok, all right. So, I’m going to ditch this stuff. I’m just going to look at this one. [indicates diagram in middle of paper]
(5) And um, all right. So now, um, in the cup, all right. So this, for that reason I’m just going to look at this long cup. And so now, I’m going to measure height from the tip of the cup. So this is h, and here is h equal to zero, and here is going to be h equal to 12. [artifact-labeled diagram] All right. So halfway up the cup then is going to be, I’m going to worry about the point where h equals 9, because that’s the point where the height in my imaginary cup corresponds to halfway up the actual cup. [artifact-modified labeled diagram]
As Adam continued to struggle with how to measure the height of the cone, he
appeared to have a revelation about how the problem situation worked. (Line 2) He then
stated how he would it explain it to students, “well instead of worrying about this kind of
cup um think about this kind of cup just imagine that you already filled it up with coffee
to here um because who cares about the total volume all you want to know is how fast the
coffee is rising when you’re at this volume.” (Line 3) He realized that it did not matter if
the cone had been truncated. Adam’s mental model appears to have undergone several
changes as he developed his formula for the volume of the cone and what each variable
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meant in terms of the problem situation. It would seem that each time he played out the
situation in his mind he gained a little insight until he finally understood how the
situation worked. It is at this point that Adam completed his step of construct an algebraic
relationship between the variables. An important part of completing this step was that he
understood exactly what the relationship between the variables was. (Line 3) Thus, his
algebraic equation had meaning in the context of the problem.
Adam then proceeded to simplify his volume equation as is seen in the following
excerpt.
Excerpt 37
Adam: All right, um, ok. So we’re going from a radius of 4 at the top to a radius of zero at the bottom over, over a space of 12 cm. So that means that the radius changes, uh, as the height over three, and that’s what we’re going to go with. So this is one half pi. Let’s see. This is a kind of small piece of paper. So, all right, so this is one half pi times the radius, which is one
third h, all squared, times h. [artifact-volume equation hhV
2
3
1
2
1
= π ]
Um, so we get a ninth times a half. So this is 18. So this is pi over 18 times h cubed. That’s the volume of my reshaped cone here, [artifact-volume
equation 3
18
1hV π= ]
After simplifying his volume equation, Adam continued to his next step of relate
the rates:
Excerpt 38
Adam: And, um, we’re interested in dh dt. So dv dt is dv dh times dh dt. [artifact-
general chain rule form equation dt
dh
dh
dV
dt
dV⋅= ] Um, and that means that
dv dt is 20 cm per second cubed, cm per second, and dv dh, uhh is what it’s going to be, pi over 9 times h squared times dh dt. [artifact-
differentiated equation dt
dhh ⋅= 2
920
π]
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Adam identified dt
dhas the rate he wanted to find and then expressed a
relationship between the rates using the chain rule. He did not operate on the volume
equation he constructed. Rather, he expressed a relationship between the rates using the
chain rule and then associated each rate with a numeric value or a function.
Adam completed his solution:
Excerpt 39
Adam: (1) All right. So when you’re halfway up the cup, that’s when h is equal to 9.
So we have pi over 9, times 9 squared, times dh dt equals 20 [artifact-substituted in values into differentiated equation]
(2) And this is all going to be cubic cm per second because you have 9 squared, 9 square cm here, and we have cm per second here. So, our units are consistent. [check]
(3) Um, all right. So this 9 cancels one of these 9s. So I have 9 pi times dh dt, and so I would say then that the height dh dt is 20 over 9 pi, uh, cm per second [artifact-solution]
(4) Looks right. So, the reason I’m worried about the units here… let’s see. We have 20 on the left, here we have cubic cm per second, over here we have squared cm, and we have cm per second. Um, so when I do the division here, I have a one over cm squared so over here, and therefore my answer here is in terms of cm per second, which is consistent with the idea of the height rising. [check] Um, so that’s the answer, I think.
After he substituted the known values into his differentiated equation (Line 1),
Adam quickly obtained an answer to the problem. (Line 3) He performed a unit analysis
to check his answer for reasonability. (Lines 2, 4)
In Adam’s solution to the coffee cup problem, he frequently referenced and
interacted with the numerous diagrams he drew. This suggests that Adam is regularly
manipulating and modifying his mental model of the problem situation. While Dan
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constructed his solution to the coffee cup problem, he also regularly referenced his
diagram.
An interesting aspect of Adam’s solution is that even though he expressed the
volume equation in terms of height, he did not operate on the function he constructed to
relate the rates. Instead, when he carried out the relate the rates step, he wrote out the
chain rule as dt
dh
dh
dV
dt
dV⋅= . (Excerpt 38) He then proceeded to associate each rate, such
asdh
dV, with either the appropriate numeric value or function. He did this when solving
the trough problem, too. Neither Dan nor Bob related the rates in this manner on any
problem they solved. They operated on the algebraic equation they constructed.
In solving the coffee cup problem, the mathematicians used steps similar to those
employed when solving the trough problem and the plane problem. These steps generated
solution artifacts. Adam and Dan both drew and labeled diagrams that were frequently
accessed and used. They also constructed and combined algebraic relationships as needed
to relate the volume and the height. After relating the volume and the height, each
mathematician differentiated his algebraic relationship with respect to time and known
values were substituted into this equation. Adam and Dan then performed algebraic
manipulations to yield a solution which was checked for reasonability.
6.1.4 Summary of the Mathematicians’ Solution Process
After reading the problem statement and identifying the problem as “a related
rates problem,” the mathematicians appeared to move through the following phases to
construct their solutions: draw a diagram and label it, construct a functional relationship,
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relate the rates, solve for the unknown rate, and check the answer for reasonability.
Completion of each phase resulted in the creation of one or more solution artifacts, a
written piece of information that conveys information about the solution process and
serves as an additional resource for the problem solver. An example of a solution artifact
is Dan’s diagram for the trough problem. Dan drew two perspectives of the trough, but
chose to use only one during his solution process (a cross section). He justified his
approach by saying, “I don’t necessarily need to show this part because the relationships,
I can capture just here.” He labeled this perspective with constants and variables. Later,
he modified his diagram by extending axes through his diagram and used the axes to
determine a relationship between the height and width of the water. He would later be
able to this relationship between the height and width of the water to eliminate a variable
(width) in the volume function. He referenced his diagram yet again when he determined
he had an error. He stated, “Um, as opposed to a constant, which didn’t make me happy
because I was pretty sure that as it filled up the volume changed.” While making this
statement, he gestured with his hand in an upward motion in relation to his diagram to
indicate the volume increasing. In determining the relationship between the height and
width of the water in the trough, Dan created another solution artifact. While he wrote
down the proportional relationship, 3
1=
w
h, based on his diagram, he used an alternate
form of the relationship ( hw 3= ) to further his solution process. Thus, I have labeled
every written piece of information in the transcripts as an artifact. However, certain
artifacts become additional resources for the problem solver and are used to advance the
solution process. Accurate diagrams allowed the problem solver to identify important
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functional relationships, these functional relationships allowed the mathematicians to
eliminate variables, and a functional relationship in two variables facilitated their relating
the rates by an application of the chain rule. These solution artifacts became key for
advancing the solution process.
The mathematicians appeared to access their content knowledge of geometry to
represent the problem situations with diagrams. When solving the trough problem, their
knowledge of isosceles triangles and prisms allowed them to correctly interpret the
problem statement that was reflected in the diagrams they drew. Dan used this knowledge
when considering multiple perspectives that allowed him to draw one diagram which
captured all the important relationships. Similarly, when solving the plane problem, their
content knowledge of geometry allowed them to construct a right triangle to represent the
relationship between the tower and the plane. When solving the coffee cup problem, they
also accessed their knowledge of cones and represented the known relationships of a cone
in the form of a diagram. In both of these situations, the diagrams provided a useful
artifact for advancing their solution process.
Their content knowledge of variable was also accessed as they proceeded to label
their diagrams. Quantities that were constant were usually labeled with the appropriate
numeric value, and quantities that changed were assigned a letter. The letter that was
chosen to represent a variable was frequently chosen in a manner that allowed the
problem solver to remember what it represented, for example, h represented the height of
the water in the trough.
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After the diagram had been labeled, the problem solver proceeded to construct an
algebraic relationship between the variables labeled in his diagram. To do this, the
mathematicians appeared to access their content knowledge of geometry and functions.
Their content knowledge of geometry provided a general relationship to find the volume
of the trough or cone. Then their interpretation of the problem statement informed them
that they wanted “volume as a function of height.” Their content knowledge of functions
informed them that composition could be used to eliminate unnecessary variables. Thus,
they accessed their content knowledge of geometry to identify a similar triangles
relationship which provided the second function needed for the composition.
In the plane problem, their content knowledge of geometry allowed them to
identify a right triangle in their diagram and provided the Pythagorean Theorem as a
means to relate the variables. Their content knowledge of functions allowed them to think
about each variable as a function of time.
Once a satisfactory algebraic relationship had been obtained, the mathematicians
related the rates in the problem. They appeared to access their content knowledge of
function and derivative. At some point in the problem solving process, the
mathematicians interpreted the given rate which had time as part of its units to mean that
they needed to find the derivative “with respect to” time. To take the derivative “with
respect to time,” each mathematician said he would use the chain rule or implicit
differentiation.
To solve for the unknown rate, the mathematicians substituted the known values
into their differentiated equation and applied algebraic manipulations to compute an
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answer. This answer was then checked using a unit analysis or covariational reasoning.
When the answer was checked using covariational reasoning, the mathematicians either
compared their answer to another known value such as the average rate of change or they
made statements that indicated they were playing out the situation in their minds. As they
played out the situation in their minds, they talked about how changes in one variable
affected changes in another variable.
6.2 A Framework for Solving Related Rates Problems
The data from the mathematicians suggested that after identifying the problem as
a related rates problem, certain knowledge and procedures should be employed to solve
the problem. The manner in which content knowledge and procedures are accessed to
produce solution artifacts may be described as phases of the solution process. Evidence
that one is engaging in these phases of the problem solving process for related rates is
provided by one’s verbal statements and the generation of solution artifacts. These phases
are: draw and label a diagram, construct a meaningful functional relationship, relate the
rates, solve for the unknown rate, and check the answer for reasonability. It should be
noted that these phases are similar to the seven steps previously presented in the literature
review.
At the beginning of this chapter, an artifact was identified as a written piece of
information that is the result of one’s mental activities. A solution artifact is more than a
piece of information that has been written down as part of the solution process. A
solution artifact is a written piece of information that is referenced or used by the
problem solver to continue the solution process; it is the creation of an additional
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resource. The solution artifacts generated by mathematicians as they carry out their
related rates problem solving strategy may include the following: a diagram, a labeled
diagram, a functional relationship between the variables labeled in the diagram, possibly
a second relationship between some of the variables in the diagram that was used to
eliminate one or more variables through function composition, an equation that has been
differentiated with respect to time, algebraic manipulations of the differentiated equation,
and a final solution. To generate each solution artifact, one likely needs to access relevant
heuristics and content knowledge in addition to creating a mental model of the problem
situation. A description of possible solution artifacts generated by each phase and the
possible access of other heuristics, content knowledge, and mental model of the problem
situation is now presented.
Draw a diagram. A mental model is a structural analogue of the world as defined
by Johnson-Laird (1983). That is, a mental model is a representation in the mind of a real
or imaginary situation (Byrne, 2006). Evidence that a mental model is being formed or
has been formed may come from verbal statements about what is being imagined or from
a drawing or diagram. An individual’s interpretation of the words in the problem
statement is usually reflected in his mental model. Constructing an accurate drawing or
diagram of the problem situation may be evidence that an individual has constructed a
mental model.
Recall that Booth and Thomas (2000) made the distinction between a picture of
reality, a drawing (closely models the picture of reality), a diagram (a more abstract
representation of reality), and a theoretical mathematical figure (the most abstract
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representation of reality). The distinction between a diagram and a theoretical
mathematical figure in the context of related rates is likely that a diagram represents
exactly one instance of the problem situation (the instance in which the water level in the
trough is half full is represented by drawing a line approximately halfway up the
diagram). A theoretical mathematical figure, on the other hand, represents any possible
configuration of the problem situation and contains associated conceptual knowledge
within a single diagram (while the water level in the trough may be drawn at about
halfway up, h could represent any height of the water).
The ability to manipulate a mental model may be important to understanding how
the problem situation works. Evidence that individuals may manipulate their mental
models may be found in their ability to draw multiple perspectives of the same object.
The ability to imagine the system changing continuously and to consider one static state
of the problem may also suggest that one can manipulate one’s mental model. An ability
to easily move between considering the dynamic system and the static system is likely
necessary to be able to successfully solve a related rates problem. Verbalizations about
the relationships between the changing quantities in the problem may be described by the
mental actions of Carlson et al.’s (2002) covariational framework as described in Chapter
3. Table 12 provides a summary of the Draw a Diagram Phase.
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Table 12: Summary of the Phase: Draw a Diagram
Phase
Draw a Diagram
Solution
Artifacts • Diagram that accurately represents the problem situation
• Diagram that has been labeled with constants and variables
• Other diagrams representing different perspectives of the problem statement
Mental
Model • Describe what one is imagining or picturing in one’s mind
• Anticipate relationships that may exist
• Attend to the nature of the changing quantities o Attend to the direction of the change in the variables o Attend to the amount of change in the variables o Attend to the average rate of change in the variables
• Attend to continuous changes in the variables
Content
Knowledge • Geometry o Accurately interpret terminology o One diagram may represent any of the possible states of the
problem situation
• Variable o Label constants and variables appropriately
Heuristics • Ask or consider, “What is ______?”
• Restate the problem or parts of the problem in one’s own words
• Ask or consider, “Which perspective of the geometric shape will provide the most information?”
• Ask or consider, “Do I need to draw more than one perspective of the problem situation?”
Construct Meaningful Functional Relationships. Known geometric relationships
provide a means to begin relating the variables in the problem situation. It is important to
be able to understand the role of the independent and dependent variable in a functional
relationship. To express one variable “in terms of” or “as a function of” another variable
suggests that the first mentioned variable is the dependent variable and the latter is the
independent variable. In an attempt to express one variable “in terms of” another,
function composition may be utilized to eliminate one or more variables in a functional
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relationship. Sometimes it is not possible to use function composition to eliminate one or
more variables, but the relationship may be solved for the desired variable. There is also
the possibility that one may do neither of the aforementioned possibilities and the
relationship is parametric in nature. Table 13 summarizes the phase of Construct
Meaningful Functional Relationships.
Table 13:
Summary of the Phase: Construct Meaningful Functional Relationships
Phase
Construct a Meaningful Functional Relationship
Solution
Artifacts • Algebraic equation (s) to relate the variables in the diagram
Mental
Model • Interact with the mental model to determine which variables need to
be related
Content
Knowledge • Understanding the nature of functional relationships o Use a diagram labeled with variables and constants to identify
relationships o Understand the role of the independent variable and the dependent
variable in a functional relationship o Understand what relationship between the independent and
dependent variables is determined by the phrase “in terms of” o Understand that function composition (or substitution) allows one
to construct a new function from two or more smaller functions, eliminating one or more variables
Heuristics • Relate the variables representing the known rate and the unknown rate
• Eliminate variables if possible
Relate the Rates. To relate the rates in the problem situation, one differentiates the
algebraic relationship with respect to time using the concepts of implicit differentiation
and/or the chain rule; this is the only place one actually writes out time, t, as a variable in
the problem. The chain rule affords one a way to compute the derivative of composite
functions easily by computing the derivative (rate of change) for the individual functions
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that are being composed and multiplying them. The rates are multiplied because the
independent rates of change are happening in the order in which the functions are
applied. Hence, the first function provides a scalar to the next function applied. The
phrase “with respect to” should have meaning. It may be important to think about “the
rate of change of height with respect to time” meaning height is the dependent variable
and time is the independent variable. It may be the case that the individual is computing
the derivative by acting on the abstract function representation of f to represent f(t).
However, the individual is likely not consciously thinking about the function as being
represented by f(t). Table 14 summarizes the phase of Relate the Rates.
Table 14:
Summary of the Phase: Relate the Rates
Phase
Relate the Rates
Solution
Artifacts • Differentiated algebraic equation
• Chain rule equation
Mental
Model
Content
Knowledge • Understanding the nature of rate of change o Understand what relationship between the independent and
dependent variables is determined by the phrase “with respect to” o Interpret from the given rate that time is the independent variable in
the functional relationship o Imagine each variable in the functional relationship as a function of
time o Differentiate the functional relationship “with respect to” time
• Perform differentiation operations on an implicitly defined function
Heuristics • Differentiate “with respect to time”
Solve for the Unknown Rate. After relating the rates by computing the derivative
with respect to time, one may substitute in the known values and use algebraic operations
to find the unknown rate. The computation of the unknown rate relies primarily on the
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problem solver’s algebraic manipulation skills. Table 15 summarizes the phase of Solve
for the Unknown Rate.
Table 15:
Summary of the Phase: Solve for the Unknown Rate
Phase
Solve for the Unknown Rate
Solution
Artifacts • Algebraic manipulations of the differentiated equation
Mental
Model
Content
Knowledge • Algebraic Knowledge o Substitute in known values for variables o Apply algebraic operations to the equation to calculate the unknown
rate
Heuristics • Perform the same operation on both sides of the equation
Check the Answer for Reasonability. Asking the question, “Is my answer
reasonable?” should be a natural part of the problem solving process. Performing a unit
analysis provides a certain measure of checking that the computations are correct. The
ability to manipulate the mental model and attend to the nature of the changing quantities
as described above may also allow an individual to answer this question successfully.
Table 16 summarizes the phase of Check the Answer for Reasonability.
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Table 16:
Summary of the Phase: Check the Answer for Reasonability
Phase
Check the Answer for Reasonability
Solution
Artifacts • Notation of units
• Other calculations
Mental
Model • Manipulate the mental model o Attend to the amount and direction of change in the variables o Compare the answer to another known quantity such as the average
rate of change
Content
Knowledge • Measurement units
Heuristics • Ask “Is this answer reasonable?”
• Perform a unit analysis, i.e. check that the units work out or match up
As each solution artifact is generated, the mind may access heuristic knowledge,
content knowledge, previous solution artifacts, and use the mental model of the problem
situation. Figure 2 provides a schematic that attempts to capture these relationships in a
meaningful way. This figure was developed from examining the mathematicians’
solutions to the trough problem, the plane problem, and the coffee cup problem which
were detailed in sections 6.1.1, 6.1.2, and 6.1.3 above.
This framework focuses on the relationships that exist between the columns of
resources and heuristics in Carlson and Bloom’s (2005) multidimensional problem
solving framework which was presented in Chapter 3. The problem solver’s resources
and heuristics are likely supplemented with a mental model. Resources are considered to
be one’s mathematical content knowledge, heuristics are strategies an individual chooses
to apply to the problem situation, and a mental model is a representation in one’s mind of
the problem situation. The resources, heuristics, and mental model reside in the mind of
the individual, and as an individual accesses this knowledge and applies it to the problem
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situation, solution artifacts are generated. The problem solving cycle of orient-plan-
execute-check is happening in the background and in some instances it is natural to use
the language of those phases to describe what is happening, but this is not the focus of the
following framework.
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Figure 2: The solution process for related rates problems.
Problem Statement
Solution Artifact
Mental Model
PS
Heuristics Content Knowledge
Previous Solution Artifacts
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The top box labeled problem statement represents the physical statement of the
problem situation on a piece of paper or in a book. The process of finding a solution to a
given problem begins when an individual reads the problem statement. The octagon in
the middle of Figure 2 represents what may be happening in the individual’s mind as he
constructs each piece of his solution. It represents the collection of resources available to
the problem solver. Each of these resources may be accessed individually or in various
combinations. As the individual reads the problem, he begins interpreting the words and
constructing a mental model of the problem situation. It is not possible to see inside a
person’s mind and what he may be visualizing, but it is possible to make conjectures
about his mental model based on statements he makes. Thus, the mental model of the
problem situation is represented by a thought cloud in Figure 2. An integral part of the
mental model is the individual’s interpretation of the problem statement and this is
represented in Figure 2 by the PS in the star shape. It would appear that the problem
solver holds much of the problem statement information in his memory (either accurately
or inaccurately), but may occasionally return to the original problem statement to check
the accuracy of his mental model. Reading the problem statement also likely causes the
problem solver to access additional problem solving heuristics, some of which may be
specific to related rates problems and others more general. Furthermore, mathematical
content knowledge may be called upon. The mind of the problem solver accesses
heuristics and content knowledge and manipulates the mental model to produce a solution
artifact. The third box in Figure 2 represents a written piece of the solution, a solution
artifact, which is the result of the interactions in middle section of the framework. The
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solution artifacts appear to be generated as a result of completing one step of the solution
process. The problem solver then cycles back to the problem statement (either the
original or the one represented by his mental model) and goes through the process again
to obtain the next solution artifact. As each solution artifact is generated, it becomes an
additional resource for the problem solver. It should also be noted that after a solution
artifact has been constructed, the problem solver may choose to check that part of his
solution process. This cycle continues until the problem solver reaches a solution.
Content knowledge of geometry, variable, function, and derivative is necessary at
different points in the problem solving process. Knowledge of geometry is necessary to
correctly interpret the words in the problem, draw an accurate diagram, and choose an
appropriate relationship between the variables. Knowledge of the concept of function is
necessary to apply composition to eliminate unnecessary variables and successfully apply
the chain rule. Knowledge of the concept of derivative is necessary to successfully
compute derivatives.
The heuristic knowledge that is accessed in the second box is likely related to
one’s content knowledge that has been accessed recently. For example, reading “how
fast” in the problem statement may cause one to rely on the heuristic that “how fast”
indicates we are talking about a rate. This heuristic may also be part of one’s content
knowledge. Many of the heuristics employed are likely related to how one interprets
phrases from the problem statement and are difficult to discern from content knowledge.
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6.3 Using the Framework
To apply this framework to data, the first thing to do is to identify where solution
artifacts are generated in the problem solving process. This may be done by marking
where the solution artifacts are generated in a transcript file while watching the video of
the problem solver. It is also useful to have a copy of the solution available for reference.
Then, try to identify which segments of the data correspond to each of the identified
problem solving strategies. These will likely be tied to the solution artifacts. One may
then consider making a summary table utilizing the phases presented in Tables 12-16.
Identify the problem solving phases that guide the solution process and their
corresponding artifacts. This will segment the data into pieces for analysis. Record them
in the first two rows of the summary table. Then examine the thought process for each
step to determine: 1) what their mental model consists of, 2) how, when and what content
knowledge they appear to access, and 3) any other heuristics that may be used.
Summarize this data in the appropriate rows. See Table 17 for a sample summary table.
Use the table summary to explain how the student is thinking as he works though each
step of the problem. Proceed to the next data segment.
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Table 17: Sample Summary Table
Phase
Relate the rates (lines 122 -214)
Solution
Artifacts ( )( ) 2
12
2
6009
600
t
t
t
z
+
=∆
∆
Content
Knowledge
Derivative
• Rate of change
o Interpret the given rate as 600=∆
∆
t
x.
Mental
Model • The students appeared to focus on the equations they just created.
• The students do not appear to reference their diagrams at all in this segment suggesting they may not be utilizing their mental models of the problem situation.
Heuristics • Write a “delta equation.”
This is how the framework will be used in Chapter 7 to analyze the teaching experiment
data.
Chapter 7: Teaching Experiment Results
In this chapter, I present the results of the teaching experiment described in
Chapter 5. Before the start of the teaching experiment, each student was interviewed
individually to ascertain their understandings of variable and function. The results of the
individual interviews are presented in section 7.1. The instructional intervention was
designed to foster the students’ ability to reason about related rates problem situations by
using a custom computer program to model problem situations. The computer program
was also designed to call attention to what it means for a variable to be a function of time
and the implications this may have when considering rates of change. The results of the
students’ interaction with the computer program are presented in section 7.2. In section
7.3, the results of a discussion about the chain rule are presented. This chapter concludes
with the results of students solving related rates problems in section 7.4.
7.1 Students’ Individual Interviews
Each student was interviewed individually the day before the teaching experiment
began. The concepts of variable and function appear many times in the descriptions of the
content knowledge section of the framework. Thus, the students’ understandings and
misconceptions of variable and function may influence what happened in the teaching
experiment. In this section, I present the results of the individual interviews on the
students’ understandings of variable, function, and covariational reasoning.
7.1.1 Initial Understandings of Variable
In the individual interview, students were asked to circle all of the items in a list
that they believed describe the meaning you give to the letter x when it is used in
mathematical expressions. They were given the options of: meaningless mark on the
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paper, changing quantity, generalized number, letter that stands for a number, name of
something, place holder, unknown, or other and asked to please specify. None of the
students thought that it was a meaningless mark on the paper. Ali believed it was a
changing quantity and that it could be a generalized number while Ann and Ben did not.
When asked why a variable would not be a changing quantity Ann said, “it’s not a
changing quantity because you would need a different variable for each unknown
quantity so x has to be constant valued.” Ann also stated “um a generalized number um
I’m not going to say that one just because I don’t think it’s generalized I think it has to be
a specific number because it can only represent one number in an equation” when asked
why it was not a generalized number.
All three of the students believed that a variable is a letter that stands for a number
and an unknown quantity. Ali and Ben thought that it would not be the name of
something. Ann thought that a variable could be the name of something and used the x-
axis as an example. On deciding whether a variable could be a place holder the votes
were split three ways: Ben did not think a variable was a place holder, Ali thought that it
was, and Ann said maybe. In Ali’s own words, “it just like makes sure it’s there, you
know it leaves a space for um a value to be plugged in.” and Ann said, “maybe it would if
it’s a place holder it would be holding the place of the unknown variable… you know like
if you have 3 x it would be holding that place cause you know three’s going to be
multiplied by something but you don’t know what’s in that place.”
Another question in the individual interview asked the students:
Consider the function given by ( ) ( )( )uuh sincos= . Let ( )uhy = .
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The students were asked to identify the independent variable, the dependent
variable, and the name of the function. The students had little difficulty identifying the
independent and dependent variables as is evidenced in the transcript excerpts below.
Excerpt 40
1. Ali: (Reads question) Um, the independent variable is u
2. NE: How do you know that?
3. Ali: Because the value of u does not depend on what the other, um what, like what number you have to plug in, whatever the output is to plug it in um
4. NE: Ok
5. Ali: Which letter indicates the name of the dependent variable? Um, I’d say y is the dependent variable because depending on what you plug in for u it gives you different y
Note that Ali used the term output to describe the dependent variable in Line 3. She did
not use the term input for the independent variable. Ali actually made very little use of
this language in talking about functions. In contrast, Ann used the terms input and output
frequently when talking about functions.
Excerpt 41
1. Ann: (Reads question) Ok it would have to be u
2. NE: Why?
3. Ann: Because h of u of y will change as u changes
4. NE: Ok
5. Ann: Right, and the independent variable changes before the dependent one
6. NE: Ok
7. Ann: And the dependent variable would be y because, because when you’re, yeah because as u changes the same one [points to previous problem in which she had also used the terms input and output to describe the
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parts of the expression and the relationships that exist in the given expression: ( )wur = ] as this one for different inputs you get different outputs
Ann talked about the variables changing (Lines 3, 7) even though she did not think they
were changing quantities when asked about the definition of variable. She used the terms
input and output as she clarified for herself the distinction between the independent and
dependent variables, “for different inputs you get different outputs.” (Line 7)
When responding to the items in the individual interview, Ben never used the
language of input and output to describe the independent and dependent variables. While
he is able to identify the independent and dependent variables correctly, he appears to
have a difficult time articulating the relationship that exists between them as is evidenced
in Lines 3 and 7 of the transcript excerpt below.
Excerpt 42
1. Ben: Number 3 (reads question) Which letter indicates the name of the
independent variable? u
2. NE: How do you know that?
3. Ben: Cause (shrugs)
4. NE: It’s one of those things you just know?
5. Ben: It is yeah
6. NE: Ok
7. Ben: Which letter indicates the name of the dependent variable? And that’s y cause that’s the other variable
8. NE: Ok
9. Ben: And because y is determined by whatever u is. So if u is a different number, y would change so it’s dependent on u, and u doesn’t need anything else to change
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It is interesting to note that Ben talked about the variables being able to change even
though he did not think they could be changing quantities in the first question (Line 9).
Thus, each of the students was able to correctly identify the independent and dependent
variable in the given expression.
The students appear to believe that a variable is a letter that stands for a number
and an unknown quantity. Only Ali reasoned that it may be a changing quantity when
presented with definitions. All the students were able to identify the independent and
dependent variables. However, Ann and Ben did talk about the variables changing when
describing the relationship between the independent and dependent variables. This
suggests that their understanding of variable is primarily static. While each variable may
take on different values, the students appear to primarily think about it one at a time and
in the context of something to be solved for.
7.1.2 Initial Understandings of Function
During the individual interviews, discussing the relationships that are represented
by a given function was more problematic than identifying the parts of the expression.
This suggests that the students’ understanding of the concept of function may be weak.
Using the term input to describe the independent variable and the term output to describe
the dependent variable has been shown to be useful for students in understanding the
concept of function. However, the students did not employ this language to their
advantage.
The majority of the time in the individual interviews was spent on the following
problem:
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Suppose that a genie gave you magical envelopes, each of which will change your
ATM deposits. Envelope A will divide your deposit by 3 and subtract 5. Envelope
B will square your deposit. You must put one envelope inside the other. When
will it be best to put B inside A and when will it be best to put A inside B?
The purpose of this question was to illuminate how students deal with function
composition in a contextual situation and whether they were able to move flexibly
between function representations. The problem was difficult for each of the students in
different ways. Ali and Ann struggled to solve a quadratic, while Ben had difficulty with
the composition. Each of the students was also dependent on their calculator to provide
graphs of the functions.
Ali began the problem by writing out functions for a of x and b of x and
immediately wanted to set them equal to each other so she could solve for x, but decided
that she did not know how to do that. She proceeded to write down the appropriate
composite functions and then the following exchange occurred:
Excerpt 43
1. NE: What are you thinking about?
2. Ali: I’m thinking I can picture it in my mind, how this happens, but I can’t figure out a way to solve it so that I could figure it out. Unless maybe take the derivative of it. I don’t know.
3. NE: What kind of things do you usually try when you kind of get stuck on a problem?
4. Ali: Just plug in different numbers, I guess. Um I dunno. I like to have it like visual kind of so
5. NE: Is there a way you can visualize this?
6. Ali: Not really. So we’re trying to, uh, maximize it I guess. So we could take the derivative of both the equations then set them equal to each
149
other and that would give us the value that it switches from one being best to the other being best, maybe.
Ali appears to be drawing on recent knowledge of derivatives and knows that they may
be used to find maximums. However, in Line 6, she again wanted to set the functions
equal to each other so she can solve for x. It would appear that setting two equations
equal to each other and solving for x is a primary problem solving strategy for Ali.
After completing the calculations for this approach (using the derivative), Ali
decided it does not help her. Ali then proceeded to try some numbers as a result of being
asked what she does when she gets stuck on a problem, leading to the following
exchange:
Excerpt 44
1. NE: So what does that tell you?
2. Ali: It tells me absolutely nothing except for that if the value is 5 that sticking it in envelope a first gives you more money. I guess you could do that with all of them but that would take too long to figure out.
3. NE: Ok, when you say with all of them
4. Ali: With every x value
5. NE: Ok how many x values are there?
6. Ali: A lot (laughs) um, you can’t have negative deposits so you have to restrict your domain to um zero to infinity, I guess. Just whatever x value
7. NE: So could you include zero or not?
8. Ali: No, because if you’re deposit something you wouldn’t deposit zero
9. NE: Ok
10. Ali: It would just, you know, um maybe you could just set the two functions together, not the derivatives of them, and then that could possibly give you an x value that you could use. So, let’s see, so x squared divided by 3 minus 5 equals x divided by 3 minus 5 squared.
150
Hmm, I guess we’d take the natural log of both sides and see how the natural log of x squared minus 3 minus natural log of 5 equals two natural log of x divided by 3 minus 5. Hmm.
Ali recognized that she cannot possibly try every possible x-value and subsequently
restricted the domain to be only positive values in Line 6. Finally, in Line 10, she decided
that she can set her original composite functions equal to each other and solve for “an x
value that you can use.” She incorrectly solved for x using natural logarithms rather than
the quadratic formula, completing the square, or an alternate acceptable means. After
determining where the functions are equal, she quickly decided that she may use a line
graph to check values on either side of her calculated value to determine the best
outcome.
In summary, Ali’s primary problem solving strategy was to set two functions
equal to each other and solve for an x-value. She applied this heuristic to the individual
functions of a and b, the derivative of the composite functions, and the original composite
functions. This suggests that Ali knows how to perform algebraic operations on an
equation to find a value, but she does not know what the functions represent. Thus, Ali
blindly applied this strategy to each pair of equations she generated. Another difficulty
arose for Ali when she needed to solve a quadratic. She used logarithms, an inappropriate
solution method. In her homework write-up of this problem, Ali used only the algebraic
representation to show her computation of the x-value and wrote the following verbal
explanation:
To find out which envelope should go in which to get the maximum amount of
money, you must write two functions, one with envelope a on the inside and the
other with b on the inside. To figure out where these two functions are equal, you
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set them equal to each other and solve for x. This value of x is the point at which
the output value switches from being maximum in one to the max of the other.
Ali’s description is lacking detail about what the input and output of the functions
represent. See Figure 3 below for Ali’s solution to the genie problem.
Figure 3: Ali’s solution to the genie problem.
Ann’s difficulties with genie problem were primarily related to algebraic errors.
She immediately wrote down the composite functions without labels which prompted the
following exchange:
Excerpt 45
1. NE: Explain to me what you’re doing.
2. Ann: I’m composing the functions.
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3. NE: Ok
4. Ann: Um, I just called a g of x and b f of x .
5. NE: Ok
6. Ann: And then this one says I put f of x into g
7. NE: Ok
8. Ann: So I put the x squared which is f of x inside the x over 3 minus 5
9. NE: Ok
10. Ann: And then I did the opposite and labeled f of g of x because g of x is inside of f of x
Ann stated that she is composing the functions, and she does do accurately. However, it is
interesting that she appears to have some compulsion to use the function names f and g
rather than a and b. This may be evidence of White and Mitchelmore’s x-y syndrome
described in section 2.5.2. Ann’s next strategy was to try a few numbers which prompted
the following exchange:
Excerpt 46
1. Ann: If you put in 12, you’d have 145 minus, then divide that 3 that’s like
maybe 5 times 4 minus 5, and so it would be better to have that one. Graph it.
2. NE: Graph it, how do you think graphing would help you?
3. Ann: Well, cause then I could get the maximums and minimums. If I could see it, I think it would be easier to figure out.
Ann requested a calculator so that she “could get the maximums and minimums, if I
could see it, I think it would be easier to figure out.” One was provided, and she
proceeded to graph her equations but had some difficulty interpreting what the graphs
represented.
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In Line 2 of the transcript excerpt below, Ann stated that her graph is telling her
“a lot of confusing stuff.” When asked to clarify what is confusing, she appears to have
trouble reconciling the fact that one of her functions decreases while the other increases.
Excerpt 47
1. NE: So what is your graph telling you?
2. Ann: A lot of confusing stuff.
3. NE: What’s confusing about it?
4. Ann: Well, the problem is, I’m not sure exactly what we’re looking for cause we’re looking for a positive x value. So I have to start this one, at least I think this one can start right there, but this one cause technically, hmm. I’m not sure cause this one’s decreasing from here until it gets to zero which is um 14ish.
5. NE: Ok
6. Ann: That’s decreasing and that one’s increasing from zero at the origin all the way up. So technically the top one should be good until around 14 15.
7. NE: Ok and how do you know that?
8. Ann: Because, I don’t know. See, because that’s when this one goes, and it’s wider, so it should go farther that way, yeah because this one goes off more in the x direction, maybe. I’m not good at these types of problems.
9. NE: Ok, so at some point your graphs cross?
10. Ann: Correct
11. NE: Do you have a way to find that intersection?
12. Ann: Probably set them equal to each other
13. NE: What do you think?
14. Ann: It’s complicated I don’t know
15. NE: What makes it complicated?
16. Ann: The x squared and the x
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17. NE: Do you have a way of dealing with things that have an x squared and an x?
18. Ann: Yeah, you could do the quadratic equation.
19. NE: Ok
20. Ann: Or you could factor it out, but I don’t know what to do with this x squared over here.
21. NE: Well in order to use the quadratic formula
22. Ann: It has to be equal to zero doesn’t it?
In Line 9, the interviewer brought the intersection point to Ann’s attention. The exchange
continued, and Ann believed that she could find the intersection point by setting the
equations equal to each other. In contrast to Ali, Ann identified in Lines 18 and 20 that
the quadratic equation and factoring were ways to solve a quadratic equation. After
finding the intersection point, Ann had some difficulty interpreting the output of her
functions and answering the question. Ann wanted to pick points, but was unsure about
how to go about it.
After some prompting to look at each side of the intersection point, Ann
recognized that she only needs to determine “which one’s higher” in Lines 11 and 13
below.
Excerpt 48
1. Ann: Oh, you know what?
2. NE: What?
3. Ann: That one goes negative. It goes down instead of coming up, and they’re both going to be going up here, but this one’s going sharp more up
4. NE: Ok
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5. Ann: So, this one would be good from zero to, from zero until the intersection and then this one would be better because it goes up more faster.
6. NE: Ok
7. Ann: So you could put, do the, um put in something lower and higher than the uh intersection.
8. NE: Ok
9. Ann: And then see what’s on either side and see
10. NE: What are you looking at on either side?
11. Ann: Um which one’s going to be higher.
12. NE: Ok what do you mean by higher?
13. Ann: Which one’s going to have the higher value
Subsequently, Ann completed the problem by providing intervals for when it is better to
put envelope a inside of b and vice versa.
When Ann wrote up this problem in her homework, she was the only student to
use multiple function representations. Ann wrote out algebraic expressions for the
individual functions and represented their compositions with algebraic representations, a
labeled graph, and a table. Next to her graph she wrote, “The slope of AB o is
decreasing, but its y-values are higher than those of BA o so it is preferred. After the
intersection (the point at which they are equal), AB o ’s y-values become smaller than
BA o ’s, so BA o is preferred.” At the bottom of her write-up, Ann also stated that
“ BA o will produce higher outputs.” See Figure 4 below for Ann’s solution to the genie
problem.
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Figure 4: Ann’s solution to the genie problem.
In summary, Ann was the only student to make significant use of the input/output
terminology in her descriptions of the role of each variable as described in section 6.1.1.
She used multiple function representations in her write-up of the genie problem
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suggesting that she may have a more mature concept of function than the other students
in the study.
Ben’s approach to the genie problem was different from that of Ali and Ann. He
first had difficulty deciding how the order of the envelopes worked. He finally decided
that the outside envelope happens first, as is seen in Line 2 of the transcript excerpt
below.
Excerpt 49
1. NE: Ok so which way do you think you should do them?
2. Ben: I think it should be outside envelope first.
3. NE: Ok
4. Ben: I changed my answer, yeah.
5. NE: So explain to me how that works.
6. Ben: How the envelopes work?
7. NE: Yeah.
8. Ben: So, you got the a envelope inside the b envelope. So you open up the b envelope, and you see that it’s going to um square your deposit, and you’re like, all right. Then you open up the a envelope, what’s inside that, and it’s going to divide by three and subtract 5. So that’s how it works.
9. NE: Ok
10. Ben: So, then to answer the actual question (long pause). I think what did, I say outside then the inside, it would be best to have b inside of a if you’re…
11. NE: Why?
12. Ben: Because when everything’s done it’s squared, and you’ll have a positive number.
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Notice that in Line 10 above, Ben has stated that is would be best to have envelope b
inside of envelope a because you will end with squaring which makes everything
positive. When asked to justify this statement, he reconsiders and thinks that maybe he
should try using formulas, but he doesn’t know what formulas or how it would help him.
Ben ventured to guess that somewhere around 4 is where it would switch which prompted
the following exchange:
Excerpt 50
1. NE: Ok, is there any way you could find a cut off point?
2. Ben: Yeah, I probably could somehow. Envelope a would be x divided by 3 minus 3, and b would be x squared. Assume x is the amount of money you have. So, set these equal to each other and solve for x will give you values you want.
3. NE: What do you think? What does setting them equal to each other have to do with this problem? How does it help you?
4. Ben: I only think when the two equations would cross on a graph would be when they’re, when the advantage changes from having one before the other
5. Ben: How would I solve it?
6. NE: um hm
7. Ben: Probably use my calculator. Is that an option?
8. NE: Sure, you can use your calculator.
9. Ben: We could use the quadratic formula, too. If I moved the x squared over to that side.
At this point, Ben constructed functions for the envelopes (occurs in Line 2 above). It
would appear that Ben understood how a graph could be useful to him (Line 4) and
requested a calculator. Also note that Ben recognized the quadratic formula as a
potentially useful tool (Line 9). Ben proceeded to put the equations into his calculator,
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graphed them, and stated it doesn’t help because “they never cross.” Ben was then
prompted to explore how he may relate a and b through function notation and the
following exchange occurs:
Excerpt 51
1. NE: Well, let’s say we want to put b inside of a, what’s happening to the money then?
2. Ben: It’s getting divided by 3 and subtract 5, and then it’s getting squared.
3. NE: So is there a way to write that from your functions?
4. Ben: Yes, there is b inside of a?
5. NE: Yeah, let’s say b inside of a.
6. Ben: Did I say to do the outside first or the inside? I can’t remember I’ll do the outside
7. NE: Ok
8. Ben: So, then you’d have x over 3 minus 5 squared.
9. NE: Can you assign some yeah notation to that?
10. Ben: So it would be b of a, so a of b. That’d be if you’re squaring it then dividing by 3, and then…Does that work? Cause x is squared then divided by 3, and you subtract 5 from it.
11. NE: Ok
12. Ben: I think it works. So if we graph those two would it cross somewhere?
13. NE: I don’t know what do you think? Do you think they will?
14. Ben: Um, I don’t know. I don’t know what this one would look like, and I actually don’t know what that would look like either. So should probably just check (uses calculator) yeah, they crossed 6.33.
When using the calculator to graph the functions the second time, Ben found an
intersection point. After this, Ben explained that you look at which curve has the higher
y-values. Ben was also asked if 6.33 was the only intersection point he needed to be
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concerned about and found the negative intersection point as well. The idea that you
could (or could not) deposit negative money did not seem to phase Ben and his answer
had three intervals that covered the entire real line.
In Ben’s homework write-up of the solution to the genie problem, he used only a
verbal description of what he did. He also reversed the order in which the envelopes were
applied (during the interview he said the outside envelope happened first and in the write-
up he had the inside being done first). In his text, he labeled his functions A=x/3 + 5,
B=x^2, B(A)=((x/3)-5)^2, and A(B)=(x^2/3)-5. It is interesting that Ben stated he “found
where they intersected each other” and then that “it could also be found by setting the two
equal to each other.” This suggests that using the calculator and using an equation to find
an intersection point may be two distinctly different things to Ben. See Figure 5 below for
Ben’s solution to the genie problem.
Figure 5: Ben’s solution to the genie problem.
The students were asked to identify the parts of the following expression,
( )wur = . Ali interpreted this as multiplication and possibly as a function u in terms of w,
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but quickly dismissed that option. She was asked about this and the following exchange
occurred.
Excerpt 52
1. NE: You said something about if it was a function, what if it was a function?
2. Ali: If it was a function, then you would actually need a w on this side so then it would be u in terms of w. Unless you had another r equals something with a w in it, possibly.
3. NE: Can you give me an example?
4. Ali: Like if r equaled w plus 2, then er, let’s see, um I don’t know. I just lost my train of thought. Yeah, if it was like w plus 2 equals r, then you could just have the r substituted for the w plus 2. So that could be a function of u in terms of w.
Ann stated, “um the meaning of each of the parts um r equals the y value u equals the
function and w equals the x value” and then she used the input and output terminology to
describe what is happening:
Excerpt 53
1. NE: That’s all the relationships?
2. Ann: Ummm, well when you put w into this, you get r. So that’s the input and that’s the output.
3. NE: Ok
4. Ann: So, what it’s really saying is that when you plug w into the function you get r.
Ann described the independent variable w as the input and the dependent variable r as the
output. Ben stated, “I guess it’d be a function r equals the function u with w plugged into
it.” For each of these students, the notion of function carries an association of “plugging
in” a value.
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It would appear that Ann has the most mature concept of function of the students
who participated in this study. In Ann’s solution to the genie problem, she used detailed
verbal descriptions together with three different function representations: algebraic
functions, graphs, and tables. This suggests that Ann may have a process view of
function. Ali may have the weakest concept of function as she was only able to represent
the functions with algebraic expressions and her descriptions lack detail about the
input/output nature of the composite functions. Ben is somewhere between Ali and Ann
with his concept of function. He used a more detailed description than Ali to support his
algebraic manipulations and claimed that he graphed the functions. However, he did not
provide multiple representations of the functions. It would appear that Ben and Ali have
begun the transition to having a process view of function, but they are not as advanced as
Ann.
7.1.3 Evidence of Engagement in Covariational Reasoning
The second most difficult problem in the individual interview appeared to be the
following:
Imagine this scenario. Water is poured into an empty bottle until the bottle is full.
As you imagine the water filling the bottle shown below, what are the changing
quantities that come to mind? Can you sketch a graph that shows the height of
water as a function of volume as water is poured into the bottle?
This question is similar to one that had appeared on the Precalculus Concept Assessment
instrument at the beginning of the semester and is designed to ascertain the extent to
which the student will engage in covariational reasoning.
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In this problem, Ali’s initial thought was that she did not know the formula for the
volume of a sphere. She then proceeded:
Excerpt 54
1. Ali: So, you have that. So they are filling the bottle out to a certain point and then going to the next point. So the cir, the circle that you’re filling up is increasing to the middle of it, and then it decreases. Um the changing quantities are like just the area of the circle that’s the, uh the uh, section of the bottle. Um, hm. Can you sketch a graph that shows the height of the water as a function of volume? Hm, the height equal, see I don’t know.
2. NE: What are you thinking about?
3. Ali: I’m thinking that, like you pour in the water. It’s gonna fill up pretty slow. It’ll start pretty fast with the height gaining height, then the, as the bottle gets wider, then the height’s going, the increase in height’s going to decrease, until like the widest point that it’s, then it’s going to be at it’s lowest, then it’s going to start, um, getting more height for the same amount of volume we put in as the bottle comes back up to it’s close, then it will just become constant right there.
Ali identified the circular cross section of the sphere as the changing quantity and
appeared to be engaging in covariational reasoning as she talked about how the height is
changing as the water is poured into the bottle. Her statements suggest she is at an MA 2
level. A few moments later, she also identified the volume of the water inside the bottle
and the height as the changing quantities. As the interview continued, Ali was focused on
having a formula to be able to draw a graph which prompts the following discussion:
Excerpt 55
1. NE: So, what does the height of water as a function of volume as water is poured into the bottle mean? Height as a function of volume?
2. Ali: Height as a function of volume, um, just that the height of the water depends on the volume of water that’s in there, I guess.
3. NE: Ok
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4. Ali: So, hm, so you’d have to do h of v possibly, that’d still be the same thing.
5. NE: What do you mean by that?
6. Ali: Well, it’s just writing it in a different way, I guess. Um, um, unless you, you’d have to have a function for the height to be able to plug the volume into it, and… No, um, you would have volume with the height as a variable so
7. NE: How do you know it’s the variable?
8. Ali: Because the volume is the function, and it says, shows the height of the water as a function. So, you have to have the height in the fun, in the equation for it to be related to the volume.
9. NE: Ok
10. Ali: See, I’m thinking somehow I need to multiply this first part by height, four thirds by pi r cubed, somehow get height in there to be able to solve for height, because radius isn’t really help with the height. Hm, hm, so this one, so the volume of the entire bottle, so somehow we need to find the volume of the water.
Ali correctly interpreted the function language in Lines 2 and 4 above, but then in Line 6
she considered changing her mind to reverse the relationship. In Line 10, she was looking
for a way to eliminate the r in the formula for the volume of a sphere because she wanted
a formula that just involved volume and height.
The interviewer proceeded to ask Ali some questions that were met with a series
of “I don’t know” responses and led to the following question:
Excerpt 56
1. NE: In your ideal world, what would I give you to make this easier to solve?
2. Ali: Um, I guess, how much is being poured in, in a certain amount of time, or um well that wouldn’t help.
3. NE: You don’t think that would help? Why not?
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4. Ali: Well, it would, but I don’t know how. Hm, I’m thinking of problems I’ve done like this in the past, and I’ve done them with cones, and that was the radius was increasing at a constant rate. This one the radius is not so it’s a little more difficult.
5. NE: Ok, so tell me more about this cone problem you’ve done in the past.
6. Ali: Um
7. NE: How does it work?
8. Ali: The two, like, I would just think of it as a regular triangle, and not worry about the volume, and just that later, and just the, um, just try to get the area of the triangle with the height, and then I could multiply by to give me the volume hm
9. NE: So what would you be trying to find in those kinds of problems?
10. Ali: Just, well, it would try to find how, how long it took to fill up the, the cone uh
11. NE: So, you would want to see how long it took the cone to fill?
12. Ali: If you put it in at a constant rate, yeah, but if you just were filling it to a certain point, then you wouldn’t need to hmm
13. NE: Did you ever sketch a graph of those functions?
14. Ali: Yeah, it’s, it starts, the height starts at zero. Then it gradually decreases. So, if this was this would be the height (labels vertical axis), the variable function, the variable would be dependent, this would be the volume (labels horizontal axis) hm
15. NE: And did you have an equation for that?
16. Ali: Yes, there was. Um, set it up like, 3 times the height equals the, the area equals base times height for triangles, so you’d have, so then the height would be the area divided by the base. So, if you were actually doing a cone, it would be the height equals the volume divided by the area of the circle on top.
In Line 4 above, Ali tried to relate this problem to a similar problem involving a cone that
she had seen before. She claimed that in those problems the radius was increasing at a
constant rate and in this situation it is not, so the problem is more difficult. The
interviewer should have pursued this line of reasoning further to understand what she was
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interpreting as the changing radius for the bottle problem. As Ali explained the cone
problem starting in Line 8, she described being able to take the area and multiply by
height to get the volume. This suggests that Ali did not know what the volume of a cone
was. When prompted about graphing this similar situation, Ali again wanted to represent
it with an algebraic representation. (Lines 14, 16) She attempted to continue the bottle
problem in a similar fashion with little success. After allowing Ali to explore this option
for some time, the interviewer prompted:
Excerpt 57
1. NE: Is it ever possible to sketch a graph without knowing the equation?
2. Ali: Yeah
3. NE: Can you give me an example?
4. Ali: Well, the, it’d be kind of like this one. This would be, the height would be small as the hold on it would start like this, but increase, and then it would gradually go down to, but it doesn’t really increase very much because it’s so wide, then it would start to increase again because it gets smaller at the top
5. NE: So what is that the graph of?
6. Ali: The, oh, I guess the heighth
7. NE: So can you label your axes?
8. Ali: That’s volume cause the volume still increases
Now, Ali was able to construct an appropriate graph of the situation with the axes labeled
appropriately. She was not willing to construct the graph of the function without an
algebraic representation unless specifically asked to do so. This was also true for Ann and
Ben.
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Ann immediately noted that the graph should be increasing, but then immediately
began searching for an algebraic representation for the volume of the bottle. When
prompted about graphing without a formula, Ann initially thought the graph would be
linear, as is evidenced in Line 6 of the transcript excerpt below:
Excerpt 58
1. NE: Do you have to know the formula to be able to sketch a graph?
2. Ann: Yup an accurate graph probably like a really good one
3. NE: Ok
4. Ann: But basically it’s going to be increasing, and like it should be just be like constant, unless the water’s going in at a like a constant rate
5. NE: Yeah the water’s going in at a constant rate
6. Ann: Constant speed, so it should be constant so it should just be a line
7. NE: All right, so if you were to put some axes on that graph and label them
8. Ann: That would be um
9. NE: So what’s y and what’s x?
10. Ann: The height (y) and the volume (x)
11. NE: All right
12. Ann: Right, because it’s getting, they’re both increasing. So but
13. NE: So what happens down here at the bottom of our bottle? Will it be the same as what happens at the top of our bottle?
14. Ann: No, that’s not true. Yeah, it would because the volume’s still increasing.
15. NE: Ok
16. Ann: Even though the shape is different, the volume is still getting bigger, and the height is still getting higher. So it shouldn’t change anything.
17. NE: Say I pour in just two inches of water (indicating height). Then I put in two more, and two more what happens?
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18. Ann: It’s going to get larger
19. NE: Ok, so height’s going to increase um do I always pour the same amount?
20. Ann: No.
21. NE: Why?
22. Ann: Because of the shape
23. NE: All right, can you explain that to me?
24. Ann: Well cause at the bottom, at the middle it’s going to need more liquid to make it rise in the middle because there’s a larger area to fill.
25. NE: Ok
26. Ann: And at the bottom it’s smaller to fill so it goes faster, then slower, then faster.
In Line 10, Ann correctly identified and labeled the axes for her graph. In Line 14, Ann
still believed that her graph would be linear. She quickly changed her mind after she was
asked to consider what happens if the bottle is filled by two inch increments in Lines 17
and 19. After this revelation, she was able to construct an accurate graph.
Similar to Ali and Ann, the first thing that Ben wanted to solve this problem was
an algebraic formula for the volume of the bottle. After he was provided with the formula
for the volume of a sphere, he did not make much progress as was the case with Ali and
Ann. Ben also had to be prompted to consider sketching a graph without knowing an
algebraic formula as is evidenced in the transcript excerpt below.
Excerpt 59
1. NE: Ok. Is it ever possible to sketch a graph without knowing a formula?
2. Ben: Yes
3. NE: How could you do that?
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4. Ben: Well, it’s just like, um, a velocity graph, acceleration, position. Do like the position is increasing, so, so is the velocity, not much of an explanation
5. NE: Ok
6. Ben: So, if the height’s getting larger, so is the volume
7. NE: Ok
8. Ben: Since they’re related
9. NE: Ok. So if you were to sketch a graph of this what would x equal?
10. Ben: I have height is the variable and the other is the dependent variable
11. NE: Ok
12. Ben: So (starts drawing) and volume, but it wouldn’t be, it wouldn’t be linear cause it’s a sphere.
13. NE: Ok
14. Ben: So when the height increases a little, volume increases faster than when it’s increasing in the middle
15. NE: Ok
16. Ben: And it’s up to this part it will be increasing
Ben constructed an appropriate graph but had a more difficult time explaining what is
happening than Ali and Ann did.
The students did not appear to rely on covariational reasoning to solve this
problem. When prompted, the students’ responses suggested that they could think about
the problem at an MA2 or possibly an MA3 level of covariational reasoning. MA3 level
reasoning was usually evident after the students were prompted or questioned by the
researcher.
In all three individual interviews, the students wanted to have an algebraic
formula to represent the volume before they could consider sketching a graph. It would
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be interesting to know if there is such a strong correspondence between the algebraic
formula and the graph due to the prevalence of graphing calculators in today’s classroom.
If the students in this study had been able to construct an algebraic formula for this
situation, they likely would have plugged it into their calculators and then merely
sketched what was on the screen. This may be detracting from the students’ ability to
think about the relationships that exist not only between function representations but the
relationship that the function represents.
7.2 Addressing Time as a Variable and Rate of Change
The first teaching episode was designed to have the students investigate rates of
change for two problem situations using a custom computer program. Only data from the
first problem will be presented as it contained the richest dialogue which highlights the
students’ development of the concept of rate. When the students worked on the second
problem, they used the language and notion of rate developed when working the first
problem.
The first problem presented to the students was the following:
Suppose we have a plane that is flying over a RADAR tower, TA, and is on
course to pass over a second RADAR tower, TB. Let u be the distance between
the plane and TA, and let v be the distance between the plane and TB. What is the
rate of change of u in relation to v?
The students had the computer program open in front of them with a visual
representation of the problem situation as seen in Figure 6. No numeric values were
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given, only a pictorial representation. The students were instructed that they could click
on the plane and drag it to observe what happens.
Figure 6: Computer simulation of the plane problem 1.
The students were prompted to discuss how they intended to find the rate of
change of u in relation to v in the following transcript excerpt.
Excerpt 60
1. NE: Ok what information do you think you need?
2. Ali: Maybe the distance between the two towers
3. NE: Ok
4. Ann: And maybe the time it took to get there
5. NE: The time it took to get there ok anything else you might want?
6. Ali: Um
7. Ben: Hmm
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8. Ali: Maybe how high the plane is so we can get an angle of the triangle that it makes with the initial point and where it is
9. NE: Ok
10. Ben: Yeah
11. NE: You like the idea of an angle?
12. Ben: Yeah it couldn’t hurt, maybe velocity too
13. NE: Maybe velocity too
14. Ben: Um hm, just in case
15. NE: Ok so how about we go up to activities and go to plane 2 so what information do you have now?
16. Ali: We have the distance in miles
17. Ann: The length of u and the length of v. So now you can find the distance between the two towers because you can use the Pythagorean theorem, if you assume the plane is right above the first tower it makes a 90 degree angle
18. Ali: Yeah
19. NE: Ok
20. Ann: Which it look like it does so
21. NE: Ok so what do you think would happen?
22. Ann: Um
23. Ali: To the lengths of the sides or
24. NE: Ok, well ultimately we want to answer the question: what is the rate of change of u with respect to v
25. Ali: I think as one increases, the other one decreases about the same length not quite so much [goes back to the program]. Up until the point that’s right in the middle, it the length of v increases a lot quicker or decreases a lot quicker than u increases
26. NE: So what do you guys think about that Ben and Ann? What Ali said
27. Ann [nods]
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28. Ben: What’d you say again?
29. Ali: The length of v decreases a lot faster than the length of u increases as it approaches the exact middle between the two towers. Then after that u increases quicker than v decreases
30. Ann: Right
31. Ben: [nods] sure
32. Ann: Are we allowed to assume that it’s a 90 degree angle?
33. NE: Ok if we assume that’s a 90 degree angle, how does that help us?
34. Ali: We can find the distance between the two towers
35. NE: And if we find the distance between the two towers how does that help us?
36. Ann: Well, I was just saying, really what I was saying was it would increase or decrease sharply until about 60 degrees. When if, what she was saying, because that would be the center and then it would be isosceles triangle, or maybe, I don’t know. It would be a different kind of triangle. I’m just confusing myself.
37. NE: So do we have a way to measure that rate of change?
38. Ben: There’s always a way
39. NE: Ok so there’s always a way
40. Ali: I think we need time too
41. NE: Why do you think you would need time?
42. Ali: Because if you don’t have time, the plane could be moving really slow, so the rate would be different I guess no
43. Ben and Ann: yeah
44. NE: You guys are nodding in agreement?
45. Ben: Or velocity
46. NE: Velocity
47. Ann: Velocity would work yeah
48. Ben: Because that’d be rate
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49. NE: Or time
50. Ann: Yeah you need a passage of time to have a rate
The students began brainstorming what information they needed to determine the rate of
change of u in relation to v. Ali suggested the distance between the towers (Line 2), and
Ann suggested the time it took to get there (Line 4). Then, Ali suggested the height of the
plane so that they could find an angle (Line 8). After Ben agreed that finding an angle
would be good, he suggested velocity (Line 12). The students were then prompted to
open plane 2 in the program.
In plane 2, the program provides the measurement of the distances in the problem
situation as seen Figure 7. As they moved the plane, the slider bars moved and provided
measurements for the variables u and v.
Figure 7: Computer simulation of the plane problem 2.
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Ann suggested that the Pythagorean Theorem will allow them to find the distance
between the towers (Line 17). When prompted to discuss what they thought was
happening with regard to the rates, Ali stated “the length of v decreases a lot faster than
the length of u increases as it approaches the exact middle between the two towers then
after that u increases quicker than v decreases” (Lines 25, 29). Ali described what is
happening in the problem using some informal notions of rate as is evidenced by her use
of the terms increase and decrease in conjunction with quicker and faster. Ali and Ann
then revisit the idea of using the Pythagorean Theorem to find the distance between the
towers and find an angle. Ann appears to be recalling parts of her content knowledge of
geometry and trying to make them fit the problem situation. However, they could not
articulate how they believed this would help them. It would appear that this may be
related to their view of mathematics which involves setting up an equation and solving
for the unknown.
In Line 37, the researcher redirects their conversation to focus on rate of change.
Ali suggested that time is needed to calculate a rate in Line 40 above; however, her
reason for wanting time is weak at best (Line 42). She appears to have the notion that the
passage of time is connected to rate, but she cannot articulate what she means. Ben and
Ann agreed with her, but Ben proposed velocity as an alternative and noted that velocity
is a rate in Line 45. Ann concluded that “you need a passage of time to have a rate” in
Line 50. Henceforth, time was a student generated variable and an important part of the
problem.
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The students were then told to open plane 3 in the program which provided them
data for distances and time as seen in Figure 8.
Figure 8: Computer simulation of the plane problem 3. This lead to the students calculating the average velocity for x as is seen in the transcript
excerpt below.
Excerpt 61
1. NE: Ok so how does that help you find your rate?
2. Ann: Well, you can measure how far the plane is going, and what time. So then you can calculate velocity right? [looks to group members]
3. Ben: Right
4. NE: All right
5. Ann: And then
6. NE: So how do you calculate this velocity?
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7. Ann: And then it would be distance over time
8. NE: Distance over time
9. Ben: Yep
10. NE: Any old distance over any old time?
11. Ann: The x distance over the time the plane has traveled.
After determining how to find the average velocity, another part of the problem came to
light. Ann reminded the others that they were looking for the rate of change of u in
relation to v. Ali suggested, “I think we’re going to find the rate of change u with respect
to x then the change in v with respect to x then we can solve it for x and compare the two
possibly.” This again is evidence of Ali’s primary problem solving strategy – set up an
equation and solve for x – that emerged in her individual interview. Ann requested that
they go back and discuss one more time how they are going to calculate average velocity.
Ann advocated final minus initial length over time while Ben suggested trigonometry. It
was decided that they could use Ann’s method to calculate the rate of change for u. To
aid them in this endeavor, they were instructed on how to construct a table of values
using the program. The students then created a table of values and calculated the average
rate of change of u with respect to time, v with respect to time, and x with respect to time.
The students were then prompted to assign notation to their calculations. Their
first response recalled the εδ − definition of derivative. It was eventually recalled that
u∆ could represent the change in u and that t∆ could represent the change in time which
allowed them to represent the average rate of change of u with respect to t as t
u
∆
∆. Then,
the following exchange occurred when the students wanted to relate u and v:
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Excerpt 62
1. NE: Right so how might we relate u and v now?
2. Ann: Well I guess [mumbles] we could set delta u over delta t equal to delta v over delta t
3. NE: Ok what does that what does that give us?
4. Ben: A false statement
5. Ann: Yes
6. NE: A false statement ok
7. Ann: Yes because we can’t
8. NE: Well, what is we want we want the rate of change of u with respect to v.
9. Ben: So we want delta u over delta v
10. NE: Ok so if we want delta u over delta v
11. Ben: Just flip this t delta t over delta v
12. NE: Just flip it over?
13. Ben: Yeah, then we’ll multiply
14. NE: Then we’ll multiply
15. Ben: And cancel the delta t’s making delta u over delta v
Note that Ann’s first thought was to set up an equation (Line 2), but Ben pointed
out that only gives a false statement and she agreed (Lines 4, 5). Ben stated that he
wanted v
u
∆
∆ (Line 9) and suggested that this could be achieved by flipping the
t
v
∆
∆ and
multiplying it with t
u
∆
∆ (Lines 11, 13, 15). This suggestion appears to be based on his
knowledge of algebra rules. It was then discussed and decided that in order for this
approach to work, the time intervals must be the same.
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We concluded with a discussion of what happens as one takes smaller and smaller
time intervals. Ali stated that they become “more accurate.” Ann then elaborated:
Excerpt 63
1. Ann: Well, it won’t give you like a number. It’ll give you the answer at that time, but it won’t necessarily be any more any closer
2. NE: Any closer to what
3. Ann: Well, because if you’re taking one at five, and you’re taking one ten, the one at five is going to be the one at five, and the one at ten is going to be the one at ten
4. NE: Ok
5. Ann: So, just because you’re doing it on shorter intervals doesn’t mean you’re going to change the difference between, because at 10 seconds it’s going that fast, and 5 seconds it will be going a different speed because at 20 seconds it’s going a different speed
Ann appears to think that the average rate of change is a constant rate for the
whole interval. The researcher probably should have pressed them to relate this to their
understanding of the mean value theorem. It was eventually decided that smaller and
smaller intervals led to the derivative, an instantaneous rate of change.
This brief summary of the first teaching episode highlights important aspects of
the problem which were carried over to subsequent teaching episodes. The computer
program provided a visual representation of the problem situation that could be virtually
manipulated. This allowed the students to attend to the nature of the changing quantities.
Time was generated as an important variable in the problem by the students, and it
allowed them to construct and relate rates. Notation was then developed to represent the
quantities that were being discussed. Finally, the idea of taking smaller and smaller
intervals of time resulted in defining the instantaneous rate change. Thus, the students
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generated time as a variable and represented the average rates of change using delta
notation: t
x
∆
∆ .
7.3 Revisiting the Chain Rule
To understand the multiplicative nature of the chain rule, consider the following
problem:
A ball is thrown into a lake, creating a circular ripple that travels outward at a
speed of 3 cm per second. How fast is the area of the circular ripple growing with
respect to time?
It is possible to express the area of the circle in terms of the number of seconds that have
passed by using function composition with the functions 2rA π= and sr 3= . Doing this,
we obtain ( ) 2293 ssA ππ == . From this we could directly compute the rate of change in
the area with respect to the number of seconds: sds
dAπ18= . This tells us that the area
changes sπ18 as much as the number of seconds. However, we may also observe that
rdr
dAπ2= and 3=
ds
dr, which tells us that the radius is changing 3 times as much as the
number of seconds that have passed and that the area is changing rπ2 times as much as
the radius. Thus, it would make sense that the area is changing at 32 ⋅rπ as much as the
number of seconds that have passed. We can then find
( ) ssrrds
dr
dr
dA
ds
dAππππ 1836632 ===⋅=⋅= , which is the same value that we had
calculated directly. It is important to note that at any given time, we generate a radius as
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the output of the function sr 3= which in turn becomes the input for the
function 2rA π= .
The most important aspect of the chain rule that appears to be useful for solving
related rates problems is the ability to coordinate relationships in the amounts of change.
To help students understand the multiplicative nature of the chain rule and coordinate
relationships in amounts of change, we can ask them to consider the rate of change of the
radius with respect to time over some (given) very small time interval: computing s
r
∆
∆.
After computing the change in the radius for the given time interval, we can then ask
them to find the rate of change of the area with respect to the same change in the radius
that they just calculated: computing r
A
∆
∆. Thus, we constructed a relationship between
area and seconds by coordinating the amount of change in the radius with the amount of
change in the area and the amount of change in the seconds. Because the given interval is
small, the values that they calculate will be close enough to the derivative. This will be a
glossing over of the limit part of the proof of the chain rule, but it should allow the
students to convince themselves that the chain rule is multiplicative. This may foster
more of a procedural understanding in that the students may think about canceling the
r∆ ’s to determine the overall rate of change, s
A
∆
∆. However, as long as they recognize
that the value for r∆ must be the same, this may not necessarily be a bad thing. It may
allow them to successfully solve problems before they have developed rich conceptual
understandings.
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An alternate way to think about why the chain rule is multiplicative in nature is to
consider that s
r
∆
∆ represents an increase of 3 units for every second increase and in turn,
r
A
∆
∆ represents an increase of rπ2 units for every increase of one unit of radius. Thus, for
every second that passes, the radius is three more than before. Then for each unit increase
in the radius, the area is increase by rπ2 . Since the radius was increased by 3, we must
increase the area by rπ2 for each of the three unit increases, or three times. Thus, the
overall rate of change of the area will be 32 ⋅rπ . Since we can define the radius in terms
of seconds, we can then rewrite the rate of change function as previously stated.
In a related rates problem, it is the ability to abstractly think about each variable
as a function of time and realize that while only 2h may be written as part of the
relationship, what is actually being conveyed is the relationship ( )[ ]2th . This requires that
one recognize that h(t) is the “inside” function in the composition and that squaring is the
“outside” function in the composition. It is then the ability to perform derivative
operations on a general function form in conjunction with recognizing time as the
independent variable that allow the derivative to be computed: ( )dt
dhth ⋅2 .
The focus of the second teaching episode was the multiplicative nature of the
chain rule. The students were given three problems to investigate. Only the third problem
will be presented as it illustrates the students’ development of the circle problem stated
above. (See Appendix C for all problems used in the teaching experiment.)
Ann started the problem:
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Excerpt 64
1. Ann: Well you could do delta a over delta r
2. NE: Ok
3. Ann: Times delta r over delta t
4. NE: Yeah, why do you think that?
5. Ann: Because you need delta a over delta t, because um you’re looking for the growth of the area, the change in the area with respect to the time, and then as the radius is changing the area is changing, and the radius causes the area to change, and then, but the radius also changes, um 3 centimeters per second so
Ann stated that she wanted t
r
r
A
∆
∆⋅
∆
∆ because “you need delta a over delta t
because um you’re looking for the growth of the area the change in the area with respect
to the time and then as the radius is changing, the area is changing, and the radius causes
the area to change and then but the radius also changes um 3 centimeters per second.” It
would appear that she has internalized how the growth rate of one variable affects the
growth rate of a second variable. Ben proceeded to question Ann about her approach:
Excerpt 65
1. Ben: Is the rate, do you need to know the rate which it’s going out?
2. Ann: Um hm
3. Ben: So is area delta r or is the radius delta r?
4. Ann: We’re trying to find the rate, aren’t we?
5. Ben: Yes, but I mean like the rate at 3 centimeters per second. What’s that? Is that a rate?
6. Ann: That’s the rate of delta r over delta t
7. Ben: Whoa
8. Ali: That’s the change in the radius
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9. Ben: That’s the radius change
10. Ali: Yeah
Ben questioned whether area or radius was represented by r∆ (Line 3) and then
whether 3 centimeters per second is a rate (Line 5). Ann explained that t
r
∆
∆represents the
rate at which the radius is expanding (Line 6). Ali rephrased it as “the change in the
radius” which Ben repeated as “the radius change” (Lines 8, 9). It would appear that Ben
is still struggling to think about rate as one quantity instead of two while Ann and Ali
understand that the symbol t
r
∆
∆ represents the given rate.
Ben proceeded to question Ann further in the following exchange:
Excerpt 66
1. Ben: What’s delta a over delta r?
2. Ann: The change in the area over the change of radius.
3. Ben: Oh yes
4. Ann: Yeah, I know, but I don’t know what it is. That’s where I’m stopped.
5. Ben: How should we find that, can we do it from zero to 1 second?
Ann was not sure about what r
A
∆
∆represented. Ben asked about doing some
computations. They were encouraged to investigate what r
A
∆
∆ may represent by
performing calculations similar to those of the previous day. After they had performed
numerous calculations, they had the following revelation:
185
Excerpt 67
1. Ann: And so you’d have pi r squared, and like for the first one it was r squared, and it was 9 pi over 3. So, you just, it would be the square of the original number, and the second time it was 36. So it would be 36 over 6, so you have the square root of the original number again.
2. Ben: It wasn’t 36 it was 6
3. Ann: It wasn’t.
4. Ben: No.
5. Ann: Oh, it was over 3 damn.
6. Ben: Yes, but it is equal to 2 pi r.
7. Ann: Is it equal to 2 pi r
8. Ben: Um hm
9. Ann: Ok, it’s equal to 2 pi r
10. Ben: You get 6, then you get 12 pi, then you’d get 18 pi.
11. NE: Does it always seem to be 2 pi r [should have had them make a table]
12. Ben: Yes
13. Ann: It would be.
14. NE: So, delta a over delta r can be represented by
15. Ann: 2 pi r
16. All: 2 pi r…
17. NE: But that 2 pi r happens to be the derivative of a with respect to r
18. Ann: Ok, right, I got it. Cause the deltas can be d’s, da dr is the derivative. Ok, I got it.
Ann observed a pattern in the values she was calculating, rπ2 , and she observed
that rr
Aπ2=
∆
∆. (Lines 1-9) However, she struggled to make the connection that this was
the derivative. (Lines 11-17) After it was pointed out to her, she quickly realized that her
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r
A
∆
∆ was equivalent to
dr
dAwhich she had been calculating all semester. This suggests that
while the students were adept at computing derivatives using accepted symbols, those
symbols had little conceptual knowledge associated with them. However, it does appear
that the symbols are beginning to hold more meaning for the students. The students then
proceeded to construct an algebraic representation for the rate of change of the area with
respect to time: t
rr
t
A
∆
∆=
∆
∆π2 . Hence, they related the rates. This problem set the stage
for the following teaching sessions. Thus, the students began constructing what they
would eventually call “delta equations.”
7.4 Students Solve Related Rates Problems
In teaching episodes three through six, the students solved related rates problems.
The problems given to the students in teaching episodes three and four were stripped of
most of the numeric data in an effort to have them attend to the dynamic nature of the
variables and the relationships that existed between them. More traditionally stated and
novel related rates problems were solved in teaching episodes five and six.
The students’ solutions to three problems are presented in the order in which they
were solved. These problems were also solved by the mathematicians: the plane problem,
the trough problem, and the coffee cup problem. This allowed for comparison between
the expert and student solutions.
7.4.1 The Plane Problem
During the third teaching episode, the students were given the plane problem in
the following form:
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A plane flying horizontally at an altitude of 3 miles and a speed of 600 mi/hr
passes directly over a radar station. Let the distance from the plane to the radar
station be represented by z. What is the rate of change of the distance from the
plane to the radar station with respect to time?
This version of the plane problem differs from the mathematicians’ version in that most
of its numeric data has been removed. It was hoped that this would encourage the
students to focus on constructing the necessary functional relationships before attempting
to substitute in values.
Following Excerpt 68, Table 18 summarizes the students’ mental activities as they
carried out the first phase, draw a diagram.
Excerpt 68
1. Ben: She said horizontically 2. Ann: Oh, pshaw, ok. 3. NE: Ok. 4. Ann: I feel like drawing a picture. 5. Ben: Me too 6. NE: You feel like drawing a picture? 7. Ann: I need to see it. 8. Ali: My plane. 9. [long pause] 10. Ben: Oh, I couldn’t draw the radar right below the plane. That’s ridiculous
that the radar 11. [long pause] 12. NE: What did you just label on your picture Ben? 13. Ali: Where’s the radar? 14. Ben: Speed of the plane, which is not 6 miles per hour. It’s 600 miles per
hour. 15. Ann: That’s a slow plane. Ah, yes. 16. Ben: So, I labeled the speed of the plane. 17. NE: Ok 18. Ben: So I should have the radar somewhere, not directly below it, or should
it? I don’t think that’s important. What’s important is there is radar. 19. Ann: Ok
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20. NE: Ok 21. Ben: Ooh, beautiful, can 22. NE: All right. So once you’ve got your picture drawn, then you’ve labeled
the velocity of the plane going across the top, you’ve also labeled the velocity of the plane going across the top, have you labeled? [to Ann]
23. Ann: I have it here. It’s going that way. 24. NE: Ok. Now they both, now you’ve got a full triangle. 25. Ann: I just have it in my head 26. NE: You have it in your head, and you had a partial triangle. So what’s the
picture in your head? 27. Ann: Well, technically, it’s like this. [draws a complete triangle] 28. Ben: What’s you vector, vector? So this is z [points to hypotenuse] 29. NE: So, what have you labeled on your picture there? 30. Ann: I have a 3, and a z 31. NE: A 3 and z, ok. 32. Ann: And this would be z. 33. NE: Ok, so 34. Ann: But it’s kind of traveling, so 35. Ben: So we know delta p, delta 36. Ann: What’s delta p 37. Ben: Plane. The rate the plane is going with respect to time is 600 miles per
hour. 38. Ann: I have a question 39. NE: Ok 40. Ann: Can we call z, um, would that be the distance? 41. Ali: z is the hypotenuse of the triangle 42. Ann: Right, right. What I’m asking is, um, we’re looking for delta z over
delta t, but I keep wanting to bring in an x. 43. NE: ok. Where do you want to bring in an x? 44. Ann: At the, uh, 600 miles per hour. I want to make it delta z over delta or
delta x over delta t. 45. NE: Ok 46. Ann: But then you have to have an x somewhere else 47. NE: Ok 48. Ann: It would have to be delta z over delta x 49. NE: Ok 50. Ann: So I didn’t know if you could make that z 51. NE: Let’s write that down for right now, and where would x be in your
picture? 52. Ann: That’s what I need to know. That’s why I didn’t want to put an x 53. NE: Well, you’ve labeled two sides of your triangle 54. Ann: Well, x could be the distance traveled 55. NE: x could be the distance traveled there. Does that relate well to having a
delta x delta t, that rate of change of the plane with respect to time?
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56. Ann: Yeah, seems like it. 57. NE: Why does it seem like it? 58. Ann: Because it’s traveling up there at 600 miles per hour 59. Ben: Is that a z? 60. Ann: Yes 61. Ben: It looks like a 3 62. Ann: It’s not a 3. I think, I think, yes it works. 63. NE: All right. So first let’s agree on the labeling of our pictures 64. Ann: Ok 65. NE: Before we go any further. 66. Ann: This is a 3 67. Ben: So you got the plane. 68. NE: So you got the plane, and when it first goes over the tower, I think
we’re all agreed that’s 3. 69. Ben: So we can say it’s 3 miles. 70. NE: We can say that’s 3 miles. I think the other thing we can all agree on is
the hypotenuse being z. 71. Ben: Um hm 72. NE: Is that a changing quantity or constant quantity? 73. All: Changing.
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Table 18: Summary Table for Draw a Diagram - Plane Problem
Phase
Draw a diagram (lines 1-73)
Solution
Artifacts
Labeled diagram
Content
Knowledge
Geometry
• The students know what constitutes a right triangle. (evidenced in their diagrams) Variable
• The students labeled the constant altitude with a three.
• The students labeled the distance between the plane and the tower with z.
• The students labeled the horizontal distance the plane was traveling with another variable.
o Ann identified 600=∆
∆
t
x and thought that she needed an extra x
somewhere. (lines 42, 44, 46, 48, 52) o Ann thought x could be the distance traveled. (line 54) o Ann did not recognize that the variable name she assigned to a changing
quantity actually represented a function of time. She consequently did not
appear to recognize the relationship between x and t
x
∆
∆.
Function
• The students were not thinking about the variables they labeled in their diagrams as functions of time which may be the cause of Ann’s difficulty with the relationship
between x and t
x
∆
∆.
Mental
Model • Ann stated, “I feel like drawing a picture” then “I need to see it.” (lines 4, 6)
• Ann said she was imaging a triangle in her head, but she initially only drew the leg from the tower to the plane and the leg along which the plane is traveling. (line 27)
• Ali asked, “Where’s the radar?” which suggests she may have some difficulty interpreting the problem statement. (line 13)
• Ben questioned whether the radar station should be directly below the plane, but thought it was not important. (lines 10, 18)
• Ann drew two diagrams, as did Ben.
• The students eventually correctly interpreted the problem situation as was reflected in their diagrams.
• When prompted, the students can identify the changing quantities. (lines 72, 73)
Heuristics • Label your diagram.
• Check your work against your neighbors’.
This table suggests that the students accessed their geometric content knowledge
of right triangles to construct a mental image of the problem situation. The diagram the
students drew on their paper was the only indication of what their mental image was. Ali
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and Ben questioned where they should put their radar station in relation to the plane and
looked to Ann for guidance. The students’ also appeared to have a heuristic that caused
them to label their diagram. They labeled the constants in their diagrams and then
assigned letters to any quantities that were unknown. They did not talk about any
quantities changing. This suggests that they held a static mental image of the problem
situation. They did not appear to be imagining the quantities changing nor did they
provide any evidence that they were engaging in covariational reasoning to explore and
understand the problem situation. As they labeled their diagrams, Ann experienced some
difficulty. She was confused by the fact that she knew the rate of change for the
horizontal distance, to which she assigned the label of t
x
∆
∆in her diagram, but she also
thought she needed to “bring in an x.” The relationship between x and t
x
∆
∆ is still
developing as part of her content knowledge. This suggests that she had not thought
about each variable as a function of time, but only as an unknown value to be solved for
later. It would appear that her understanding of how to label a diagram involved
assigning only one letter to each unknown. Thus, the diagrams that the students drew to
represent the problem situation appear to provide only superficial information.
Following Excerpt 69, Table 19 summarizes the students’ mental activities as they
complete the second phase, construct a functional relationship.
Excerpt 69
74. NE: All right, and now what about the leg across the top of our triangle? 75. Ben: 600 miles per hour. 76. Ali: 600 t. 77. NE: You get 600 t.
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78. Ali: Yeah 79. NE: Ok. 80. Ben: What’s t, time? 81. Ali: Time, yeah. 82. NE: All right. 83. Ben: Do you have, no. 84. NE: All right. So you started saying that we can say that the thing across
the top is 600 t. So we can actually write it as a function of time? 85. Ali: Um hm 86. NE: Ok. Can we do that for the other legs of the triangle? 87. Ann: You can write that one, the one at the top, the one I labeled x. 88. NE: Ok. 89. Ann: Because it’s changing as z is changing. It’s getting larger as z is getting
larger. 90. NE: Ok 91. Ben: What? 92. NE: So can you write each leg of the triangle as a function of time? 93. Ann: I don’t know about the 3 miles one because it’s at a constant altitude. 94. Ben: So it’s always just going to be 3 miles 95. NE: All right 96. Ben: So, if you call that like d for distance. Now, delta d over delta t equals
3 miles. 97. NE: Ok. Is it changing though? 98. Ben: Because it’s supposed to be a function? No. 99. NE: So let’s start with just writing the functions of time. So the first one,
whatever we’re calling this top leg, we agreed was 600 t. So can you write that using function notation?
100. Ann: [sighs] Would it be? 101. Ben: p of t equals 600 t 102. NE: So a lot of times we write things like f of x equals something with x’s
in it right? 103. Ann: Ok 104. NE: So if you’re calling that top leg, what does it depend on? 105. All: Time. 106. NE: So we call that t. So x of t equals 107. Ben: 600 t 108. NE: 600 t. Would you agree that represents what’s happening along the top
leg of the triangle? All right. So now your constant leg, this one that’s the 3 over here, could you write that as a function of time?
109. Ben: You just won’t have t in the function. Could you just have like d of t equals 3?
110. NE: Sure, why not? Would you agree with that? 111. Ben: I would agree with it. I said it. 112. NE: Have we ever seen constant functions before?
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113. Ben: Twice 114. NE: Twice. Ok. So how about that z? Can you write z of t? 115. Ben: Um 116. Ali: Pythagorean theorem 117. Ben: Can you do that? 118. NE: Why not? Is that a valid 119. Ben: 600 t squared, the square root of 600 t squared, plus 3 squared 120. NE: Sure. 121. Ben: I’ll give it a whirl.
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Table 19:
Summary Table for Construct a Functional Relationship - Plane Problem
Phase
Construct a functional relationship (lines 74-121)
Solution
Artifacts ( ) ttx 600=
( ) 3=td
( ) ( ) 223600 += ttz
Content
Knowledge
Functions
• Express the variables as functions of time. o Ali quickly represented the horizontal distance with 600t, but she and the
other students needed guidance to use the function notation of ( ) ttx 600= .
(line 76) o Ann was not sure about what to do with the constant 3, but Ben suggests that
it is just a constant function. (lines 93, 94) This is further evidence of the students understanding of function hindering them.
o Ben resolved the issue of how to deal with the constant with, “you just won’t have t in the function could you just have like d of t equals 3” (line 109).
o Use the Pythagorean Theorem to relate the variables. Derivative
• Ben confused the constant altitude of the plane with a constant rate, “so if you call that like d for distance now delta d over delta t equals 3 miles.” (line 96) This was later resolved as the discussion of functions continued. It would appear that this may be the result of Ben trying to apply key word strategies to the problem statement.
Geometry
• The Pythagorean Theorem o Ali suggested that it will allow them to relate their variables. (line 116) o Ben asked, “can we do that?” and then applied it “600 t squared the square
root of 600 t squared plus 3 squared” (lines 117, 119)
Mental
Model • When prompted about the horizontal leg of their triangle, Ali interpreted it as 600t.
This is likely related to her heuristic of “write an equation” that emerged in her individual interview. (line 76)
• Ann stated, “because it’s [x] changing as z is changing it’s getting larger as z is getting larger” when asked if she could use a function to express the horizontal distance the plane travels. (lines 87, 89)
Heuristics • Ali appears to be relying on “write an equation and solve for x” strategy which was prominent in her individual interview. It this case, she would probably eventually want to solve for t.
• Possibly, distance equals rate times time is a heuristic. This could also be considered part of one’s content knowledge.
• Ben may have employed a key word strategy when he initially thought the 3 mile altitude represented a rate. However, it is not clear from the statements he made, and he was not prompted to further explain why he thought the 3 represented a rate.
• Compare your work to your neighbors’.
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The first thing that stands out about this step is that the students expressed each
variable as an explicit function of time. It would appear that Ali quickly employed her
heuristic of “write an equation” and her content knowledge of functions to express the
horizontal distance as a function of time. This may be related to her knowledge of
distance equals rate times time as part of her content knowledge. There was some
question about what should be done with the plane’s altitude which was constant. While
they had seen constant valued functions before, it was not obvious to the students that
they could write a constant valued function as part of a problem. Yet another question
arose when the students used the Pythagorean Theorem to relate the variables. Even
though they had performed algebraic operations on functions previously, the idea of
combining functions in this situation seemed inappropriate to the students. This suggests
that their notion of function is still developing. It is also evident from the solution
artifacts that were generated that the students accessed their content knowledge of
geometry to apply the Pythagorean Theorem to relate the variables.
When prompted about whether she could express each leg of the triangle as a
function of time, Ann stated that you could, “because it’s [x] changing as z is changing
it’s getting larger as z is getting larger.” Thus, it would appear that Ann can imagine the
plane flying over the radar tower and what happens to each of the legs of the triangle as
she does so. However, the students did not refer to their diagrams or make statements that
indicated they used their mental image to help guide their solution process. As was seen
with Ann, the students could make observations about the nature of the changing
quantities when prompted. This suggests that they may have the ability to imagine the
196
problem situation changing and attend to the nature of the variant quantities, but it is not
a natural part of their solution process.
In Table 20, following Excerpt 70, the students’ mental activities related to the
phase of relate the rates are detailed.
Excerpt 70
122. NE: I mean, have we done anything that seems wrong? 123. Ali: Um hm [shakes head] 124. NE: Really, all we’ve done is use the other two functions to write the z
function, right? 125. Ben: Um hm 126. NE: We’ve done things like that before. So now, what do you think we
might do? 127. Ali: Take the derivative of the p of t times the derivative of z of t 128. Ann: Well, first you have to write what functions we’re going to use, to use
to make delta z over delta t 129. NE: Ah, we want to get delta z over delta t cause that’s the rate of change
of the distance between the plane and the tower with respect to time. So how can we get delta z over delta t?
130. Ben: I think it’ssss delta p over delta t divided by delta d over delta t. 131. Ann: Why? 132. Ben: Because that doesn’t have a z in it all. I lied. 133. Ann: I have delta z over delta x times delta x over delta t 134. Ali: I have basically the same thing. Which one is your x? the 3? 135. Ann: This one. 136. Ali: Yeah, that’s the same thing I have. 137. NE: So what is the rate of change of the plane’s horizontal distance with
respect to time? 138. Ann: What change? 139. NE: What is the rate of change of the plane’s horizontal distance with
respect to time? 140. Ann: 600 miles per hour. 141. Ali: Rate of change, yeah, 600. 142. NE: Ok. So how can we represent that? 143. Ben: Delta 144. Ann: Delta x over delta t 145. Ben: Can we use p? 146. Ann: Why? x stands for distance. 147. Ben: Because we both have p. 148. NE: Well, you can use p, and she can use x.
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149. Ben: I just don’t understand what she says 150. NE: How about the rate of change of the plane’s vertical distance with
respect to time? 151. Ali: That’s 152. Ann: It doesn’t have any 153. Ben: Delta d over delta t 154. Ann: It doesn’t have a change of vertical distance. It’s just 3. 155. NE: Ok 156. Ali: Zero 157. NE: That’s zero 158. Ali: That should be zero, right? 159. Ben: But can it be represented as delta d over delta t? 160. NE: Sure 161. Ben: We probably don’t need it, but 162. NE: But we might 163. Ann: Change that, too. So I know what you’re talking about 164. Ben: See, I told you. 165. NE: Still, that delta z over delta t is what we need to figure out. 166. Ann: Right, I have delta z over delta x. 167. NE: You have delta z over delta x. 168. Ann: To figure out delta, yeah, but I have to figure out how to do that 169. NE: All right. So right now we’ve got z equals something with just t in it.
So could we just compute delta z over delta t? 170. Ali: Yeah 171. NE: How? 172. Ben: Just the derivative. 173. Ali: Take the derivative. 174. NE: Take the derivative, ok. 175. Ben: But that can’t be the answer 176. Ali: I think it is 177. NE: You think it is 178. Ali: Yeah 179. Ben: I’ll do it, then you’ll tell us 180. NE: What are you thinking 181. Ann: I’m thinking that I got a different answer than Ben 182. Ben: I know why 183. NE: You know why. Why? 184. Ben: Because I did it wrong 185. NE: How did you do it wrong 186. Ben: I forgot to put the negative one half, but that wouldn’t change it much 187. Ann: You have to square the 600 [points to Ben’s work] 188. Ben: What? No, you don’t 189. Ann: Yeah 190. Ben: You multiply by two when you find the derivative of the inside
198
191. Ann: No, but it’s 600 t [motions with hands] squared 192. Ben: What? 193. Ann: The whole thing’s squared. 194. Ben: Oohhh, stupid parentheses. Wait, but then it’d just be 2 times 600 t
times 600. 195. Ann: Huh? what? 196. Ben: Yeah, two times 600 t times 600. So that makes it twelve hundred.
You don’t have to square it cause it’s the quantity. So you have to do the chain rule. So you have to do 2 times 600 t times 600
197. Ali: But you can cancel out the 2 with the one that you brought down from the derivative of the whole.
198. Ann: What did you get? 199. Ali: I got 600 squared t over 9 plus 600 t squared the quantity to the one
half. 200. Ann: That’s what I got 201. Ben: What? [Ali shows him her paper] 202. Ali: Where did you get thirty six hundred? 203. Ann: I added not enough zeros 204. Ali: Oh 205. Ben: That’s, yeah that’s what I have. I just have to go do this like that so
you have thirty six thousand t 206. Ann: Three hundred sixty 207. Ben: Three hundred sixty thousand t 208. Ali: Yeah 209. Ben: How many zeros do I need here 210. Ali: More zeros 211. Ben: Times 600 t squared plus 9 to the negative one half 212. Ann: Yeah 213. Ben: Parentheses
199
Table 20:
Summary Table for Relate the Rates - Plane Problem Phase
Relate the rates (lines 122 -214)
Solution
Artifacts t
x
x
z
t
z
∆
∆⋅
∆
∆=
∆
∆
( )( ) 21
2
2
6009
600
t
t
t
z
+
=∆
∆
Content
Knowledge
Derivative
• Rate of change o Ali wants to, “take the derivative of the p of t times the derivative of z of t.”
This suggests that Ali knows how to compute derivatives, but she may be using a heuristic she developed in the previous sessions. (line 127)
o Ann replied to Ali’s suggestion, “well first you have to write what functions we’re going to use to use to make delta z over delta t.” This suggests that Ann’s heuristic is based on figuring out how the rates are related before computing any derivatives. Again, this may be tied to the discussions of the previous sessions. (line 128)
o Ben responded with, “I think it’ssss delta p over delta t divided by delta d over delta t.” When questioned why, he explained because it did not have a z in it.(line 130)
o Ann came up with, “I have delta z over delta x times delta x over delta t.” (line 133) Ali then wrote a similar statement using her variables.
o Ann questioned how she could find x
z
∆
∆(lines 166, 168) which suggests that
her understanding of derivative is focused on rules of computation rather than relationships between the independent and dependent variable. It did not occur to her that she could use implicit differentiation to compute this. Thus, implicit differentiation is likely a separate derivative rule for her.
o Ben suggested derivative (line 172) after they were reminded by the researcher that they had an equation with just z and t in it.
o The students were capable of applying their known derivative rules to their equation, including the chain rule. (lines187-214) The student checked their results with those of the other students.
Mental
Model • The students appear to focus on the equations they have just created.
• The students do not appear to reference their diagrams in this segment.
Heuristics • Compute the derivative was suggested by Ali. (line 127)
• Write a “delta equation” was suggested by Ann and appears to be related to the discussion of the chain rule. (line 128, 133)
• Check your answer against your neighbors’.
Relating the rates was a difficult step for the students as their understandings of
rate and derivative are still developing. The students appeared to be trying to make the
ideas from the discussion of the chain rule in the previous session fit the present situation.
200
They each wrote down a “delta equation,” an equation that related the rates using the
chain rule. This may be a heuristic that they developed in the previous sessions. Ali and
Ben each wrote down an equation that will not work which suggests that they may be
randomly trying to make the chain rule fit. Ann stated, “I have delta z over delta x times
delta x over delta t,” which will work. She may have a better understanding of how the
chain rule applies to this situation due to some of the questions she asked in the previous
session. However, she is not certain how to find x
z
∆
∆. She did not think about using
implicit differentiation, and the researcher did not push her to explore this option. This
suggests that the notation x
z
∆
∆ still does not convey information about the relationship
between the independent and dependent variables for her. Thus, the students’
understanding of derivative and rate of change is still developing. Computing derivatives
appears to be a set of rules applied to an equation rather than providing informing about a
rate of change for these students. After the researcher commented that the equation
related z and t, Ben suggested that they could find t
z
∆
∆directly by taking the derivative.
This may have been just another attempt to use a heuristic he had developed in the
previous sessions.
The students were then verbally provided with the numeric data the
mathematicians had been given as part of their problem statement. They used this
information to compute an answer and their mental activities are described in the
summary table following Excerpt 71.
201
Excerpt 71
214. NE: all right so now what if I said 215. Ben: it’s wrong 216. NE: no what if I said we also now that z is equal z of t is equal to 5 217. Ann: then you could set those two equations equal to each other right 218. NE: which two equations 219. Ann: the z of t equals equation and 5 220. NE: ok and what would that get for us 221. Ann: I have no idea z of t I don’t know 222. NE: well if z of t is equal to 5 and if I actually wanted to know the rate for
change of the distance from the plane to the tower with respect to time 223. Ali: then you would have to plug in 5 for t in this first equation 224. Ann: but it’s not saying t equals 5 its’ saying that the z of t the whole
function equals 5 225. NE: saying z of t right 226. Ali: ok so then we’d have to solve it for it 227. NE: solve it for what 228. Ali: 5 solve it z of t for 5 229. Ben: and so 230. NE: so what do you mean by solve z of t for 5 231. Ali: set it equal to 5 and then solve for t equals 232. NE: ok and then if you have t what can you do 233. Ali: plug it into the the the delta z over delta t 234. Ann: yeah 235. NE: ok and then we would actually know the rate of change of that distance
with respect to time Table 21:
Summary Table for Find the Unknown Rate - Plane Problem
Phase
Find the unknown rate (lines 215-236)
Solution
Artifacts ( ) ( ) 53600 22=+= ttz
Content
Knowledge
Function
• Understand that the value of 5 that was given is an output and may be represented by
( ) 5=tz . (line 220)
• Solve for t. (line 220, 232)
• Substitute the value of t into the derivative equation. (line 234)
• Perform appropriate algebraic manipulations
Mental
Model • The students did not refer to their diagram, suggesting that they were not using their
mental representation of the problem situation.
Heuristics • Compare your answer with your neighbors’.
202
In this step the students primarily accessed their content knowledge of function to
substitute in the values the researched provided them and perform appropriate algebraic
manipulations to compute an answer.
Finally, the students were asked to reconsider their functional relationship without
using the representation in which time was an explicit variable. That is, they were asked
to express everything using only ( )tz , ( )tx , and ( )td .
Excerpt 72
236. NE: Do you agree with that? So it’s kind of like the other problem. Now let’s step back a minute to where we constructed our z of t function. Here, instead of writing 9, is there another thing we could have put it place of that?
237. Ali: The height squared 238. NE: Ok. The height squared, or what did you call it? um d of t. 239. Ali: Yeah 240. NE: You could put in d of t squared. Similarly, what could we have put in
for the 600 t? 241. Ann: x of t squared. 242. NE: x of t squared. So let’s do that. 243. Ann: Ok 244. [writing things out] 245. NE: Now, could we find the rate of change of that with respect to time?
Leaving in those general function names? 246. Ann: Yeah 247. NE: Yeah, how would you do it? 248. Ben: Just find the derivative. 249. NE: Ok. So, if you do that, what happens? 250. Ben: Uh oh… oh stupid. 251. Ali: I don’t think I did that right. 252. NE: Why not? 253. Ali: I don’t know. It just looks 254. NE: All right. How does it relate to what you did before? Because they
should be equivalent answers, right? 255. Ali: Yeah. Oh, I have to take the derivative of that one. 256. NE: What did you come up with Ann? 257. Ann: I got x of t plus a of t over the square root of x of t squared plus a of t
squared. 258. Ben: That’s what I got, and I obviously used different letters.
203
259. Ann: It’s the altitude. 260. Ben: Altitude. 261. Ann: It starts with an a 262. Ben: It’s also the distance above the ground 263. NE: Ok. So when you took, [pause for Ali] when you took the derivative of
the inside 264. Ann: Um hm 265. NE: Um, so when you take like the derivative of this d of t squared 266. Ann: d of t squared 267. NE: Yeah 268. Ann: 2 d of t 269. NE: That’s it 270. Ali: No 271. NE: You say no. Why not? 272. Ali: Because it is a function to a power, so you have to use the chain rule
whatever it is, so have you do 2 d of t times d prime of t. 273. NE: What do you guys think about that? 274. [58:00] 275. Ben: I wasn’t listening. 276. NE: You weren’t listening. 277. Ben: No. 278. NE: All right. 279. Ben: What’s she saying? 280. NE: So when you take the derivative of the inside of that d of t squared 281. Ben: Um hm 282. NE: What’s the derivative of that? 283. Ben: 2 d of t plus 2 x of t. It’d be one half that those cancel with 284. NE: But when you take the derivative of d of t quantity squared, would you
agree that that’s a composition? 285. Ali: Um hm 286. Ben: What’s a composition? Oohhh 287. NE: A composition of two functions 288. Ben: So you have to multiply that by d of t afterwards is that what you’re
saying? 289. Ann: Times the rate of change of d of t 290. NE: Right. You have to multiply it by the rate of change of d with respect
to time, right? 291. Ali: Um hm 292. Ben: What? 293. NE: The rate of change of d with respect to time 294. Ali: d prime of t 295. Ben: Ok, yes.
204
Table 22:
Summary Table for Relate the Rates, Alternate Approach – Plane Problem
Phase
Relate the rates: Alternate approach (lines 237-296)
Solution
Artifacts ( ) ( )( ) ( )( )22tdtxtz +=
( ) ( )( ) ( )( )( ) ( ) ( ) ( ) ( )( )tdtdtxtxtdtxtz ′+′+=′−
222
1 21
22
( )( )( ) t
x
tz
txtz
∆
∆⋅=′
Content
Knowledge
Function
• Composition may be used to express the relationship with only general function names.
Derivative
• The students initially computed the derivative without the chain rule. (lines 249-259)
• Ali suggested that you need d prime of t, i.e. apply the chain rule to ( )( )2td . (lines
266-295)
Mental
Model • The students did not appear to utilize their mental model of the problem situation in
any way.
Heuristics • Compare your answer with your neighbors’.
The purpose of this discussion was to guide the students to be able to see that even if they
were unable to explicitly express the functions in terms of time, they could still construct
a relationship into which they could substitute known values. This discussion likely
would have been more meaningful if the researcher had pressed Ann to figure out a way
to represent x
z
∆
∆ when she brought it up when the students first attempted the phase of
relate the rates. Had she been pushed to explore this, she would have discovered that
xx
zz 22 =
∆
∆which could then be solved for
x
z
∆
∆ to find that
z
x
x
z=
∆
∆. Hence, another form
to relate the rates is given by t
x
z
x
x
z
∆
∆⋅=
∆
∆. This was eventually determined by
performing numerous substitutions.
205
The students had difficulty computing the derivative when using the general
function notation. The students used the power rule to compute the derivative of the
outside function (the squared part), but neglected to take the derivative of the inside. Ali
stated that you needed a ( )td ′ . This is interesting because the instructor normally used
Leibniz notation to represent derivatives in class. However, Ann and Ben were quick to
use the same notation.
7.4.2 The Trough Problem
The students were given the trough problem during the fourth teaching session. It
was stated as follows:
A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3
ft across at the top and have a height of 1 ft. Assume that the trough fills at a
constant rate. Let V represent the volume of the trough, h represent the height of
the water in the trough, and b represent the length of the base of the water in the
trough. What is the rate of change of the height of the water, h, in the trough with
respect to time?
This version of the trough problem has been stripped of all data except the dimensions of
the trough. Again, the purpose of removing the numeric data was to encourage students to
focus on building general relationships before substituting in values.
The students read the problem and immediately engaged in the phase of draw a
diagram which is summarized in the table following Excerpt 73.
Excerpt 73
1. NE: So what are you guys doing in your first steps? 2. Ann: Drawing a picture and labeling stuff.
206
3. NE: And labeling stuff? 4. Ann: Yeah 5. Ali: I’m getting an equation for the volume, and yeah 6. Ben: Oh, you’ve gone three dimensional. 7. NE: What’d you start with? 8. Ben: I drew a very rough sketch. 9. NE: Ok. 10. Ben: Write some numbers down, did some erasing. That’s it. 11. NE: So, how were you labeling things Ann? 12. Ann: Um, I labeled I labeled b equals 10, which I’m not sure if that’s right,
but I figured b is the length of the base of the water in the trough, and water’s going to fill it all.
13. NE: Ok 14. Ann: So I figured that would be 10, so I labeled b 10. Um, I labeled h 1
because it’s 1 foot high, and then labeled 3 across. 15. NE: Ok 16. Ann: And now I’m gonna put the volume equation, which I think is going to
be, um, one half, um, 3 times h times 10. Well, times b because the area of the base times the heighth.
17. NE: Ok 18. Ann: And, um, the area of the base is one half base times height, which is 3
times h, cause I didn’t really label 3 as anything. It’s just there, so 19. NE: Ok 20. Ann: But, yeah, so I have the volume, and I’m looking for um delta h over
delta t 21. NE: Ok 22. Ann: Yeah, [long pause] and so 23. [6:40] 24. NE: And what have you done Ali? 25. Ali: I have the volume for the trough, but the total volume that it can hold 26. NE: And how have you labeled your diagram? 27. Ali: I have the, the length is b, that’s 10, and the width is 3, that’s the top,
and height is 1. 28. NE: Ok 29. Ali: And 30. NE: How about you? [to Ben] 31. Ben: Oh, I’ve done the same thing, only less. 32. NE: Only less? 33. Ben: Yeah, I didn’t know the equation for volume. One half base times
width times height. Is that it? 34. NE: What’s base and width and height? 35. Ben: the height’s 1, the base’s 10, the width’s 3. 36. NE: Ok and ok 37. Ben: Is that right?
207
38. NE: Ok. Yeah. 39. Ben: So the maximum volume is 15 cubic feet. 40. Ann: Yeah 41. NE: That’s the maximum volume, yes. 42. Ben: I don’t know if I need to know that at all. Probably not because I found
it. 43. NE: So do you think that everything you find is something you don’t really
need? 44. Ben: Probably 45. NE: Probably 46. Ben: Unless it’s money. I always need money. 47. [8:50] 48. NE: All right. 49. Ann: So 50. NE: So what have we just done Ann? 51. Ann: Well, I think I’m gonna, the heighth, the height of the water is gonna
be 1 minus a variable. 52. NE: Ok 53. Ann: Because it’s gonna go up. So that’s the height. 54. NE: That’s the height. 55. Ann: And I’m not sure. Um, I don’t know what to really put cause it’s not
the height, cause the height is 1, and just call it h 1 equals 1 minus x, I guess, cause x is a good variable.
56. NE: x is a good variable? 57. Ann: It’s just a variable. 58. NE: It’s just a variable. 59. Ann: I just said it’s good cause it’s one that’s there. 60. NE: Ok 61. Ann: Um 62. NE: All right. So in your picture, what are the things that are changing? 63. Ann: the height of the water. 64. Ben: Volume height and height. 65. NE: Height and height. 66. Ben: I didn’t mean to say height twice. 67. Ann: The volume and the height and time. 68. NE: The volume and the height and time. Ok. 69. Ali: The height, time, the volume, and the length of 70. Ann: That’s 71. Ali: The side 72. Ben: Length of the side 73. Ali: Yeah, the side of the trough. Like whenever you fill it up, if you go
only halfway there it’s to here. So this length is going to be changing. It’s kind of like the plane problem where as instead of, well instead of
208
just having a constant height, this has a changing height, which makes, which changes the height of the, the top, which changes the side.
74. Ben: I think the side stays the same. It’s just included in the volume changing.
75. Ali: It’ll change when like whenever the height goes up. Like if it’s here, this length is going to be different from if the height’s here. It would be only here.
76. Ann: h represents two things in this problem. We have h representing the height of the water trough, and then we’re trying to find, yeah you’re trying to find the rate of change of the height of water as h, so can I call this h1?
77. NE: Sure 78. Ann: Ok so 79. Ben: You can in fact call it h 1 80. Ann: So the height’s changing, the volume’s changing, and the time 81. NE: Actually, I think the first sentence, that should read let h represent the
height of the water in the trough 82. Ann: Oh, ok. 83. [12:06] 84. Ben: So that one’s supposed to be something else 85. NE: That one 86. Ben: Yeah, when we have h equals one, that should be a different variable. 87. NE: The height of the whole trough can still be 1 foot 88. Ben: Yeah, but should we label that something else if h should be the height
of the water? 89. NE: You can call it whatever you want. 90. Ben: I’m going to use q. 91. NE: q. 92. Ben: It’s a really unused variable. I think it feels sad. 93. Ali: It’s a quantity. 94. Ben: I don’t deal with quantities. 95. NE: Can I make a suggestion that maybe when we look at our trough, we
look at it just from the end [draws picture] and say we know that’s going 10 feet back into the paper or coming out of the paper? We’re just looking at an end view.
96. Ann: I can try. 97. NE: Because does the length of our trough ever change? 98. Ali and Ann: No 99. NE: So that’s one of the few things that’s actually constant. [they draw] So
how does that change your labeling? 100. Ben: The labeling, in what sense do you mean? 101. NE: So if you’re going to be labeling things in your picture now that we’re
looking at it just straight on, what kind of things do you think are important and should be labeled?
209
102. [14:39] 103. Ben: Base and the height 104. NE: Ok why would you label the base and the height? 105. Ben: Because we need it to solve things. I don’t know what you’re asking
for sure. 106. NE: All right. So you started out, you drew this picture. Now I’ve
suggested an alternate way to draw the picture. And when you first drew your picture, you started labeling things.
107. Ben: Um hm 108. NE: So I guess I want to know, now that we’re looking at this drawing that
you’ve done again, you’ve drawn a diagram. If you’re going to label things, what things are you going to label in this picture and why?
109. [15:30] 110. Ann: Ok. I have a 3 across the top because that’s how wide the top of the
thing is, and the height labels 1, and I have, I have an h drawn somewhere in the middle, and then, or somewhere on the side that’s h, then I have an x.
111. NE: Ok 112. Ann: Then I have a 1 minus x equals h 113. NE: Ok 114. Ann: So then I labeled whatever’s underneath h is the volume 115. Ben: Cause you used different labels 116. NE: So why did you choose to label those things in your picture? 117. Ann: Because the height is changing, and because the height of the entire
trough equals 1. So when it reaches full capacity, it’s gonna, the height h is gonna equal 1.
118. NE: Ok 119. Ann: And, and I put x because it’s a variable, but x is the difference between
h and the full capacity. 120. NE: Ok 121. Ann: So, and then 3 is the top. So I just labeled it so we know how far it is
across, because if you’re going to calculate the volume, you need to know the top.
122. NE: Ok 123. Ann: In theory 124. NE: Ok how have you labeled things Ali? 125. Ali: I’ve got the top, I’ve got it labeled base b, and the height which is h,
and I haven’t worried about the entire triangle. I just, yeah 126. NE: Ok 127. Ali: So keeping it general 128. NE: Ok [long pause] All right. You guys discuss among yourselves and see
what you think you should do? 129. Ali: Ok 130. NE: Try to agree on one labeling scheme between the three of you
210
131. Ann: Well, I don’t feel comfortable calling, um, the top the base, because the base is b, which is 10, which is the base of the whole trough
132. Ali: Um 133. Ann: That’s just why I didn’t label that b 134. Ali: Should we call it length? 135. Ann: That would work 136. Ben: Or width 137. Ann: Width 138. Ben: Length 139. Ann: [mumbles something – long pause] 140. [19:01] 141. NE: So what have you decided to call your variables? 142. Ali: Um, w and x I guess 143. Ann: The top is w, the height is q, the height of the whole trough, the height
of the water is h, and the distance between the height of the water and the height of the trough is x
144. NE: Ok 145. Ben: That’s what I have 146. Ali: The height of the water is h 147. Ben: Um hm 148. Ali: Ok
211
Table 23: Summary Table for Draw a Diagram - Trough Problem
Phase
Draw a diagram (lines 1-148)
Solution
Artifacts
Labeled diagram
Content
Knowledge
Geometry
• Know what an isosceles triangle is.
• Volume of triangular prism. Variable
• Label everything that is unknown with a variable. o Each student labeled the height, width, and length.
o Ann labeled the height of the water with x−1 . (lines 51, 53, 55)
o Ann believes that h represents two things, the height of the water and the height of the trough. (line 76)
o The researcher clarifies the problem statement that h represents the height of the water in the trough. (line 81)
o The students label the height of the trough with one variable which is equal to 1, but the height of the water in the trough with h.
Function
• Ali wrote down a formula for the volume of the trough. (line 5) She also computed the total volume that the trough can hold. (line 25) Ann and Ben followed her lead and computed it for themselves. (lines 26-41) When asked how that will help them, they did not have an answer.
Derivative
• Ann identified that she is looking for delta h over delta t. (line 20)
Mental
Model • Ben initially drew a 2-dimensional triangle, but noticed “oh you’ve gone three
dimensional” when he looked at Ann’s and Ali’s diagrams. (line 6) Then, he changed his diagram to look like Ann’s. However, the researcher suggested that they consider just the cross-section in line 95.
• Ann identified that she wanted to find t
h
∆
∆. (line 20)
• When prompted to identify which quantities are changing, the students quickly identified, volume, height, and time. Ali then suggested the width. (lines 62-75)
Heuristics • Label everything that is unknown with a variable.
• Check your work against your neighbors’.
From the summary of the students’ mental activities during the draw a diagram
step, we can see that Ann and Ali drew 3-dimensional figures while Ben initially drew
only the 2-dimesional cross-section. It would appear that Ann and Ali created a mental
model of the trough which was then drawn on their paper. It is possible that Ben
212
imagined the whole trough and chose to draw only the cross-section, or that Ben’s
diagram may have been based on key words in the problem statement, specifically
isosceles triangle. However, Ben did not make any statements to suggest that he had
considered the whole trough. Since Ben modified his diagram to match what Ann and Ali
had drawn, it would seem more likely that his diagram was based on key words in the
problem statement. Each student labeled the dimensions of the trough with the
appropriate numeric value. The students accessed their content knowledge of geometry to
write down the formula for the volume of the trough and computed the maximum volume
of the trough. However, they could not elaborate on how they thought that this would
help them. As they proceeded to label their diagrams with variables, Ann appeared to
have the same difficulty she had in the plane problem. She could not let h represent the
height of the water in the trough and the height of the trough. Thus, she labeled the height
of the water with x−1 , where x represented the difference between the height of the
water and the top of the trough.
As the students proceeded, they engaged in the phase of construct a functional
relationship which is summarized in Table 20, following Excerpt 74.
Excerpt 74
149. Ann: So you can set an equation for the volume, but we’re not really looking for the volume, are we? I know how to find the volume.
150. NE: Ok. So what’s your volume equation? Because that seems to be something that ties all these variables together. Would you agree with that? So can you set up a formula for the volume of the water in the trough?
151. Ann: Yeah I have b equals 15 h 152. NE: How did you get that? 153. Ann: Um because v equals one half 3 times h times 10, because um it’s one
half base heighth times or it would be one half w h b?
213
154. NE: Ok 155. Ann: With the width of the top times the height times the base of the whole
trough, or the base, I’m sorry, and then you know the values for w and b so just plug those in, and it’d be15 h. You can also change those into 15 times 1 minus x.
156. NE: Do you agree with that? 157. Ali: No 158. NE: Why not? 159. Ali: Because the 3 is the width for the top of the trough, and that’s not
necessarily how the width of the water as you’re filling it. 160. Ann: That’s true. 161. Ben: So you need another variable? 162. Ann: I think it, yeah, probably 163. NE: Another variable. All right. So you’ve got a volume for the volume of
the water in the trough. What’s yours? 164. Ali: 5 w h 165. NE: 5 w h 166. Ann: [mumbled] 167. NE: What was that? 168. Ann: [mumbled] 169. Ben: Is w the width of the water? 170. Ali: Yeah up at the top 171. Ben: Ok 172. Ann: I thought 173. Ben: Oh, because of one half base times. So you took one half and then the
ten always 174. NE: All right. So if we agree that the volume can be represented by 5 times
the width times the height 175. Ann: So if we look at the width of the top, so it should be w 1, or w, not or
something because 176. Ben: What? 177. Ann: Because w is up here 178. Ben: Oh, we said that was l 179. Ann: I thought you changed it 180. Ben: No [mumbled] I need to make bigger drawings. Becoming
increasingly difficult 181. NE: Ok, so in effect [draws diagram with labels] you guys are calling this
w, is that right? 182. Ali: Um hm 183. NE: And you’re calling that h 184. Ali: Um hm 185. Ben: Yeah 186. NE: Ok 187. Ali: Similar triangles. We can compare the two possibly
214
188. NE: All right. So how can we use this idea of similar triangles to compare those two?
189. [23:18] 190. Ali: We compare like the volume to like the whole trough maybe [it’d even
be two?] 191. Ben: So 192. NE: Well, right. Now we’ve got three variables. Have we ever really had
three variables like that before? 193. Ali: No, since it’s three dimensional, not two dimensional. 194. Ben: We haven’t even brought time into this at all. 195. NE: Not yet, have we? 196. Ali: Um hm 197. Ben: It hurts 198. NE: So how could we maybe bring time in? 199. Ali: The change in volume with respect to time 200. NE: Volume with respect to time ok 201. Ben: I think you put time into an equation, because you can’t just make
things up 202. NE: You can’t just make things up ? 203. Ali: But it’s in the rate at which the water’s coming in, that is constant 204. Ben: But we want numbers for like the final answer of change in height
with respect to time 205. NE: In the end we’ll get to some numbers 206. Ben: Oh you have the answers 207. NE: All right. So let’s go with Ali’s idea of how to relate w and h, and see
if we maybe can eliminate one of our variables. Is that a reasonable thing to do? Have we ever done things like that before?
208. Ali: Yes 209. NE: How about in our maximization and minimization problems? We were
frequently trying to eliminate one variable or another, right? So if we were going to eliminate one of these three variables, which do you think we should eliminate?
210. Ben: w 211. NE: You wan to eliminate w. why? 212. Ben: Um, I don’t know. Cause you think with the length of the top and the
height of this thing it would be some way you could eliminate w 213. NE: Ok 214. Ben: But I don’t know how 215. NE: Well, what are we ultimately trying to find? 216. Ann: The change in height 217. Ben: Change in height with respect to time 218. NE: The change in height, so we want the change in height with respect to
time 219. Ben: So we don’t want to eliminate h
215
220. [26:06] 221. NE: So we don’t want to eliminate h. We want to keep that around, right?
That would seem to be a useful thing to keep 222. Ali: Um hm 223. NE: So, can we eliminate that w? 224. Ali: Could we find the, the angle using the whole trough? The angle of this
so we can use a different equation to like, so we can find w with height in the other value?
225. NE: Ok 226. Ann: You could have um 3 over w equals one over h for the equation, the
ratios 227. NE: Ok 228. Ann: Because l is the length of the trough, and w is the height of the, or the
width, then you have height of the whole thing over the height h 229. NE: Ok 230. Ann: Then you cross multiply and you get w equals 3 h 231. NE: Ok, but what about that l? 232. Ann: l equals 3 233. NE: l is three. Does that ever change? 234. Ann: No 235. NE: No 236. Ben: So w equals 3 h 237. NE: Ok 238. Ben: Wait a second. That can’t be right, could it be? 239. NE: Why do you think it wouldn’t be right? 240. Ben: Well, that would be right because then when it got to the maximum the
width would be three 241. NE: Ok 242. Ben: It’s 3 h, huh? Crazy l over w equals 1 over h, so l h equals w equals 3 h 243. Ann: So now the volume equation’s like all in complete terms of h. Not that
that’s anything, but it is. 244. NE: So, yes, the volume is in terms of h 245. Ben: The volume’s 246. Ann: So it’s 15 h squared 247. Ben: 5 times 3h times
216
Table 24:
Summary Table for Construct a Functional Relationship - Trough Problem
Phase
Construct a functional relationship (lines 149-247)
Solution
Artifacts Ann: hV 15= (line 151)
Ali: whV 5= (line 164) 215hV =
Content
Knowledge
Function
• Ann substituted in 3 and 10 for their variables in her first attempt to write an equation. (lines 149-155)
• Ali did not agree with Ann’s equation. She appears to recognize that there are two changing quantities that need to be accounted for in her equation. (lines 156-170)
• Ann has the issue with the width that she had with the height. She thinks they should be assigned two different variables. (line 175)
• Ali suggested similar triangles, and thinks she should relate the volume of the whole trough to the volume of the water. (lines 187-190)
• Ben notes that we have not brought time in and suggests bringing time in: “I think you put time into an equation because you can’t just make things up.” (lines 194-201)
• The researcher suggested going with Ali’s idea and referred them to how similar triangles were used in the max/min problems they had done previously. They were also prompted to discuss which variables they think they want to eliminate. (lines 207, 209)
• Ben noted that with length and height, it would seem there should be a way to eliminate w. As a result of some prompting, he further stated that we do not want to eliminate h because we are trying to find the change in height with respect to time. (lines 210-221)
• The students used composition to eliminate w from their volume equation. (lines 226-247)
o Ben questioned, “well that would be right because then when it got to the maximum the width would be three.” (line 238, 240)
o After setting up her ratio, Ann stated, “then you cross multiply and you get w equals 3 h.” (line 230)
o Ann used substitution and stated, “so now the volume equation’s like all in complete terms of h not that that’s anything but it is.” (line 243)
Geometry
• Rectangular prism o The students know what the volume of a rectangular prism is.
• Ali asked, “could we find the the angle using the whole trough the angle of this so we can use a different equation to like so we can find w with height in the other value?” This did not lead to any discussion. (line 224)
• Ann suggested, “you could have um 3 over w equals one over h for the equation of the ratios.” (line 226)
Mental
Model • Ali again identified that the width of the water was changing, but now has a context to
discuss why it is important. (line 159)
Heuristics • Ali appears to have some heuristic that causes her to always think about angles (line 224). She also brought up angles in the plane problem.
• Compare your work to your neighbors’.
217
The students spent the majority of their time for this problem determining an
appropriate functional relationship. Ann substituted 3 and 10 into her volume formula
with which Ali disagreed. Ali again stated that the width of the water is a changing
quantity, so one cannot substitute in that value. She had noted that the width of the water
was changing earlier, but did not have a context in which to discuss why it was important.
Relating her mental model of the problem situation to the algebraic representation gave
Ali the context to discuss why the width of the water was an important changing quantity.
She could now identify that the width of the water was part of the algebraic
representation of the volume of the trough which could not be replaced by a numeric
value because it was not constant. Ali suggested using similar triangles to relate the
volume of the whole trough to the volume of the water in the trough. Thus, Ali accessed
another part of her content knowledge of geometry. Using similar triangles may be a
heuristic Ali developed as a result of solving max/min problems earlier in the semester.
However, it is not clear if she understands how they are useful and when they are
appropriate because she later suggests that they will allow her to determine angles. Ben
thought that time had to be brought into the function which may be the result of how the
plane problem was solved in the previous session. Recall that the students were able to
solve the plane problem by explicitly expressing each function in terms of time. The idea
that each function must be explicitly represented in terms of the independent and
dependent variables suggests that the students’ understanding of the concept of function
is still developing. They do not appear to be able to think about each variable as a
function of time unless they have expressed it the form ( )tf .
218
The researcher encouraged the students to pursue Ali’s similar triangle approach
and referred them to the max/min problems they solved earlier in the semester. It was
noted that in solving extreme value problems, extra variables were frequently eliminated.
This prompted them to think about which variables should be eliminated. Ben concluded
that w should be eliminated because we were interested in the rate of change of the height
with respect to time. Ann used similar triangles to set up a proportional relationship and
determined that hw 3= . Thus, the students used composition to express the volume of
the water in the trough in terms of height. However, it was not clear to them that this
would be useful. (Line 243)
After they had successfully related the variables in the problem with an algebraic
representation of the functions, the students continued with the phase of relate the rates.
Excerpt 75
248. NE: So, now you guys agree that this is changing over time? 249. Ali: Um hm 250. NE: So, how have we been relating everything else to time? 251. Ann: Taking derivatives 252. NE: All right. Taking derivatives. What does that mean? 253. Ann: Well, cause delta h over delta t is, you can take the derivative of the
that, and you’ll get the change in height with respect to time 254. NE: Ok 255. Ann: Taking the change in height with respect to time 256. NE: ok. So when you take the change in height with respect to time, how
does that come out of this? Can you show me? 257. Ann: No 258. [28:57] 259. NE: Do you agree that if we want the rate of change of volume with respect
to time, that’s the same thing as rate of change of volume with respect to height times the rate of change of the height with respect to time?
260. Ali: Wait, say that again. 261. NE: The rate of change of volume with respect to time is equal to the rate
of change of the volume with respect to height times the rate of change of the height with respect to time
219
262. Ali: Um hm, yeah 263. NE: Ok. So if you write that out, so what have you got? 264. Ali: Uh, I just lost myself. I don’t know what I was thinking. Um, rate of
change of volume with respect to height, the height with respect to time
265. Ben: So, it’s the rate of change of height with respect to time equal to the rate of change of volume with respect to time divided by the rate of change of volume with respect to height?
266. NE: What would dividing buy you? How would that help you? 267. Ben: What do you mean? Then you’d be multiplying by the reciprocal, and
you get delta h over delta t, which is what you want, right? 268. NE: Um hm 269. Ben: [mumbled] 270. NE: All right. So, what did you write Ann? 271. Ann: I have v of h equals 15 h squared because the volume is in terms of h
now. So it should be delta h or delta v over delta h 272. NE: Ok 273. Ann: And so I took the derivative that, and it’s 30 h 274. Ben: What’s the derivative equal to? 275. Ann: Delta h over delta v or delta v over delta h 276. Ben: Ok. So we still have to find delta v over delta t 277. Ann: Well, the change in volume over the change in time is constant, isn’t
it? The rate at which it fills is a constant. 278. NE: It is a constant 279. Ann: So when it fills the rate is constant 280. NE: So then what would be constant 281. Ann: If the rate at which it fills is constant, then the rate of change of the
volume with respect to time would also be constant 282. NE: Um hm. All right. Are you still having a hard time writing that down
Ali? 283. Ali: Yeah 284. NE: So think about the rate of change of the volume with respect to time.
So write that first 285. Ali: Ok with respect to time [Ann and Ben have small side conversation
hard to hear] 286. NE: That is equal to 287. Ali: Equal to the rate of change of 288. NE: Just put an x through this for right now 289. Ali: Ok 290. NE: And let’s just start all over down here 291. Ali: Ok 292. NE: We’ve got lots of paper. So, so that is equal to, you want the rate of
change of the volume, but what is the volume related to in your equation?
220
293. Ali: Height 294. NE: All right. So the rate of change of the volume with respect to the
height, no leave that as time the rate of change of the volume with respect to time. Now, we’re going to look at the other side of our volume equation. So this is volume. We want to know the rate of change of that with respect to time. Over here volume is in terms of
295. Ali: The height 296. NE: So we want the rate of change in the volume 297. Ali: With respect to the height 298. NE: With respect to the height, right, but that’s not quite there yet because
we also want to know the rate of change of the height with respect to time
299. Ali: So, would I multiply that? 300. NE: Um hm, because like when we wanted to know the rate of change of
the temperature on the mountain, right, how did that work? 301. [33:35] 302. Ali: You had to multiply the rate of change in temperature with respect to
height times the rate of change in height 303. NE: The rate of change of height with respect to 304. Ali: Time 305. NE: With respect to time right 306. Ali: Yeah 307. NE: Is this the same thing 308. Ali: Um hm 309. NE: Because the time in this instance time is not explicit. We have volume
and height that are related 310. Ali: Right 311. NE: Right. Ok, so now the rate of change of volume with respect to the
height, what’s that? 312. Ali: Um, with respect to the height, 30 h 313. NE: That’s the 30 h. So, how does that change your equation now? 314. Ali: Now, that is change in volume with respect to height equals 30 h times
the rate of change in height with respect to time 315. NE: Right
221
Table 25:
Summary Table for Relate the Rates - Trough Problem
Phase
Relate the rates (lines 248 -315)
Solution
Artifacts t
h
h
V
t
V
∆
∆⋅
∆
∆=
∆
∆
hh
V30=
∆
∆
h
t
V
t
h
30
∆
∆
=∆
∆
Content
Knowledge
Derivative
• Rate of change
o The students knew they wanted to find t
h
∆
∆ but were unsure of how to go
about it. The researcher reminded them of how they have used the chain rule before. (lines 248-268)
o Ann noted that she had 215hV = which she took the derivative of to get
that hh
V30=
∆
∆. (lines 271, 273)
o Ann proceeded to explain to Ben, “if the rate at which it fills is constant then the rate of change of the volume with respect to time would also be constant.” (line 277)
o Ali struggled to understand this, and the researcher discussed it with her again. (lines 282 -315)
Mental
Model • The students appear to focus on the formula they had just created.
Heuristics • Check your answer against your neighbors’.
After being reminded about the chain rule, Ann applied the chain rule to write
t
V
V
h
t
h
∆
∆⋅
∆
∆=
∆
∆ , a relationship that relates the rates. She then quickly noted that since
she knew volume in terms of height, she could use the derivative to find that hh
V30=
∆
∆,
and explained her reasoning to Ben. Ben agreed that what she had done was correct. They
222
continued by substitutingh30
1 for
V
h
∆
∆ in their delta equations. Thus, Ann and Ben
completed the phase of relate the rates.
Relating the rates was particularly difficult for Ali. She was proficient with
computing derivatives. However, she struggled to understand the connection between the
symbol h
V
∆
∆ and the terminology of “the rate of change of the volume with respect to
height.” These did not appear to fit with her understanding of derivative. It seemed that
Ali would have preferred if the equations could have been expressed explicitly in terms
of time. A problem from the second teaching session in which the multiplicative nature of
the chain rule was discussed was revisited to allow her to identify how it is not necessary
to be able to explicitly express the functions in terms of time.
Thus, the students had created a general algebraic representation for the
relationship between the rates in the trough problem as stated. The researcher then
provided them with numeric values for some of the unknown quantities which they
successfully substituted into their equation. This is summarized in the following table.
Excerpt 76
316. NE: All right. So now, what if you actually knew that the rate at which the water was coming in was 12 feet cubed per minute?
317. Ann: Then you would just put 12 feet cubed per minute over 30 h 318. NE: Ok 319. Ben: Um hm 320. Ann: So it would be 12 over 30 h 321. NE: Ok, and what if I told you that the height was 6 inches? 322. Ann: Then it would be with the 12, with the 12 feet 323. NE: Right. So we’re still calling it 12 feet cubed 324. Ann: 12 divided by you just plug h in with the 12 325. NE: So, and that h being 6 inches, does that make a difference?
223
326. Ann: Oh, it’d be point 5 feet. So, it’d be different, it’d be like point 5, 30 times point 5.
327. Ben: 15 328. NE: Ok. Now what, are you there Ali? 329. Ali: Um hm, yeah.
Table 26:
Summary Table for Find the Unknown Rate - Trough Problem
Phase
Find the unknown rate (lines 316-330)
Solution
Artifacts ht
h
30
12=
∆
∆
)5.0(30
12=
∆
∆
t
h
Content
Knowledge
Function
• The students were told they knew that, “the rate at which the water was coming in was 12 ft3/min”, which they successfully interpreted as the rate of change of the volume and substituted it into their equation.
• The students were given the height of 6 inches which they appropriately substituted into their equation.
• The students carried out appropriate algebraic manipulations to solve for the unknown rate.
Mental
Model • The students appeared to focus on their equations and did not refer to their diagram,
suggesting that they were not using their mental representation of the problem situation.
Heuristics • Compare your answer with your neighbors’.
After the students had completed the problem, they were asked how the problem
would have been different if they had been asked to find the rate of change of the width
of the water with respect to time. Without hesitation, they all answered that they would
have solved for h instead of w in the proportional relationship that they had defined.
At the end of this session, the students were asked how they would tell someone
else how to solve a problem like this. Ann summarized the problem solving process as:
draw a picture, label everything, find a formula for the volume, eliminate variables if you
can, and draw yourself a delta equation. Ann further explained the delta equation:
224
Excerpt 77
1. Ann: They show you the composition of what, what you’re trying to find is, like it shows you the steps of what you’re actually trying to find
2. NE: Ok
3. Ann: Like delta h over delta t is delta h over delta v times delta v over delta t, and those are like the steps you have to get to, and the things you have to figure out before you, so you can find out what the rate of change is
This suggests that Ann uses the chain rule to relate the rates in an equation which then
guides her solution process.
The students were asked to write up the solution to this problem as homework.
Ben did not turn it in as he did not complete the study beyond this point. Ali and Ann
turned in their solutions which are presented in Figures 9 and 10.
225
Figure 9: Ann’s solution to the trough problem.
Notice that Ann’s solution begins with a diagram, “to get a better understanding
of what the problem was asking me.” She drew the 3-dimensional trough, labeled with
constants and the 2-dimensional cross-section labeled with variables. The next piece of
the solution is the “delta equation” with the known and unknowns identified. She labeled
226
V
h
∆
∆ as what she wants to find and
t
V
∆
∆as a known quantity. The solution then focused
on identifying the volume in terms of height and its derivative. She noted, “in terms of h
because I am looking for ( )hVV
h
′=
∆
∆ 1 because it is the reciprocal of the derivative.” The
first part of her statement suggests that her concept of rate may now include some sort of
relationship between the independent and dependent variables. However, the second
because indicates that the relationship is not fully developed.
227
Figure 10: Ali’s solution to the trough problem.
Ali’s solution is not as clear as Ann’s. It seems that Ali is just trying to remember
the steps in a procedural manner to obtain a numeric solution. Ali attempted to recall the
numeric values that were used to illustrate how a numeric solution might be found. She
228
incorrectly recalled that 5.0=∆
∆
t
h(0.5 had actually been given as the height of the water
in the trough). Thus, she solved for the height. This suggests that she had not yet
recognized the basic pattern of the problems, solve for the unknown rate. In contrast, Ann
stopped when she had successfully answered the question as stated in the original
problem.
Another interesting difference between Ann and Ali’s solutions is the voice they
chose to use. Ann wrote in the first person, “I did…” and “I know…” This suggests that
Ann is taking ownership of the solution and the mathematics. Ali wrote in the second
person, “You do…” This suggests that Ali is not taking ownership of the solution nor the
mathematics. Rather, it is merely a procedure that must be carried out.
7.4.3 The Coffee Cup Problem
In the fifth teaching episode, the students were given the coffee cup problem
stated exactly as it had been for the mathematicians:
Coffee is poured at a uniform rate of 20 cm3/sec into a cup whose inside is shaped
like a truncated cone. If the upper and lower radii of the cup are 4 cm and 2 cm,
respectively, and the height of the cup is 6 cm, how fast will the coffee level be
rising when the coffee is halfway up the cup?
By having the students solve this problem, it is possible to observe whether the students
approached it the same way as the mathematicians or approached it differently. Ben did
not solve this problem as he did not complete the study. Thus, the data reported is only
for Ann and Ali.
229
As in the previous problems, the students began by drawing a diagram and
labeling it.
Excerpt 78
1. Ali: Truncated cone is like a cone that the tip is cut off, right? 2. NE: Right 3. Ali: Ok 4. Ann: I can’t draw. 5. Ali: So, basically when I draw 6. NE: It’s a cup, imagine that 7. Ann: I’m going to put a handle on it 8. NE: You can put a handle on it sure 9. Ann: Ok, coffee [long pause while each work independently for a while] 10. [2:23] 11. NE: So, what are you guys thinking about? 12. Ali: Just labeling the values and think of a…. got to get the equation for
volume Table 27:
Summary Table for Draw a Diagram - Cup Problem
Phase Draw a diagram (lines 1-12)
Solution
Artifacts
Labeled diagram
Content
Knowledge
Geometry
• Know what a truncated cone is. o Ali confirmed that it was a cone that had the tip cut off. (line 1)
Variable
• Label the known and unknown quantities. Function
• Find a formula for the volume. (line 12)
Mental Model • Imagine what a cup looks like and draw what has been imagined as a diagram on the paper.
Heuristics • Compare your work to your neighbors’.
Ali confirmed that a truncated cone meant that the tip had been cut off. After
which she and Ann both drew an appropriate diagram which they labeled with the
constant values. Ali also stated that she wanted a formula for the volume of the cup.
230
In response to Ali’s request for an equation for the volume, Ann began identifying
the given rate and the rate she wanted to find. They proceeded to relate the rates with a
“delta equation” as seen in Table 28.
Excerpt 79
13. NE: Why do you need to get an equation for the volume? 14. Ali: Just so we can find the rate that it’s rising possibly 15. NE: Ok 16. Ali: Well, we have the true velocity of time, or change in volume with time 17. [3:37] 18. Ann: Hm 19. NE: Hm 20. Ann: So, we’re trying to find the change of volume with respect to height 21. Ali: Change in height with respect to volume, isn’t it? Or with respect to
time, maybe? 22. Ann: Well, it couldn’t be with respect to time because we already have it. 23. NE: Why can’t you have it with respect to time? 24. Ann: Well, you already have the change in volume with respect to time. 25. Ali: The change of height with respect to time 26. Ann: Oh, um [couple more sighs] 27. NE: Tell me what you’re thinking 28. Ann: I think we are looking for it [height?] with respect to time but…
because we’re looking for how fast, and so then we need the cm cubed over seconds to do that
29. NE: Ok 30. Ann: Cause we know how fast the coffee is coming into the cup, but we
don’t know how high, or how fast the level is rising. So, we’re not looking for volume, we’re looking for height with respect to time
31. NE: All right. So you think we’re looking for height with respect to time. What do you think Ali?
32. Ali: I think we’re looking for the change in height with respect to time 33. NE: So you agree with Ann? 34. Ali: Um hm 35. NE: Ok. So how are you going to relate those? 36. [5:22] 37. Ann: Delta h over delta v times delta v over delta t. The change in height
over the change in volume times the change in volume over the change in time.
38. NE: Ok, and what do you think Ali?
231
39. Ali: Um, let me check. I have the same thing, just with the change in volume with respect to height on the opposite side of the equation. So, you can solve for it so…
Table 28:
Summary Table for Relate the Rates Part 1 - Cup Problem
Phase
Relate the Rates - Part 1 (lines 13-39)
Solution
Artifacts Ali:t
h
h
V
t
V
∆
∆⋅
∆
∆=
∆
∆
Ann: t
V
V
h
t
h
∆
∆⋅
∆
∆=
∆
∆
Content
Knowledge
Derivative
• Identify the given rate and the rate you want to find. (lines 20-32) o Ali identified that change in volume with respect to time is given. (line
16) o Ann identified the change in height with respect to time as the rate they
wanted to find. (line 20)
• Use the Chain Rule equation to relate the rates. (lines 37, 39)
Mental Model • The students did not appear to use their mental model of the problem situation in this segment of the data.
Heuristics • Write a chain rule equation to guide the solution process.
• Compare your work to your neighbors’.
It was decided that they wanted to find t
h
∆
∆and that
t
V
∆
∆ was given. Ann and Ali
then decided that they needed to relate volume and height so that they may determine
what h
V
∆
∆ is. It may be that writing a “delta equation” is a heuristic step they have
developed to help guide their solution process.
The students then focused their attention on determining an algebraic relationship
between the volume and the height. The majority of the time spent on this problem was
spent on this step.
Excerpt 80
232
40. NE: Ok. So do you think it matters if you write it one way or the other to start?
41. Ali: No 42. Ann: No 43. NE: No, ok. 44. Ann: I need an equation for volume. 45. Ali: Do you know the equation for the volume of a truncated cone? 46. Ann: I don’t know the volume of that. 47. Ali: Would we do like the whole cone minus the part of the cone? Possibly,
for the volume? 48. NE: Why do you think you would want to do that? 49. Ali: Sssss 50. Ann: It makes sense 51. Ali: Cause I don’t know how else to do it. 52. NE: It makes sense, and you don’t know how else to do it. 53. Ali: Yeah 54. NE: Ok. So do you want to try that? 55. Ali: Yeah 56. Ann: Seems like it’s, it seems like it should work. 57. NE: It seems like it should work. Why do you say that? 58. [7:01] 59. Ann: Cause, it’s like an integral? 60. NE: It’s like an integral? 61. Ann: To me, yeah. Well, I guess, cause like taking the volume between 62. Ali: Or, we could just treat it like a trapezoid. 63. Ann: I don’t know the formula for that either. 64. Ali: Neither do I. 65. Ann: But 3, presumably if it’s half way up the radius should be 3 cm
technically. 66. NE: How can you convince me of that? 67. Ann: I can’t. That’s the thing, I don’t know. From halfway between 4 and 2
is 3, and halfway up this way 68. NE: All right. So what have you written down as your volume? 69. Ali: I have the general equation for a cone, which is four thirds pi r cubed,
I think. 70. Ann: Isn’t it one third? 71. Ali: One third? 72. NE: It’s one third. 73. Ali: Ok 74. NE: And it’s not even cubed. When you have four thirds pi r cubed that’s a
sphere. 75. Ali: Ok, never mind, [going to be easier?] what is it, one third? 76. Ann: pi r squared. 77. Ali: pi r squared.
233
78. Ann: Um hm. 79. NE: Um, that’s most of it. There’s something else you need in that formula,
pi r squared is like the circle that’s at the base, right? 80. Ann: Times the height. 81. NE: Times the height, right. You also have to account for how tall it is. 82. Ali: I don’t know how to do that. Would you do like, I’m assuming that
since the radius goes from 4 to 2 in 6 cm high, it will go from 2 to zero in another 6 cm. Is that correct in thinking?
83. NE: Ok, can you can you prove it. 84. Ali: No, I can’t. Well, I possibly could. 85. NE: How could you possibly do that? 86. Ali: Plug in the height and the radius maybe or the unknown 87. NE: Ok. So you were talking about saying it was the whole cone, and then
subtracting off the part that you chop off. 88. Ali: Um hm 89. NE: So, how would that change your diagram if you were to consider the
whole cone? Ok, now when you look at that, is there anything that comes to mind as to how you could justify that it would go from 2 to zero in another 6 cm?
90. Ali: It’s about the same. 91. NE: It’s about the same? 92. Ali: The ratios. 93. NE: The ratios? 94. Ali: If it’s 95. Ann: Could you do similar circles 96. NE: Similar circles? 97. Ann: It could work 98. NE: It could work 99. Ali: Cause if you have the 4 cm with the 6 cm height or hm 100. NE: What did we use when we were doing the maximization minimization
problems? 101. [10:45] 102. Ann: Perimeter and surface area. I don’t know. 103. NE: When we were dealing with cones there was another trick we had, so
to speak, to relate the radius and the height? 104. Ann: I don’t think I was there. 105. Ali: I don’t remember. 106. NE: Similar triangles. 107. Ali: Oh, yeah. 108. Ann: I was there. 109. NE: You were there. I thought you were cause you provided some of the
labels. 110. Ann: Oh, did I? 111. NE: Legs of the triangles.
234
112. Ann: I don’t know what I’m writing. 113. Ali: But how can I know that the whole thing would be 12 inches or 12 cm 114. NE: Well, so how much additional height do you have on the bottom? 115. Ann: Well, if you have 6 at the bot, well shouldn’t they just be equal once
you cross multiply? 116. NE: What are we cross multiplying? 117. Ann: The, wouldn’t it be the, um, the 4 and the 2, and the 6 and the 12? 118. NE: Ok, but if we don’t, that there’s an extra 6 down there. Can we call it
something else? 119. Ann: x 120. Ali: x 121. NE: x 122. Ali: So, call this whole height 6, 6 plus x 123. NE: Right 124. Ali: And would we do like the ratio like over 4 over 8? Or 125. NE: Which one 126. Ali: I think over 4 just cause that’s what I have. So x over 2 [both doing
some work] 127. Ann: This equals 6 128. Ali: Yeah 129. NE: So you’ve just proved that if you were to extend the whole cone 130. Ali: It would be 131. Ann: So, could we do the same thing with the 3 to find out what the radius is
at 3? 132. NE: Ok. So, what would you do then? 133. Ann: You, instead of 134. Ali: You’d have 135. Ann: You’d have 6 minus x, and then 6 equals 136. NE: Ok 137. Ali: Wait, what? 138. Ann: No, this would be 3, and then you’d have 139. NE: So, what is it you’re trying to find now? 140. [14:16] 141. Ann: It would be a y right there. We want this radius right here. 142. NE: So you want to find the radius? Ok. 143. Ann: So you’d use 4 then x or y [does calculation] 144. Ali: Hm [continues working] 145. Ann: [mutters] I have problems setting up the equation right. I have no idea
what I’m doing. 146. NE: Ok. What are you trying to do? 147. Ann: Figure out that radius 148. NE: Figure out that radius. Ok. So what other values do you know? 149. Ann: I know 3, 6, and 4. 3 is the height, 6 is the height of the total thing, and
4 is the radius at the top.
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150. NE: What’s 151. Ann: Or would you use the radius at the bottom? 152. Ali: You’d use like one of these two that you used, cause it used the whole
triangle. 153. Ann: Oh, ok. That didn’t work, did it? I got two. 154. NE: You got two, ok. So you take the 4, which is the radius across the top,
and compare that to the height of the whole triangle, which is how much?
155. Ann: 6 156. NE: The whole triangle is 6 157. Ann: Oh the whole thing, the whole whole thing 158. NE: The whole whole thing, right, because that 4 is the base of the whole
whole triangle. 159. Ann: Ok, um, k 160. NE: And then you know that the radius you’re interested in corresponds to
a height on a triangle of 161. Ann: 3, well 162. NE: 3 for the part that’s in the cup. 163. Ann: Oh, so it’s 9. 164. NE: But it’s right taking into account the rest of the cone 165. Ann: And that’s 3 166. NE: Right 167. Ann: So, the radius here is 3. 168. NE: Right, ok. So we have some exact values which would capture an
exact instant in time, right? However, over time what quantities are changing?
169. Ali: The radius. 170. NE: The radius. 171. Ann: And the height. 172. NE: And the height. 173. Ann: And the volume. 174. NE: And the volume. 175. Ann: And time. 176. NE: Ok. So 177. Ali: Um hm 178. NE: That radius and that height, is there a relationship between them? 179. [17:40] 180. Ann: Yes 181. NE: What is that? 182. Ann: So as the radius gets larger, so’s the height 183. NE: Ok, as the radius gets larger, so does the height 184. Ann: 185. Ali: Radius equals 3 height 186. NE: How did you get that?
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187. Ali: Because if you have the 4, radius 4 cm radius 188. Ann: Or is it one third 189. Ali: And then it’s 12, 12 inches in height. So then it’s just the, the 4 radius
equals 3 height 190. NE: Would you agree with that? 191. Ann: [nods] 192. NE: Yes? Are you sure? 193. Ann: Um hm. 194. NE: Ok. Um, so if you know that the radius is always equal to 195. Ann: Wait 196. NE: Ok, we’ll wait. 197. Ann: Well, cause at 3, the radius and the height are both 3. No, they’re not.
The height is 9. 198. NE: Right 199. Ann: So [too quiet] 200. NE: Right 201. Ali: So then we can plug the radius back into the volume equation 202. NE: Correct. Why do you want to do that? 203. Ali: So you can take 204. Ann: So we can solve for a height. 205. NE: So we can solve for height, ok. 206. Ali: You can find the volume of the, yeah, yeah. 207. Ann: The volume with respect to height. 208. Ali: So it’s, [writes] so we’re trying to find the change in height with
respect to the time, or 209. Ann: Yeah 210. Ali: Ok 211. NE: Ok
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Table 29:
Summary Table for Construct a Functional Relationship - Cup Problem
Phase
Construct a functional relationship (lines 40-211)
Solution
Artifacts Ali wrote down the volume of the cone as: 3
3
4rV π= (line 69)
Algebraic formula for the volume of the cone: hrV2
3
1π= (line 81)
Modified diagram, the whole cone. (line 89)
Relationship between the height and the radius: hr 3=
Algebraic representation for the volume of the cone in terms of height:
( ) hhV2
33
1π=
Content
Knowledge
Function
• May use subtraction of functions to create a new function that represents the volume of the cup. (line 45)
• Students prompted about the relationship between the radius and the height. (lines 178 - 193)
o Ann stated, “as radius gets larger so’s the height.” (lines 182) o Ali provided, “radius equals 3 height.” (line 185)
• Use composition to eliminate variables. (lines 197-211) o Ali stated, “so then we can plug the radius back into the volume
equation.” (line 201) o Ann added, “so we can solve for a height.” (line 204)
Geometry
• The students did not know a formula for the volume of a truncated cone. (lines 44-36)
o Ali suggested taking the volume of the whole cone minus the volume of the little cone that has been cut off. (lines 45-57)
o Ann likened it to an integral. (lines 58-61) o Ali suggested that they could treat it as a trapezoid, but neither Ann nor
Ali know a formula for that either. (lines 62-64)
• Know what the volume of a cone is. o Ali had the incorrect volume of a cone written down; she had the
volume of a sphere written down. Ann thinks it should be one third, not four thirds. The researcher asked questions that helped them recall it. (line 69-81)
• Ann stated, “I’m assuming that since the radius goes from 4 to 2 in 6 cm high it will go from 2 to zero in another 6 cm is that correct in thinking?” (line 82)
• Know what similar of figures are. (lines 92- 167) o Ali suggested ratios. (line 92) o Ann suggested similar circles. (line 95) o Ann suggested perimeter and surface area. (line 102) o The researcher suggested similar triangles. (line 106) o Students had trouble setting up the ratio. The need to create another
variable was problematic. (lines 113-127) o Students incorrectly used similar triangles to generate the relationship
between the radius and the height. (lines 185-191)
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Mental Model • The students imagined the whole cone with the tip cut off. (line 45)
• The students imagined a cross-section which they determined was a trapezoid (line 62).
• The students were prompted to modify their diagram to show the whole cone. (line 89)
• The researcher asked if there was a relationship between the radius and the height to which Ann replied, “so as the radius gets larger so’s the height.” (line 182)
Heuristics • Similar figures may be helpful. (line 95)
• Eliminate variables if you can.
• Compare your work to your neighbors’.
The students began by accessing their content knowledge of geometry to see if
they had a formula for the volume of a truncated cone. When the students realized they
did not know a formula for this, they considered other options. Ali suggested that they
could find the volume of the cup by considering the whole cone and subtracting off the
volume of the tip of the cone. She also considered viewing it as a trapezoid. Neither
student could accurately recall the volume of a cone formula and needed guidance from
the researcher.
The researcher asked them about how the whole cone minus the small cone would
work. This caused them to mention ratios and similar figures. However, there is no
evidence that they knew what that would do for them. It was the researcher who then
suggested that similar triangles might be useful and again asked them to recall their
experience with max/min problems from earlier in the semester. The students needed
guidance on how to set up an appropriate proportional relationship. Ali in particular was
not sure how she could deal with the missing height without knowing how much was
missing. This suggests that the students are uncomfortable with assigning variables in the
middle of a problem. The students discovered that the missing height was 12, and then
proceeded to determine the relationship between the radius and the height. They made an
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error in determining the relationship between the radius and the height. Ali stated that
four times the radius is equal to twelve times the height, and Ann agreed with her. This
was incorrect, but the students did not catch the error. It would appear that they were not
consistent with their application of the properties for similar triangles.
After they had constructed their functional relationship, the students revisited
their delta equation.
Excerpt 81
212. NE: So, now that we’ve got our nice formula for 213. Ann: Don’t we take the deriv, don’t we take the derivative? 214. NE: And what do you 215. Ann: And make it a reciprocal 216. Ali: Ok 217. NE: Take the derivative, and make it a reciprocal. So what do you mean? 218. Ann: Well, because we’re looking for delta h over delta v, but we have v of
h which would give us delta v over delta h 219. NE: Ok 220. Ann: So, we’d have to flip it, and that should give us um the delta h over
delta v. 221. NE: Ok 222. Ann: So, then you have 3 pi h squared times delta h over delta t. 223. NE: Ok 224. Ann: Um
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Table 30:
Summary Table for Relate the Rates Part 2 - Cup Problem
Phase
Relate the rates - Part 2 (lines 212-225)
Solution
Artifacts t
h
h
V
t
V
∆
∆⋅
∆
∆=
∆
∆(from before)
t
hh
t
V
∆
∆⋅=
∆
∆ 23π (line 223)
Content
Knowledge
Derivative
• Interpret the symbols in your delta equation. o Ann stated that she wanted to take the derivative and make it a
reciprocal “well because we’re looking for delta h over delta v but we have v of h which would give us delta v over delta h.” (lines 213, 215, 218)
Mental Model • The students appear to be focused on their equation.
Heuristics • Interpret the symbols in your delta equation.
• Compare your work to your neighbors’.
The students identified what each symbol in their delta equation represented and
replaced it by either an algebraic expression or numeric value. Ann stated, “because
we’re looking for delta h over delta v but we have v of h which would give us delta v over
delta h.” Thus, it appears that she better understands the relationship between the symbol
h
V
∆
∆which represents the rate of change of the volume with respect to height and her
function for the volume in terms of height.
Finally, the students solved for the unknown rate.
Excerpt 82
225. Ali: What I don’t get is, I have change in height with respect to time, and on the other side I have the variable is height.
226. NE: But, what do we know at this very specific instant? 227. Ali: The height is 3 228. NE: The height in the cup is three 229. Ali: Yes 230. NE: But everything we’ve been doing has been with this whole cone 231. Ali: Right
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232. NE: So, our height at this exact instant is actually? 233. Ali: 9 234. NE: 9, do you agree with that? 235. Ali: Yes 236. NE: So, we actually know what h is in this case. 237. Ali: It’s a constant 238. NE: Um hm 239. Ali: Ok 240. Ann: And we’re looking for the h at 3, or is it at 9? 241. Ali: 9 242. NE: Why is it at 9? 243. Ann: Because of what you just said, but 244. NE: What did I just say? 245. Ann: That because the cone ends down here, that that’s 6 plus 3, which is 9. 246. [22:05] 247. NE: Ok. So, is it ok to think of the cup as the whole cone? 248. Ali: Um hm 249. NE: Why? 250. Ali: Because that’s the shape that it comes from, and even if you don’t use
the lower part, then that doesn’t matter, cause it just, like it’ll still be the same rate of change no matter if you cut off the bottom, or if you just leave it
251. NE: Ok 252. Ali: Cause it’s still the same width around. So, it will still rise the same. 253. NE: Ok. So are there any units on 254. Ann: Cm cubed per second 255. Ali: I don’t think there’d be cm cubed. I think it just be cm per second.
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Table 31:
Summary Table for Find the Unknown Rate - Cup Problem
Phase
Find the unknown rate (lines 226-256)
Solution
Artifacts
Solution
Content
Knowledge
Function
• Substitute in known values for variables where possible. o What do you use for the height of the cone? (lines 226-251)
• Use algebraic manipulations to solve for the unknown rate. Derivative
• Correctly interpret the given rate.
Mental Model • The students are focused on their algebraic equations.
• Ali stated, “because that’s the shape that it comes from and even if you don’t use the lower part then that doesn’t matter cause it, just like it’ll still be the same rate of change no matter if you cut off the bottom or if you just leave it.” (line 251)
Heuristics • Compare your work to your neighbors’.
As the students substituted in their known values to compute the desired rate, they
questioned where the height should be measured from. The researcher suggested that
everything they have done to this point involved the whole cone and asked what would
be the actual height of the coffee in the cup if it were the whole cone. They decided that
the height would be nine. Ali reasoned, “because that’s the shape that it comes from and
even if you don’t use the lower part then that doesn’t matter, cause it’s just like it’ll still
be the same rate of change no matter if you cut off the bottom or if you just leave it.”
Thus, it would appear that Ali may have imagined the cone filling with coffee again and
realized that the existence of the bottom of the cone is irrelevant.
7.5 Summary
The students’ individual interviews provided insight about how they understood
concepts that are used when solving a related rates problem. In these interviews, the
students were asked to solve problems involving the concepts of variable and function.
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They were also asked to solve a problem that required covariational reasoning. From
these interviews, it was determined that the students’ principle understanding of variable
was something for which they would solve or that it was a placeholder for an unknown
value. (See Section 7.1.1) When it came to their knowledge of functions, the students
could easily identify the independent and dependent variables. They also talked about the
independent and dependent variables changing in relation to each other. This is
interesting because they did not talk about variables as changing quantities when asked
about the definition of a variable. During the individual interview, the majority of the
time was spent on a word problem that required function composition. The students were
asked to write up their solution to this problem for homework. One student, Ann, used
three different function representations in her solution while the other two students relied
on the algebraic representation. One student, Ali, had difficulty solving a quadratic
equation. Their solutions suggest that their understanding of the concept of function is
still developing. (See Section 7.1.2) The last problem in the individual interview required
the students to engage in covariational reasoning by imaging a given bottle filling with
water. Each of the students believed that they needed a formula before they could
successfully graph the height as a function of volume. They had to be prompted to think
about how the change in one variable would affect the change in the other. This suggests
that the students do not think about using a mental model and covariational reasoning to
solve math problems. (See Section 7.1.3)
In the first teaching episode, the students interacted with a custom computer
program which modeled related rates problem situations. (See Section 7.2) The students
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were able to physically manipulate the problem situation and to observe the results of
those manipulations. This led the students to generate time as a variable that would allow
them to determine the average rate of change for each variable.
The students looked at problems requiring the use of the chain rule in the second
teaching episode. (See Section 7.3) In this session, the students were presented with word
problems in which one could generate an overall rate of change for the problem situation
from the given rate of change of other functions. This illuminated the connection between
the chain rule and function composition. At the end of this session, the students were
identifying what they later called “a middle man,” a variable through which they could
relate two other variables.
In the third through sixth teaching episodes, the students solved related rates
problems. (See Section 7.4) The problems presented to the students in the third and fourth
teaching episodes had most of the numeric data stripped from them in attempt to focus
the students’ attention on the relationships that existed between the variables. In the fifth
and sixth teaching episodes, the problems were stated in a traditional textbook manner.
The students appeared to engage in phases of problem solving that were similar to
those used by the mathematicians. The students appeared to create a mental image of the
problem situation as a result of interpreting the words in the problem statement. This
resulted in their drawing a diagram which was then labeled with constants and variables.
However, as the students labeled their diagrams, they appeared to be merely assigning
letters to unknown quantities rather than assigning variables to relevant changing
quantities. This was evident in the trough problem when the students decided to label the
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distance between the top of the trough and the top of the water with an x. Thus, it appears
that the students did not build a mental model to help them understand the nature of the
relevant quantities in the problem situation. After the diagram had been constructed, the
students would identify an appropriate algebraic relationship to relate the variables in
their diagram. The next step the students employed was to write a “delta equation” as
Ann called it. This delta equation then served as a guide for the rest of the solution
process. It allowed the students to identify the known and unknown rates, in addition to
the form they would like to have the functional relationship. The students then eliminated
variables if possible and substituted in either numeric values or algebraic expressions into
their delta equations. Thus, the primary difference between the students’ solutions and the
mathematicians’ solutions was that the students would relate the rates using the chain rule
before generating the functional relationship.
Recall that Adam did something similar when solving the trough problem and the
coffee cup problem. After generating his relationship, he wrote out an equation using the
chain rule and then determined what each piece represented. This is interesting because
the most time had elapsed since Adam had solved a related rates problem. Thus, it would
seem that Adam generated his solutions from first principles of calculus.
Chapter 8: Discussion and Conclusions
This chapter discusses the framework in conjunction with the results of the
teaching experiment. Furthermore, the conclusions of the study are presented. The
limitations of the study are then addressed. It concludes with directions for future
research.
8.1 Discussion of the Framework
The framework presented in Chapter 6 characterizes the approaches and ways of
thinking that emerged from the research literature and observing mathematicians when
solving related rates problems. It describes the mental activities involved in developing a
mental model of the problem situation for specific related rates problems. Its
characterization of the reasoning patterns, heuristics, and knowledge for solving a related
rates problem should be useful for both curriculum developers and researchers. Findings
from the data analysis of mathematicians and students solving related rates problems
have informed the framework refinement, and this new version of the framework will
guide the next iteration of the curriculum design.
When confronted with a related rates problem, a problem solver first reads the
problem and notices key features in the problem statement (e.g., a rate is given) to
categorize the problem as being a “related rates” type problem. After the problem solver
has identified the type of problem as “related rates,” he typically begins the solution
process by drawing a diagram. This is the first phase of the problem solving process. As
each subsequent phase of the solution process is carried out, this diagram and additional
solution artifacts such as functions to represent geometric relationships are carried
forward as an additional resource for the problem solver. As one example, the diagram
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becomes a solution artifact that is used for constructing a mental model of the problem
situation. The solver typically then uses his diagram, and various content knowledge such
as understanding of a specific geometric relationship, to construct a meaningful
functional relationship between the variables and constants labeled in the diagram. The
framework also describes the role of covariational reasoning and other content
knowledge, when generating diagrams, formulae, and other solution artifacts while
moving through the five phases of solving a related rates problem.
8.1.1 The Role of the Mental Model
An individual’s mental model of the problem situation is reflected in the
diagram(s) they draw as part of their solution. This diagram is usually constructed and
modified as the problem solver interprets the problem situation and accesses relevant
content knowledge. As a diagram is modified, it may become more meaningful to its
creator and more useful as a tool to continue guiding the problem solving process.
The mathematicians made significant use of their diagrams. Recall Dan’s solution
to the trough problem. He frequently used his mental model and diagram to make
decisions about his next step in the problem solving process. He talked about how he
imagined the trough filling with water and how that indicated to him that he needed to
find a functional relationship between the variables of height and volume. (See Excerpts
12 and 13) The mathematicians referred to their diagrams to identify which quantities
were changing versus those that remained constant. They also referred to their diagram to
note the varying quantities that were related prior to expressing the geometric relationship
algebraically. They studied the diagram and appeared to use the diagram as a means of
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staying focused on the relevant relationships while completing the problem. Finally, after
completing the problem, the mathematicians often referred back to the diagram to
confirm the reasonableness of their solutions.
In contrast, although the students were able to draw accurate diagrams of the
problem situations, they did not refer to them frequently in their problem solving process.
They labeled the diagram with constants and variables. However, analysis of students’
descriptions and utterances suggest that they did not think about these variables as a
representation of the varying quantities in the problem situation; rather they thought
about the variables as placeholders for an unknown value. When prompted, the students
could answer questions about the changing quantities in the problem situation, but their
responses to these prompts appeared to be disconnected from their solution process. The
students did not spontaneously reference their diagram once it was drawn and labeled.
That is, they did not appear to use their diagrams to advance their solution process.
References to their diagrams occurred only when prompted. (See Tables 18 and 19) This
suggests that the students in this study had created a static image of the problem situation
and had not built a mental model of the problem components and how they varied when
“running” the system. They also did not appear to spontaneously reference their diagrams
as a means of advancing their solution process; nor did they use them to check the
reasonableness of specific steps when responding to the problem.
This stark contrast in how the mathematicians and the students differed in their
use of the diagrams they constructed illustrates the primary difference between artifacts
and solution artifacts. While many things are written down during the solution process,
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only particular written pieces of information become additional resources (solution
artifacts) for the problem solver. For the mathematicians, the diagrams they drew,
modified, and referenced not only reflected their mental model of the problem situation,
but they also allowed them to identify other important aspects of the problem situation.
Recall Dan’s solution to the trough problem in Chapter 6. Dan’s diagram facilitated his
reasoning about relationships between the height and width of the water in the trough
which allowed him to create a relationship between the height and width of the water.
This in turn allowed him to eliminate a variable in his relationship between the volume of
the water and the height and width of the water. In contrast, when the students solved the
trough problem, they did not use their diagram as readily to determine such a relationship
unless prompted by the researcher. However, they were able to use the relationship to
eliminate a variable. This suggests that the students were not as adept in using what they
wrote down to advance their solution process.
8.1.2 The Role of Content Knowledge
The mathematicians accessed their understanding of variable, geometry, function,
and derivative throughout the solution process. To illustrate this point, recall that the
mathematicians were observed noticing geometric relationships in their diagrams that
may be helpful for advancing their solution process. (See Dan’s solution to the trough
problem, Excerpt 12, Line 3) The ability to notice such relationships was influenced by
his understanding of how specific geometric entities are related, e.g. two triangles are
similar when the measures of their corresponding are the same which was easily seen
when an appropriate cross-section of the trough was drawn. Their deep and connected
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understandings of fundamental concepts allowed them to anticipate how to construct a
viable solution during each phase of their solution process. For example, when solving
the trough problem, Dan recognized that as the volume changes the height changes. This
allowed him to discover an error before the completion of the problem. (See Excerpt 14)
Similarly, Bob found his error in the trough problem by comparing his answer to the
average rate of change. (See Excerpt 24)
For the students, much of their content knowledge appeared to be
compartmentalized. Recall the students’ solution to the trough problem. (See Section
7.4.2) The students needed to determine the formula for the volume of the trough, a
triangular prism. This was usually accomplished by looking at the back of the book rather
than thinking about volume as base area multiplied by height. Furthermore, the students
focused on their algebraic representation for the volume of the trough without
considering what each variable meant in terms of their mental image or diagram. The
students did not readily observe that by using similar triangles, they would be able to
eliminate the extra variable in their formula. This suggests that their geometry knowledge
is not closely connected to their function knowledge. If they made an error, it had to be
pointed out by the researcher. For example, when the students were solving the coffee
cup problem, they incorrectly used similar triangles to relate the radius and the height.
(See Table 29)
Many of the heuristics employed in solving the problems appeared to also be
related to content knowledge. For example, whenever there was a triangle in a problem,
Ali frequently suggested “using angles.” This suggests the presence of a general heuristic
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that relates triangles and angles. It was never established how she thought that finding
angles would be helpful or exactly what her heuristic was. This could also suggest that
she merely associates triangles with finding angles.
8.1.3 The Role of Time
In most related rates problems, time is only stated as part of the given rate.
However, each variable in the problem situation is actually a function of time. Yet, time
is not labeled on the diagram nor is it expressed explicitly in the functional relationships
that are written. To relate the rates, the derivative of the functional relationship is
computed with respect to time. This raises the question, “How does the problem solver
deal with time in these problems?”
For the mathematicians, interpreting the units of the given rate and restating the
given rate using the terminology “with respect to time” appears to convey that they
should be thinking about each variable as a function of time. (See Excerpts 12, 15, 19,
22) This allowed them to successfully apply the chain rule and differentiate their
functions with respect to time.
The students in this study did not interpret the given rate in this manner. While the
computer program did make time more explicit in the problem, the role of time in the
problems was slow to develop. In the plane problem, the students followed Ali’s lead and
explicitly expressed each variable as a function of time. (See Excerpt 69) While this
approach worked for this problem, most related rates problems are not as easily solved
using this approach; they require many more calculations which frequently results in
more errors. The researcher prompted them to consider using only the function names,
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such as ( )tx , in the solution to the problem. (See Excerpt 72) This highlighted their still
developing understanding of the chain rule. Each student was proficient at calculating
derivatives; they had all passed the mastery exam. However, it was not obvious to the
students that ( )[ ]2tx represented a composition of functions. When they first took the
derivative they wrote down ( )[ ]tx2 and neglected to take the derivative of the inside
function. Thus, it would appear that much more needs to be done to help the students
understand the chain rule.
When the students were working on the trough problem, Ben noted that time had
not been brought into the problem during the construct a functional relationship phase.
(See Excerpt 74) It seemed that Ben was trying to use the same approach they had used
for the plane problem to the trough problem. He wanted to write out each variable as an
explicit function of time. The researcher did not encourage the students to pursue this
solution approach. While this route would have worked, the calculations are extensive
and provide more opportunity for error. Is it possible that encouraging the students to
solve related rates problems using this method would allow them to more easily transition
to more traditional means later? It would certainly capitalize on the “distance equals rate
times time” idea, that students learn at a very early age. This is a question for future
research.
8.1.4 The Phases
While a textbook may provide a list of about seven steps to follow when solving a
related rates problem, most mathematicians and students appear to condense this to about
four or five steps which may be described as phases. These phases include: draw and
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label a diagram, construct a functional relationship, differentiate the functional
relationship with respect to time, solve for the unknown rate, and check the answer for
reasonability.
The first phase of draw and label a diagram is the same first step Martin (2000)
identified (See Section 2.5). Draw a diagram was the second step listed in Stewart’s
(1991) calculus text book, after read the problem carefully (See Chapter 1). Stewart’s
third step, introduce notation by assigning symbols to all quantities that are functions of
time, is also partly captured in this step. Is it necessary to be thinking about the changing
quantities as functions of time at this point in the problem solving process? The
mathematicians appeared to think about the quantities as changing when they assigned
them variables, but specifically thinking about them as functions of time usually occurred
later in the problem solving process (usually in the differentiation step). The students’
labeling process was directed by the quantities that were known and the quantities that
were unknown and likely not related to thinking about which quantities were changing.
The students’ labeling is similar to Martin’s second step, identify given and requested
information.
The second phase of construct a functional relationship has aspects of Martin’s
third and seventh step (identify the relevant geometric formula and solve an auxiliary
problem if necessary) and Stewart’s fifth step (write an equation that relates the various
quantities in the problem and use the geometry of the situation to eliminate variables if
necessary). The mathematicians identified the given rate and the unknown rate to
determine that they needed “one variable as a function of another variable.” Thus, they
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used their content knowledge of geometry to identify appropriate relationships and
eliminate variables if necessary by applying function composition. In contrast, the
students were able to access their content knowledge of geometry to write down an
appropriate geometric relationship, but they experienced difficulty when deciding how to
proceed from there. They had to be prompted to realize that they needed to eliminate
variables and that similar triangles would provide the necessary relationships. (See
Section 7.4.2) The reason for this appeared to be because they did not engage in mental
activities that led to their constructing a mental model of the problem situation; rather
they were merely carrying out steps that they hoped would lead to a solution.
The third phase of relate the rates captures Martin’s fourth step (implicitly
differentiate with respect to time) and Stewart’s fourth and sixth steps (express the given
and unknown rates in terms of derivatives and use the chain rule to differentiate with
respect to time, respectively). For the mathematicians, it was because they were able to
think about the variables in the functional relationship as functions of time that allowed
them to successfully apply the chain rule or implicit differentiation. The mathematicians
used the phrase “with respect to” to restate both the known and unknown rates in the
problem. It seemed that this was an extremely powerful tool for the mathematicians as
the phrase appeared to help them identify the independent and dependent variables in the
problem. Thus, they were able to compute the rate of change of each variable “with
respect to time” using the chain rule. For the students, it was the development of the
“delta equation” that appeared to help them successfully complete this phase.
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The process of “write a delta equation,” which occurred just before or just after
identifying a geometric relationship for the problem situation, appeared to develop from
the discussion of the chain rule in the second teaching episode. The “delta equation” was
a statement of the chain rule as it applied to the problem at hand. Recall the students’
solution to the plane problem. After they had expressed z as a function of x, Ann
attempted to use a delta equation to relate the rates when solving the plane problem (See
Excerpt 70, Table 20). She wrote: t
x
x
z
t
z
∆
∆⋅
∆
∆=
∆
∆. However, she was uncertain about how
to go about determiningx
z
∆
∆. After further discussion about the chain rule, this issue was
resolved. In solving the trough problem, the students immediately wrote down
t
h
h
V
t
V
∆
∆⋅
∆
∆=
∆
∆ after writing down the volume of the trough. This allowed the students to
identify that they needed to eliminate the variable w so that they would be left with
volume in terms of height. Later, when the students were asked to describe how they
would tell a friend how to solve related rates problems, they responded with the phrase,
“draw a delta equation.” It appears that the students have internalized that the chain rule
allows them to relate two variables through what they called “a middle man” variable.
Perhaps it is a strictly procedural part of the process as a result of the chain rule
discussions. However, it would appear that this understanding is related to their
understanding of function composition, although they never use the term composition.
This is interesting because the delta equation helped the students identify what was
known and unknown and consequently directed their solution process. It also reflected
thinking that was similar to the problem solving process observed in Adam, one of the
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mathematicians. (See Excerpts 19, 38). Could the development of this step be natural and
foster a better understanding of related rates problems? More research is needed to
determine if promoting the use of this step leads to improvements in students’ ability to
solve related rate problems.
The fourth phase of solve for the unknown rate captures Martin’s fifth step
(substitute in values and solve) and Stewart’s seventh step (substitute in values and
solve). This step did not cause any difficulty for the mathematicians or students aside
from a few algebraic errors.
The final phase, check the answer for reasonability, may be considered part of
Martin’s sixth step, interpret and report results. Stewart did not direct the problem solver
to check his solution. However, this step was a natural part of the mathematicians’
solution process. They also used their diagrams and other solution artifacts to check the
reasonableness of their results throughout the solution process. Performing checks after
each step frequently allowed the mathematicians to catch errors before moving forward
with their solution. However, this step was not automatic for the students. They had to be
prompted to check their work, and even then, they did not catch many of their errors.
8.1.5 The Revised Framework
The framework used to analyze the teaching experiment data relies on the
generation of solution artifacts. There are several mental activities that may occur to
produce a solution artifact. Three primary sources of mental activities were identified:
mental models of the problem situation, mathematical content knowledge, and heuristics.
Originally, heuristics were listed separate from the mental model and content knowledge.
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However, it became evident that the majority of the heuristics employed by both the
mathematicians and the students were related to their content knowledge. For example,
the mathematicians restated the rates in the problem using the language “with respect to.”
This appears to be a heuristic strategy that allowed them to infer information about the
relationship between the independent and dependent variables in the problem situation. It
facilitated their thinking about the variables as functions of time. Thus, the revised
framework includes heuristics as a part of content knowledge as seen in Table 32, and the
relationship between the categories now appears to be given by Figure 11 which follows
Table 32.
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Table 32: A Revised Framework for the Solution Process for Related Rates Problems
Phase Draw a Diagram
Solution Artifacts
• Diagram that accurately represents the problem situation
• Diagram that has been labeled with constants and variables
• Other diagrams representing different perspectives of the problem statement
Mental Model
• Describe what one is imagining or picturing in one’s mind
• Anticipate relationships that may exist
• Attend to the nature of the changing quantities o Attend to the direction of the change in the variables o Attend to the amount of change in the variables o Attend to the average rate of change in the variables
• Attend to continuous changes in the variables
Content Knowledge and Related Heuristics
• Restate the problem or parts of the problem in one’s own words
• Geometry o Ask or consider, “What is a ______?” o Accurately interpret terminology o Ask or consider, “Which perspective of the geometric shape will
provide the most information?” o Ask or consider, “Do I need to draw more than one perspective of
the problem situation?” o One diagram may represent any of the possible states of the problem
situation
• Variable
• Label constants and variables appropriately
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Phase Construct Meaningful Functional Relationships
Solution Artifacts
• Algebraic equation (s) to relate the variables in the diagram
Mental Model
• Imagine the problem situation changing
• Identify useful relationships between variables
• Modify the mental model to determine which variables need to be related
Content Knowledge and Related Heuristics
• Understanding the nature of functional relationships o Relate the variables representing the known rate and the unknown
rate o Eliminate variables if possible o Use a diagram labeled with variables and constants to identify
relationships o Understand the role of the independent variable and the dependent
variable in a functional relationship o Understand what relationship between the independent and
dependent variables is determined by the phrase “in terms of” o Understand that function composition (or substitution) allows one to
construct a new function from two or more smaller functions, eliminating one or more variables
Phase Relate the Rates
Solution Artifacts
• Differentiated algebraic equation
• Chain rule equation Mental Model
• Notice which quantities are changing in relation to each other
Content Knowledge and Related Heuristics
• Understanding the nature of rate of change o Understand what relationship between the independent and
dependent variables is determined by the phrase “with respect to” o Interpret from the given rate that time is the independent variable in
the functional relationship o Imagine each variable in the functional relationship as a function of
time o Differentiate the functional relationship “with respect to” time
• Perform differentiation operations on an implicitly defined function
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Phase Solve for the Unknown Rate
Solution Artifacts
• Algebraic manipulations of the differentiated equation
Mental Model
• Did not appear to use
Content Knowledge and Related Heuristics
• Algebraic Knowledge o Substitute in known values for variables o Apply algebraic operations to the equation to calculate the unknown
rate
Phase Check the Answer for Reasonability
Solution Artifacts
• Notation of units
• Other calculations Metal Model
• Ask “Is this answer reasonable?”
• Manipulate the mental model o Attend to the amount and direction of change in the variables o Compare the answer to another known quantity such as the average
rate of change
Content Knowledge and Related Heuristics
• Measurement units
• Perform a unit analysis, i.e. check that the units work out or match up
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Figure 11: The revised solution process for related rates problems.
Problem Statement
Solution Artifact
Mental Model
PS
Content Knowledge
Previous Solution Artifacts
Heuristics
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8.1.6 Generalization of the Framework
This framework could be developed for other categories of math problems. The
first step would be to identify the phases of the problem solving process for a category of
mathematical problems and the types of solution artifacts those phases might generate.
One could also hypothesize about the types of mental models and content knowledge that
may be useful. It would then be necessary to observe experts and students engaged in
solving typical problems to identify the mental models they construct and the content
knowledge they access. This would generate a framework that looks something like the
following:
Table 33: General Framework Template
Phase Solution Artifacts Mental Model Content Knowledge and
Related Heuristics
Phase A • Artifact 1 • Mental model 1 • Concept 1
Phase B • Artifact 2 • Concept 2
Phase C • Artifact 3 • Mental model 2 • Concept 3
For example, consider applied maximization and minimization problems from
calculus. The following is a typical problem from a calculus text:
An open box is to be made from a 16-in. by 30-in. piece of cardboard by cutting
out squares of equal size from the four corners and bending up the sides. What
size should the squares be to obtain a box with largest possible volume? (Anton,
1995, p. 217)
After examining my own solution to this problem, I conjecture that the phases for
this type of problem may be: draw a diagram and label it, construct a functional
relationship, differentiate the functional relationship, determine the extreme values of the
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function, answer the question, and check the answer for reasonability. The first two
phases are very closely related to those used for related rates problems and are typical of
many math problems. When constructing the functional relationship in this case, it is
necessary for the relationship to be between two variables and function composition is
used to eliminate extra variables. Determining extreme values requires the problem solver
to be able to not only apply differentiation rules, but to be able to find critical numbers
and interpret them. Answering the question, involves interpreting the information in the
previous step accurately. In the example above, the question asks for the size of the
squares. It also could have asked for the maximum volume of the box. Checking the
answer for reasonability is important as extreme values frequently exist outside the
interval of possible values. This means that when constructing the functional relationship,
when determining the extreme values, or when interpreting the extreme values to answer
the question the problem solver must take the domain of the function into consideration.
Thus, applying the framework to extreme value problems may produce something similar
to Table 34. Of course, this would need to be further investigated to fill in the rest of the
boxes.
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Table 34: Possible Framework for Extreme Value Problems
Phase Solution Artifacts Mental Model Content Knowledge
and Related Heuristics
Draw a diagram
• Labeled diagram • Imagine the problem situation
Geometry
Construct a functional relationship
• Appropriate functional relationship
• ( )( )xxxxV 230216)( −−=
• ( ) 32 492480 xxxxV +−=
• Imagine the problem situation changing
Function
• Composition
Differentiate the functional relationship
• Correctly differentiated formula
• 212184480 xx
dx
dV+−=
Derivative
• Derivative rules
Determine the extreme values of the function
• Critical numbers
• 3
10=x , 12=x
Derivative
• Critical numbers
• First derivative test
Answer the question
• Indicate that 12=x is outside
the domain of the function, so eliminate this possibility.
• Check3
10=x and endpoints
of the interval of possible values.
• Identify that 3
10=x is the
maximum.
Function
• Meaning of the input/output values of the functional relationship
Check the answer for reasonability
• Ask, “Is this answer reasonable?”
8.2 Conclusions
The primary purpose of this study was to gain insight about how a problem solver
might understand and solve related rates problems. A framework was developed that
characterizes how a problem solver approaches and solves a related rates problem.
Unique to this framework is a description of how solution artifacts at one phase of the
problem solving process become a resource that leads to a more powerful model of the
situation.
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8.2.1 General Conclusions
The mathematicians had a richly connected understanding of the concepts of
geometry, variable, function, and derivative. This allowed them to easily access
information and relate it to the problem situation. This abundant source of content
knowledge also facilitated their ability to build a mental model that accurately
characterized the problem situation. For example, the mathematicians could easily
identify the formula for the volume of a prism and then think about it as a function of two
variables that change in tandem. The mathematician could then note that one of the
variables could be eliminated by assessing relevant knowledge of geometry and applying
function composition. This new function that was created could then be thought about as
a function of time. They identified time as the independent variable stated in the given
rate, thus made a connection between their function knowledge and understanding of
derivative/rate of change.
For the students in this study, their conceptual knowledge appears to be
compartmentalized. They could recall geometric formulas, but they could not relate them
to their function knowledge to determine that they needed, say, volume as a function of
height. This suggests that their knowledge of geometry is segmented from their
knowledge of functions. This calls for an increased effort on the part of educators to
ensure that students build connections between concepts. Students should also be
encouraged to construct mental models rather just mental images.
One of the most interesting results was the emergence of the “write a delta
equation” as a step in the students’ solution process. Knowing the chain rule and using it
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to construct a “delta equation” may be a natural way to understand what is happening in a
related rates problem. This step appeared to help the students identify the known rate, the
unknown rate, and the functional relationship between the variables. Indeed, even Adam,
the mathematician who had not taught calculus for the longest time, used a “delta
equation” to relate the rates. Adam used basic principles of calculus to solve these
problems. He knew the chain rule and wrote out an equation that related the rates he had.
Then, he identified either a numeric value or a function for each rate and made the
appropriate substitutions. (Excerpts 19, 38) Clearly, the delta equation approach appears
to be a useful tool for the related rates problem solver.
Another particularly interesting finding of the study was the role of time as a
variable. While the mathematicians could image each variable as a function of time and
operate on functional relationships without expressing them as explicit functions of time,
the students could not. For example, the mathematicians expressed the functional
relationship in the plane problem with 222cba =+ , but they were actually thinking
about it as ( )[ ] ( )[ ] ( )[ ]222tctbta =+ . In contrast, the students needed to see time explicitly
represented as a variable. The computer program appeared to foster the students’
recognition of time as a variable in the problem. However, in the plane problem, they
wanted to express each variable as an explicit function of time.
8.2.2 Answering the Research Questions
One goal of this study was to be able to describe the necessary mental
constructions and knowledge required to solve a related rates problem. It was the
expectation that these descriptions would provide more depth to the step approaches
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described in the literature review in Chapter 2. This goal led to the posing of four
research questions. Each of the research questions posed in Chapter 1 will now be
addressed in turn.
Question 1. What is involved in solving a related rates problem? What problem
solving processes are necessary? What cognitive constructions are necessary for the
student to have an effective mental model? What formal mathematical knowledge is
necessary?
The framework that emerged from the mathematicians’ solutions to related rates
problems addresses these questions. Correctly interpreting the problem as stated begins a
process of using problem solving strategies together with content knowledge and
heuristics related to the concepts of geometry, function, and derivative. Interpreting the
problem statement should lead the problem solver to access content knowledge of
geometry to begin constructing a mental model of the problem situation. Initially, the
mental model appeared to be primarily static, but as the solution process continued the
model was refined by covarying the quantities in the situation. As the mental model was
revised in each phase, it began to incorporate information about the nature of the
problem’s variables and how they vary in tandem. This was revealed in how the
mathematicians made sense of the phenomena and advanced their understanding of the
relationships in the situation by creating and refining solution artifacts. (See Section 6.1)
The mathematicians also readily identified that they needed to relate the variables
through time.
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The students drew diagrams which they quickly labeled with variables, too.
However, as they continued in their solution processes, they did not make reference to
their diagrams unless specifically directed to do so by the researcher. The students did not
appear to imagine the problem situation playing out nor did they appear to attend to the
nature of the relationship between the changing variables in the problem situation. This
suggests that students’ mental image was static. As the students moved through the
problem solving phases, they did not appear to modify or use their mental image in the
same manner that the mathematicians did. It appeared that their mental image was
primarily constructed from their interpretation of key words in the problem statement at
the beginning of the problem solving process. They did not engage in mental activities
that led to their development of a dynamic model of the situation. This suggests that more
needs to be done to develop students’ ability to identify and vary relevant relationships in
their mind; thus leading to their creation of dynamic mental models that can be used
powerfully in solving related rate problems.
Question 2. How do students’ function conception and covariational reasoning
abilities influence their understanding of and ability to complete related rates problems?
The students in this study appeared to have a primarily static view of function.
While the students could easily identify the independent and dependent variables of a
given function, they appeared to have difficulty identifying them in the context of related
rates problems. At no point in solving the related rates problems did the students talk
about input and output or that they needed to have one variable as a function of another
variable; nor did they speak about how the two varying quantities change together. This
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difficulty may contribute to their weak understanding of function composition. Recall
that when presented with a function composition problem in the individual interview,
each of the students had some difficulty. (See Section 7.1.2) Their inability to understand
function composition as a means of linking two function processes together prevented
them from recognizing how to use function composition as a way to eliminate an extra
variable that is frequently introduced when defining a geometric relationship in the
related rates context. (See Sections 7.4.2 and 7.4.3)
The fact that the students did not readily discuss how the varying quantities
changed together suggests that they were not engaged in covariational reasoning. When
solving the covariational reasoning problem in the individual interview, each student
initially thought that he must know the formula for the volume of the bottle in order to be
able to sketch the graph of the height of the water as a function of volume. It would seem
that if they knew the formula, then they would be able to plug it into their calculator to
obtain the graph. However, without knowing the formula, the students struggled to
produce an accurate graph of the problem situation. (See Section 7.1.3) The graphs the
students sketched were not always correct. Recall that Ann initially thought that the graph
would be linear because the bottle was being filled at a constant rate. Perhaps Ann was
relying on the key word “constant” to indicate the shape of the graph or a heuristic that
relates the word constant with a linear graph. It was only after the researcher asked her to
consider what would happen if incremental amounts of water were poured into the bottle
that she explained: “Well cause at the bottom, at the middle it’s going to need more liquid
to make it rise in the middle because there’s a larger area to fill, and at the bottom it’s
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smaller to fill so it goes faster, then slower, then faster.”(See Excerpt 58 in Section 7.1.3)
This illustrates that students did not spontaneously use covariational reasoning in a
powerful way so as to help them build a mental model of the problem situation.
The computer program that was used in the teaching experiment was designed to
assist the students in manipulating and observing the dynamic nature of the variables. It
would appear this had a positive impact on the students’ covariational reasoning abilities,
although the intervention was not long enough. In later teaching episodes, the students
improved covariational reasoning was revealed in their ability to more readily answer
questions about how the variables in the problem changed in relation to each other. When
asked what was changing in the trough problem, Ann elaborated: “Yeah, the side of the
trough. Like whenever you fill it up, if you go only halfway there it’s to here. So this
length is going to be changing. It’s kind of like the plane problem where as, instead of,
well instead of just having a constant height, this has a changing height, which makes,
which changes the height of the, the top, which changes the side.” (See Section 7.4.2) It
would appear that the students are using an MA3 level of covariational reasoning rather
than the MA2 that was more evident in the initial interview. However, I did not observe
any of them exhibiting the flexibility in moving between MA3 and MA5 level
covariational reasoning that the mathematicians did. Recall that the mathematicians’
covariational reasoning abilities allowed them to recognize how they should deal with the
coffee cup problem. When the students solved the same problem, they needed prompting
to reach the same conclusions. The computer program was used to revisit the concept of
rate of change late in the semester. Given the progress that these students made in their
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covariational reasoning abilities and understanding of rate of change, it would appear that
it may be beneficial to use the computer program to introduce and develop their
covariational reasoning abilities and concept of rate of change throughout the semester.
This was beyond the scope of this dissertation, but it provides a starting point for further
investigation.
Question 3. How do students’ understandings of the chain rule influence their
ability to solve related rates problems?
At the start of the study, the students knew how to apply the chain rule in a
procedural manner to compute derivatives. However, it would appear that after revisiting
the chain rule in the second teaching episode, the students were beginning to understand
the multiplicative nature of the chain rule. (See Section 7.3) This understanding of the
chain rule led them to create the “write a delta equation” step for solving related rates
problems. The delta equation appeared to be a way for them to apply the chain rule to
write an equation relating the given and unknown rate. After writing down their delta
equation, the students then figured out what each piece represented; the given rate, the
unknown rate, and the derivative of the functional relationship. (See Section 7.4) Note
that one mathematician, Adam, used this same strategy to solve related rates problems.
(See Section 6.1)
Question 4. Does the use of a computer generated dynamic animation of related
rates problem situations impact students’ ability to solve and understand related rates
problems? If so, how?
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The computer program allowed the students to observe how the quantities in a
related rates problem situation changed in relation to each other. It also allowed them to
observe time as a variable in the problem even though time was only stated as part of a
given rate. Being able to observe the dynamic nature of the problem situation appeared to
foster the students’ ability to think about variables changing over time and be able to see
that time was an integral part of the problem. Initially, the students wanted to express
each variable in the problem as an explicit function of time. When solving the trough
problem, Ben observed: “I think you put time into an equation, because you can’t just
make things up.” He wanted to write out each variable as an explicit function of time as
they had done when they first solved the plane problem. (See Sections 7.4.1 and 7.4.2)
The researcher did not encourage the students to pursue this line of thought, but it
generates another question: is this approach more natural to students? This, together with
the discussion of the chain rule in the second teaching episode, led to their development
of the “write a delta equation” step. It would seem that being able to use the computer
program (with some modifications) during more of the differential calculus instruction
would be beneficial to students. This question needs to be further investigated in the next
phase of the study.
8.3 Limitations of the Study
The sample size for this study was small. It was conducted with only three
mathematicians and three students. While working with a small number of subjects
allowed the researcher to closely examine each individual’s utterances and solution
artifacts, the data set is restricted to that of the mathematicians and students who
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participated in this study. Thus, the problem solving behaviors observed here are
representative of this population and may not apply to all mathematicians and students.
It is not possible to know exactly what someone is thinking or imagining when
solving a problem. At best, one can interpret verbal and written statements made by the
problem solver in an attempt to understand what he meant. In fact, a problem solver may
be thinking one thing and say something different. In this study, I had to believe that the
utterances an individual made were reflective of what he was actually thinking. It is also
possible to misinterpret what someone has said or written. Since the results presented in
this dissertation are only one researcher’s interpretation of what the problems solvers said
and did, it is possible that another researcher may have an alternate interpretation of the
subjects’ artifacts.
The teaching experiment took place over the course of six days outside the regular
classroom. This was a very short time frame. It required the students to attend two math
classes everyday. As a result, the students may not have enough time to reflect on what
they had learned in each session.
8.4 Directions for Future Research
This study provided a framework through which to analyze students’
understandings of mathematical problems. The framework was used to analyze how the
students in this study solved related rates problems. Refinement of the classroom
materials has begun, and more research needs to be conducted to address new questions
that arose in the course of this study.
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8.4.1 Refinement of the Classroom Materials
The classroom materials that were developed for this study were used with a
group of three students over the course of six days. A logical next step is to begin using
them with a larger class. To do this, the students must have computer time available to
them at school. The program should be accessible in a web friendly format so that it may
reside on the instructor’s webpage for easy access. By having the program on a webpage,
it will also allow the students to access it outside of class time. Having larger numbers of
students interact with new software also creates many challenges that will need to be
worked out. There are plans to expand the number of problems in the program.
It would be beneficial to use the first two activities where they fall naturally in the
curriculum (when the students begin studying rate of change and when they learn the
chain rule). There is potential to expand these activities for use throughout the semester.
8.4.2 The Role of Affect
Affect is another column in Carlson and Bloom’s (2005) multidimensional
problem solving framework. The data suggests that affect may be strongly correlated
with the level of success in students’ solutions. Ann appeared to be fairly strong
mathematically as evidenced by her solution write-ups which suggested that her
conceptual understandings were advanced compared to her peers. However, her dialogue
is frequented with statements such as “I’m not good with numbers” and “I’m not very
good at math.” These may become self-fulfilling prophesies as she makes frequent
algebraic errors and relies heavily on her calculator.
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8.4.3 Use of the Word “Like”
Discourse analysis was also used to examine how students use the word “like” in their
mathematical discussions. This was only an exploratory study with some of the pilot
study data. There were many ways in which students use the word like in mathematical
discussions. It would appear that like serves as: a hedge, a means for taking the floor, a
means to indicate elaboration and clarification through example, an equal to, and an
indicator of motion. These uses are beyond the customary simile or expression of
enjoyment. Two large categories emerged from these data: 1) for example and 2)
indication of gesture. The dimensions of “for example” include taking the floor, hedging,
and the ‘like x but y’ construction. The dimensions of “indication of gesture” include
strictly hand motions or a combination of hand motion and symbols on the paper.
There were instances where like was used in what would appear to be an
idiosyncratic manner; the number was so small that more data would be necessary to
determine if they are used by others or just outliers in the data. I described the use of like
in place of equal to in the analysis as I think this may be more common than some of the
others. Like may have been replaceable by “as if” in a couple of instances and was also
used as what seemed to be “approximately” in a few instances. The approximation
interpretation may be unique to this class as it was a foundational idea throughout the
semester. There was one instance where like appeared in what seemed to be a conditional
statement; Laura says: “if you just flipped it, like you just looked at it…” Certainly, much
more analysis needs to be conducted to determine how the initial codes are related and to
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completely establish all possible uses of the word like. This was not the focus of this
study, but it is a possible direction for future research.
8.4.4 Dealing with Time as a Variable
One of the most interesting questions to arise from this study was, “How does one
deal with the variable of time in related rates problems?” The mathematicians’ experience
allowed them to think of each variable as a function of time and capitalize on this
understanding to their advantage. The students’ knowledge of variable and function was
still developing. Interacting with the computer program allowed the students to recognize
time as an integral part of the problem, but the power of thinking about each variable as a
function of time was not readily available to them. It was natural for them to want to
explicitly write out each function in terms of time. The idea of rate multiplied by time
may be a powerful tool available to students, but the researcher did not recognize the
potential this approach may provide. Thus, it would be advantageous to pursue this
approach in a subsequent study.
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Appendix A
Human Subjects Materials
287
Recruitment - Verbal Script
I am a graduate student under the direction of Professor Marilyn Carlson in the Department of Mathematics and Statistics at Arizona State University. I am conducting a research study to investigate how first semester calculus students understand and solve related rates problems. I am requesting that you may be videotaped while working on mathematical tasks in small groups which will occur during your regular class period. I would also like to have several volunteers to participate in a series of six one hour individual/small group interviews that will involve you working on calculus problems and answering some general questions about how calculus is taught. Individual or small group interviews will be scheduled at your convenience and will also be videotaped. You will be paid $20 per hour for the time spent in individual or small group interviews outside of class. The videotapes will be transcribed and analyzed using pseudonyms for every student. The pseudonyms may be used in any published results of this study. The videotapes may be kept in archives for up to 5 years after the completion of the study, after which they will be erased or destroyed. Your participation in this study is voluntary. If you choose not to participate or to withdraw from the study at any time, there will be no penalty; it will not affect your grade. The results of the research may be published, but your name will not be used. If you have any questions concerning the research study, please call me at ( 480 ) 727-7575.
288
Letter of Consent
Dear Student: I am a graduate student in the Department of Mathematics at Arizona State University, working under the supervision of Professor Marilyn Carlson. I am conducting a research study to investigate how calculus students understand and solve word problems. This project will take place during the spring semester of 2005. I am requesting your participation, which will involve completion of in-class activities and possibly six individual/small group interviews (about 1 hour in length). The individual/small group interviews will be scheduled at your convenience. During the interview(s), I will be asking you to explain the reasoning you used to answer specific mathematics questions and some questions about how you think calculus is taught. In addition, I am requesting your permission to videotape you while working on these tasks. Your participation in this study is voluntary. If you choose not to participate or to withdraw from the study at any time, there will be no penalty (it will not affect your grade). The results of the research study may be published, but your name will not be used. Videotapes may be kept in archives for up to 5 years after the completion of the study. While the researcher must maintain your confidentiality, when you are working on collaborative tasks, please be aware that other participants are not required to maintain this same level of confidentiality. All names will be changed in any published results. Although there may be no direct benefit to you, the possible benefit of your participation is to assist me in developing new materials for calculus instruction. If you have any questions concerning the research study, please call me at (480) 727-7575. Sincerely, Nicole Engelke Department of Mathematics Arizona State University I am at least 18 years old and give consent to participate in the above study. I give consent for you to video tape me while completing mathematical tasks and to use this video data for analyzing and reporting my results. _____________________________________________ ______________ Signature Date
289
If you have any questions about your rights as a subject/participant in this research, or if you feel you have been placed at risk, you can contact the Chair of the Human Subjects Institutional Review Board, through Karol Householder, at (480) 965-6788.
Appendix B
Pilot Study Materials
291
Review of Similar Figures for Calculus Students Using GSP
1. What does the word “similar” mean to you?
2. Open the sketch named triangle1. Measure lengths AD and AC. Compute their
ratio. Repeat for lengths AE and AB. Repeat for lengths DE and BC. What do you
notice about the ratios?
3. What happens if you drag one of the points around? What happens if you animate
the segments? What happens if you animate just one point?
4. Measure angles ACB and ADE. Measure angles AED and ABC. What do you
notice?
5. Make a conjecture about what are the properties for similar triangles. Test your
conjecture on the sketch named triangle2. If it does not work, why? Revise your
conjecture and try again.
6. Extend what you have learned about similar triangles to quadrilaterals. Test your
conjecture on the sketches named quad1 and quad2.
7. Can you generalize your definition of similar to a general polygon? Would a
mathematician accept your definition of similar?
292
Pilot Study Version - Related Rates Activity
Problem: A spherical snowball is melting in such a way that its volume is decreasing at a rate of 1 cm3/min. At what rate is the diameter decreasing when the diameter is 10 cm? 1. Draw a picture of the snowball such that you can best identify quantities that are
changing. Is the diameter decreasing faster and faster or slower and slower? 2. Determine an appropriate formula for the volume of the snowball. 3. Fill in the following table.
Quantities that Don’t Change Quantities that Change
Goal (Want to Find)
4. Implicitly differentiate your formula with respect to time. 5. Solve for your goal. Make sure to include units!
Problem: A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 1 ft. If the trough is filled with water at a rate of 12 ft3/min, how fast does the water level rise when the water is 6 in. deep? (Stewart, 1991) 1. Draw a picture of the trough such that you can best identify the quantities that are
changing. Describe how the height is changing. 2. Determine an appropriate formula for the volume of the trough. 3. Fill in the following table.
Quantities that Don’t Change Quantities that Change
Goal (Want to Find)
293
4. Determine any unknown quantities in your formula that may be related. (Hint: base and height can be related using similar triangles.) Using this relation, rewrite your formula for volume, eliminating an unknown.
5. Implicitly differentiate your new formula with respect to time. 6. Solve for your goal. Make sure you include units! 7. How would the problem have been different had you been given that the water was
at the level of 2 ft across, instead of 6 in. deep? 8. How is problem 1 different from problem 2?
294
Pilot Study Version - Related Rates Activity 2
Problem: Nikki buys an ice cream cone on a hot summer day. The ice cream is a perfect sphere of chocolate with a radius of 5 cm. However, the ice cream is melting and dripping into the bottom the cone at 2 cm3/min. If the ice cream cone is 9 cm tall and 5 cm wide, how fast is the liquid ice cream rising when the liquid ice cream is 2 cm deep? 1. Draw a diagram to represent the situation and label the appropriate variables. 2. Determine the appropriate formula to model the situation. Justify. 3. Fill in the following table.
Quantities that Don't Change Quantities that Change
Goal (Want to Find)
4. Determine what unknown quantities in your formula are related. Determine if these
relations will help you to rewrite your formula and eliminate an unknown. 5. Implicitly differentiate your new formula with respect to time. 6. Solve for your goal. Make sure you include units! 7. How could you change the problem? What other kinds of things could you have
determined? 8. Solve the trapezoidal pool problem from the book. 9. Create your own related rates problem based on a real world experience.
295
Related Rates Puzzle
Turvey for Related Rates Found at: http://www.pen.k12.va.us/Div/Winchester/jhhs/math/lessons/calculus/relrates.html
Back in 1953, Roger Price invented a minor art form called the Droodle, which he described as "a borkley-looking sort of drawing that doesn't make any sense until you know the correct title." In 1985, Games Magazine took the Droodle one step further and created the Turvey. Turvies have one explanation right-side-up and an entirely different one turned topsy-turvey (when standing on your head – or turning the paper upside down, if you prefer).
Here is the title right-side-up:
"__ __ __ __ __ __ - __ __ __ __ __ __ __ __ __ __ __ ." 5 2 6 3 13 4 7 5 2
Here is the title upside-down: "__ __ __ __ __ __ __ __ __ , __ __ __ __ __ __ __ __ __ __ __ __ __ 8 12 6 11 7 9 6 10 8 10 6 3 9 10 1 __ __ __ __ __ ." 8 7 14 6
296
To determine the titles to this turvey, solve the 14 Related Rates problems. Then replace each numbered blank with the letter corresponding to the answer for that problem. The unnumbered blanks are all vowels. You should be able to determine both titles. 1-2. A certain calculus student hit Mr. Leacher in the head with a snowball. If the snowball is melting at the rate of 10 cubic feet per minute, at what rate is the radius changing when the snowball is 1 foot in radius (Problem #1)? At what rate is the radius changing when the snowball is 2 feet in radius (Problem #2)? Answers should be expressed in terms of feet per minute. 3-4. A baseball diamond is 90 feet square (NOT 90 square feet!). Coach Jack Handley runs from first base to second base at 25 feet per second. How fast is he moving away from home plate when he is 30 feet from first base (Problem #3)? How fast is he moving away from home plate when he is 45 feet from first base (Problem #4)? Answers should be expressed in terms of feet per second. 5. Water flows at 8 cubic feet per minute into a cylinder with radius 4 feet. How fast is the water level rising when the water is 2 feet high? Answer should be expressed in terms of feet per minute.
6-7. The Monticello High School swimming pool is an inverted cone with height 20 meters and radius 5 meters. It is being filled by Mr. Lundin with a hose which pumps in water at the rate of 3 cubic meters per minute. When the water level is 2 meters, how fast is the water level rising (Problem #6)? How fast is the radius changing at this moment (Problem #7)? Answers should be expressed in terms of meters per minute.
297
8-9. A stone is dropped into Sherando Lake, causing circular ripples whose radii increase by 2 meters/second. How fast is the disturbed area growing when the outer ripple has radius 5 meters (Problem #8)? How fast is the radius increasing at that moment (Problem #9)? Answers should be expressed in terms of square meters per second (#8) and meters per second (#9). 10-11. A fish is being reeled in at a rate of 2 meters / second (that is, the fishing line is being shortened by 2 m/s) by a fisherwoman at Mill brook. If the fisherwoman is sitting on the dock 30 meters above the water, how fast is the fish moving through the water when the line is 50 meters long (Problem #10)? How fast is the fish moving when the line is only 31 meters (Problem #11)? Answers should be expressed in terms of meters per second.
12. A student at James Wood was painting the high school and standing at the top of a 25-foot ladder. She was horrified to discover that the ladder began sliding away from the base of the school at a constant rate of 2 feet per second. At what rate was the top of the ladder carrying her toward the ground when the base of the ladder was17 feet away from the school? Answers should be expressed in terms of feet per second. 13. A spherical balloon was losing air at the rate of 5 cubic inches per second. At what rate is the radius of the balloon decreasing when the radius equals 5 inches? Answers should be expressed in terms of inches per second. 14. Oil spills into Lake Winchester in a circular pattern. If the radius of the circle increases at a constant rate of 3 feet per minute, how fast is the area of the spill increasing at the end of 10 minutes? Answers should be expressed in terms of feet per minute.
298
Answers:
B. 0 G.
π2
5−
M. π180 S.
π
12
C. π2
1 H. 61
61
62 N.
2
5
T. π20
D. 55 K. 2 P.
2
105 V.
336
33617−
F. π20
1− L.
π8
5− R.
π
3
W. -1.855
Appendix C
Teaching Experiment Materials
300
Individual Interview Protocol
Purpose: To determine a baseline for the students’ understandings about variable, function, and covariational reasoning abilities
Homework: Write up 2 problems from the interview
1. Circle all of the items in the list below that describe the meaning you give to the letter x when it is used in mathematical expressions.
a. Meaningless mark on the paper b. Changing quantity c. Generalized number d. Letter that stands for a number e. Name of something f. Place holder g. Unknown h. Other: please
specify____________________________________________________ 2. Read this expression out loud:
( )wur =
a. What does this expression mean to you?
b. What is the meaning of each of the parts of ( )wur = ?
c. Discuss the relationships that are represented by the expression ( )wur = .
3. Consider the function given by ( ) ( )( )uuh sincos= . Let ( )uhy = .
a. Which letter indicates the name of the independent variable? b. Which letter indicates the name of the dependent variable? c. Which letter indicates the name of the function?
d. Complete this statement: ( ) =2xh
4. Suppose that a genie gave you magical envelopes, each of which will change your
ATM deposits. Envelope A will divide your deposit by 3 and subtract 5. Envelope B will square your deposit. You must put one envelope inside the other. When will it be best to put B inside A and when will it be best to put A inside B?
5. A ball is thrown into a lake, creating a circular ripple that travels outward at a speed
of 3 cm per second. Express the area, A, of the circle in terms of the number of seconds, s, that have passed since the ball hits the lake.
301
6. Imagine this scenario. Water is poured into an empty bottle until the bottle is full. As
you imagine the water filling the bottle shown below, what are the changing quantities that come to mind? Can you sketch a graph that shows the height of water as a function of volume as water is poured into the bottle?
302
Instructional Sequence Lesson Logic
Day 1: Rates of Change with computer program
Purpose: To have the students investigate rates of change and the relationships which exist between them. The students should also become aware that time is a variable that may be used to relate two quantities.
Homework: Write up one or two questions from the session
Lesson Logic: Rates of Change
Necessary Student Knowledge:
• Be able to construct a table of function values
• Be able to compute the average rate of change over a given interval (the slope between two points)
Problem 1: Suppose we have a plane that is flying over a RADAR tower, TA, and is on course to pass over a second RADAR tower, TB. Let u be the distance between the plane and TA, and let v be the distance between the plane and TB. How fast does v change with respect to u? Table 35: Lesson Logic for Plane Problem Using Computer Program
Step Action Reason
1 Ask the students to open the program called plane, and choose Plane1 under the activities menu.
This is the tool students will be using to investigate rates of change and the relationships between them.
2 Ask the students: How fast does v change with respect to u?
In Plane 1, there is no indicator bar for time, so this is impossible to determine. This forces the students to start thinking about other ways to relate variables.
3 Ask the students: How fast does u change with respect to v?
In Plane 1, there is no indicator bar for time, so this is impossible to determine. We want the students to recognize that we need some way to relate u and v.
4 Ask the students what they believe might be necessary to be able to relate u and v.
We want the students to recognize time as a common input variable and that we can measure changes in u and v over the same time intervals.
303
5 Ask the students to now choose Plane 2 from the activities menu.
In Plane 2, we have added the variable time which will allow the students to relate u and v.
6 Ask the students to record the distance between the plane and TA for several points. Ask the students if it matters if we take the measurements at evenly spaced points.
This allows the students to construct a table of values that will be used to calculate the average rate of change over several intervals.
7 Ask the students to record the distance between the plane and TB for several points.
This allows the students to construct a table of values that will be used to calculate the average rate of change over several intervals.
8 Ask the students: How fast does v change with respect to time?
We want the student to relate the variable v to time by computing the average rate of change for his intervals. This may not be apparent to the student at first, so be ready to revisit the car trip example: you drive from A to B in some amount of time, and the distance traveled is x miles. What was your average rate of change?
9 Ask the students: How fast does u change with respect to time?
We want the student to relate the variable u to time by computing the average rate of change for his intervals.
10 Again ask the students: How fast does v change with respect to u?
The student should now see that we may relate v and u by considering their
ratio:
t
ut
v
u
v
∆
∆∆
∆
=∆
∆
11 Again ask the students: How fast does u change with respect to v?
Similarly, we want the students to see that we may relate u and v by
considering the ratio:
t
vt
u
v
u
∆
∆∆
∆
=∆
∆
12 Ask the students: How fast does v change with respect to x?
This is just an extension of the previous questions: we now want to relate v and x.
13 Ask the students: How fast does u change with respect to x?
This is just an extension of the previous questions: we now want to relate u and x.
304
14 Ask the students: What happens as we take smaller and smaller intervals of time?
We want the student to see that taking smaller and smaller intervals is taking the limit and we get an instantaneous rate of change.
Problem 2: Suppose that you have a snowball that is 20 cm in diameter. It is a warm day, and the snowball is melting. The rate at which the diameter decreases is a function of the temperature. Let d represent the diameter of the snowball, let T represent the temperature, and let V represent the volume of the snowball. How fast does d change with respect to time? Table 36: Lesson Logic for Snowball Problem Using Computer Program
Step Action Reason
1 Ask the students to open the program called plane, and choose Snowball II under the activities menu.
This is the tool students will be using to investigate rates of change and the relationships between them.
2 Ask the students to record the length of the diameter for several points.
This allows the students to construct a table of values that will be used to calculate the average rate of change over several intervals.
3 Ask the students: How fast does V change with respect to time?
This is similar to what we did in problem 1.
4 Ask the students: How fast does d change with respect to T?
This is similar to what we did in problem 1, but now we are leading the
students to see that we can get T
d
∆
∆ by
using the chain rule.
5 Ask the students: How fast does T change with respect to time?
This is similar to what we did in problem 1, but now we are leading the
students to see that we can get t
T
∆
∆by
using the chain rule.
6 Ask the students: How fast does d change with respect to time?
We want the students to see that by applying the chain rule and multiplying
we can get that t
T
T
d
t
d
∆
∆
∆
∆=
∆
∆.
305
Day 2: The Chain Rule
Purpose: To make explicit the multiplicative nature of the chain rule.
Homework: Write up an explanation of how the chain rule works and one problem from the session Problem: A car can drive 6 times as fast as a man can run. A plane can fly 100 times as fast as a car can drive. How much faster than the man can the plane fly? Table 37: Lesson Logic for Chain Rule - Car Problem
Step Action Reason
1 How can we express the rate of change
of the car with respect to the man?
6=∆
∆
man
car
Encourage the student to use the notation developed in the previous session to express the rate of change.
2 How can we express the rate of change
of the plane with respect to the car?
12=∆
∆
car
plane
Encourage the student to use the notation developed in the previous session to express the rate of change.
3 What is the rate of change of the plane with respect to the man?
72612 =⋅=∆
∆⋅
∆
∆
man
car
car
plane
Have the student explore the multiplicative nature of the chain rule.
4 What would be the speed of the plane if the man can run at 8 miles per hour?
Apply the chain rule to find an actual value.
Problem: Suppose you are hiking up a mountain at the rate of 0.5 kilometers per hour. As the elevation increases, the temperature decreases at a rate of per kilometer. At what rate is the temperature changing with respect to time?
306
Table 38: Lesson Logic for Chain Rule - Hiking Problem
Step Action Reason
1 How can we express the rate of change
of the hiker’s altitude with respect to
time? 5.0=∆
∆
time
altitude
Encourage the student to use the notation developed in the previous session to express the rate of change.
2 How can we express the rate of change
of the temperature with respect to
altitude? 6=∆
∆
altitude
etemperatur
Encourage the student to use the notation developed in the previous session to express the rate of change.
3 What is the rate of change of the temperature with respect to time?
Have the student explore the multiplicative nature of the chain rule.
Problem: A ball is thrown into a lake, creating a circular ripple that travels outward at a speed of 3 cm per second. How fast is the area of the circular ripple growing with respect to time? Table 39: Lesson Logic for Chain Rule - Ball Problem
Step Action Reason
1 What is the formula for the area of a
circle? 2rA π=
Make sure the students know the formula for the area of a circle.
2 Can you express the radius as a function of time? sr 3=
Make time an explicit part of the problem by expressing the radius in terms of seconds.
3 What is the rate of change of the area of the circle with respect to the radius?
rdr
dAπ2=
Encourage the student to use the notation developed in the previous session to express the rate of change of the area of the circle in general.
4 What is the rate of change of the radius
with respect to time? 3=dt
dr
Use the notation from the previous session to express a relationship for the specific rate of change of the radius.
307
5 What is the rate of change of the area of the circle with respect to time?
dt
drr
dt
dr
dr
dA
dt
dAπ2=⋅=
Use the chain rule to relate the information in the two previous steps.
6 How does this relate to the snowball problem from yesterday?
dt
drr
dt
dr
dr
dV
dt
dV 24π=⋅=
Encourage students to extend the knowledge from this session to a problem from the previous session.
Days 3-4: Related Rates
Purpose: To solve related rates problems by applying the knowledge they have gained in the previous sessions.
Homework: Write up two problems from the session Lesson Logic: Related Rates
Necessary Student Knowledge:
• Geometric formulas for area, volume, etc.
• Understand rate of change
• How to compute derivatives using derivative rules Problem 1: A plane flying horizontally at an altitude of 3 miles and a speed of 600 mi/hr passes directly over a radar station. Let the distance from the plane to the radar station be represented by z. What is the rate of change of the distance from the plane to the radar station with respect to time?
308
Table 40: Lesson Logic for Related Rates - Plane Problem
Step Action Reason
1 Draw a diagram that represents the situation.
Encourages the student to begin visualizing the situation.
2 Label your diagram with variables that represent changing quantities.
The student needs to be able to recognize the difference between a variable and a constant in the problem situation.
3 What is an appropriate geometric formula for this situation?
222 zyx =+
The student needs to examine and express the relationships between the variables in his diagram.
4 Express each of your variables as a function of time:
( ) ttx 600=
( ) 3=ty
( ) ( ) ( )( ) ( )[ ] ( )[ ]( )2
1222
122
3600 tytxttz +=+=
The student needs to be able to recognize that each variable is a function of time (eventually we won’t know what the explicit relationship is).
5 What is the rate of change of the plane’s horizontal distance with respect to time?
600=dt
dx
This relates the problem to the knowledge gained in the previous activity in which the students were measuring the rate of change of every variable with respect to time.
x
y
z
309
6 What is the rate of change of the plane’s vertical distance with respect to time?
0=dt
dy
This relates the problem to the knowledge gained in the previous activity in which the students were measuring the rate of change of every variable with respect to time.
7 What is the rate of change of the distance from the plane to the radar station?
( )( ) ( )ttdt
dz000,7203600
2
12
122
−
+=
( )( ) ( )( )60060023][2
12
122 ttx
dt
dz −
+=
( )( ) ( )
+=
−
dt
dxtxtx
dt
dz23][
2
12
122
( )( )
=
dt
dx
tz
tx
dt
dz
This relates the problem to the knowledge gained in the previous activity in which the students were measuring the rate of change of every variable with respect to time. This should encourage students to look at the general relationships before focusing on a particular instant in time, thus eliminating the desire to substitute in values too soon.
8 What is the rate of change of the distance from the plane to the radar
station when ( ) 5=tz ? Make sure to
include units!
( ) 4806005
4==
dt
dz
Now, we focus on a specific instant in time and solve the problem. Note that when 5=z , 3=y hasn’t changed, and
so it must be that 4=x .
9 What would have happened if we had
just differentiated 222 zyx =+ with
respect to time?
We want the student to consider if it was necessary to know the exact formula for each variable.
310
Problem 2: A spherical snowball is melting in such a way that its volume is decreasing at a rate of 1 cm3/min. Let d represent the diameter of the snowball in cm. At what rate is the diameter decreasing with respect to time? Table 41: Lesson Logic for Related Rates - Snowball Problem
Step Action Reason
1 Draw a diagram that represents the situation.
Encourages the student to begin visualizing the situation.
2 Label your diagram with variables.
The student needs to be able to recognize the difference between a variable and a constant in the problem situation.
3 What is an appropriate geometric formula for this situation?
3
3
23
4
3
4
==
drV ππ
We want the student to examine relationships between the variables in his diagram.
4 Express each of your variables as a function of time.
( ) ( )trtd 2=
( ) ( )[ ]( ) 3
3
23
4
3
4
==
tdtrtV ππ
The student needs to be able to recognize that each variable is a function of time (eventually we won’t know what the explicit relationship is). In this case, the function is not explicit, but we can relate them through function composition.
r
311
5 What is the rate of change of the radius of the snowball with respect to the diameter?
2
1=
dd
dr
This relates to the previous activity in which the students were
6 What is the rate of change of the diameter of the snowball with respect to time?
?=dt
dd
This is what most related rates problems ask students to find for a particular instant. This should encourage the student to look at general relationships before focusing on a specific instant.
7 Differentiate your function with respect to time.
( )( )[ ]
dt
ddtd
dt
ddtd
dt
dV 2
2
22
1
23
3
4 ππ =
=
We need to do this to solve the problem.
8 What is the rate of change of the
diameter when ( ) 1.0=td ? Make sure to
include units!
Now, we focus on a specific instant in time and solve the problem.
9 How is problem 2 similar to or different to problems 1?
We want the student to start to think about the similarities between the problems and start to conjecture about a general problem solving strategy.
Problem 3: A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 1 ft. Assume that the trough fills at a constant rate. Let V represent the volume of the trough, h represent the height of the water trough, and b represent the length of the base of the water in the trough. What is the rate of change of the height of the water, h, in the trough with respect to time? Table 42: Lesson Logic for Related Rates - Trough Problem
Step Action Reason
312
1 Draw a diagram that represents the situation.
Encourages the student to begin visualizing the situation.
2 Label your diagram with variables.
The student needs to be able to recognize the difference between a variable and a constant in the problem situation.
3 What is an appropriate geometric formula for the volume of the water in the trough?
bhlV2
1=
We want the student to examine relationships between the variables in his diagram.
4 Express each of your variables as a function of time.
( ) ( ) ( ) ( )tlthtbtV2
1=
( ) 10=tl
( )( ) 1
3=
th
tb
( ) ( )thtb 3=
We want the student to examine relationships between the variables in his diagram.
5 What is the rate of change of the height of water in the trough with respect to the base of the water in the trough?
This relates to the previous activity in which the students were
b
h
313
6 What is the rate of change of the height of the water in the trough with respect to time?
This is what most related rates problems ask students to find for a particular instant. This should encourage the student to look at general relationships before focusing on a specific instant.
7 Can you construct a function that
relates that relates ( )tV and ( )th ?
( ) ( )[ ] ( )( ) ( )[ ]215103
2
1thththtV ==
Here we see that we may relate the variables through function composition. This makes the idea that the base and height are related through one relationship that may be used to determine b in terms of h which may then be used to relate V and h.
8 Differentiate your function with respect to time.
We need to do this to solve the problem.
9 If the trough is filled with water at a rate of 12 ft3/min, what is the rate of change of the height of the water when the water is h in. deep?
Now, we focus on having one rate given and know that we can find the other for any given instant.
10 What is the rate of change of the height
of the water when ( ) 6=th in? Make
sure to include units!
Now, we focus on a specific instant in time and solve the problem.
11 How would the problem have been different had you been given that the water was at the level of 2 ft across, instead of 6 in. deep?
We want the student to think about the relationship between the radius and height and what role that played in the problem.
12 How is problem 3 similar to or different from problems 1 and 2?
We want the student to start to think about the similarities between the problems and start to conjecture about a general problem solving strategy.
13 Ask the students to describe a solution procedure for solving any related rates problem.
We want the students to construct their own “solution guide” for related rates problems.
314
Days 5-6: Solving More Related Rates Problems
Purpose: To observe how the students solve unfamiliar related rates problems and identify the problem solving behaviors they employ.
Homework: Write up one or two problems from the session Problem: Coffee is poured at a uniform rate of 20 cm3/sec into a cup whose inside is shaped like a truncated cone. If the upper and lower radii of the cup are 4 cm and 2 cm, respectively, and the height of the cup is 6 cm, how fast will the coffee level be rising when the coffee is halfway up the cup? Problem: Two boats are racing with constant speed toward a finish marker, boat A sailing from the south at 13 mph and boat B approaching from the east. When equidistant from the marker the boats are 16 miles apart and the distance between them is decreasing at the rate 17 mph. Which boat will win the race?
Problem: The thin lens equation in physics is fSs
111=+ where s is the object distance
from the lens, S is the image distance from the lens, and f is the focal length of the lens. Suppose that a certain lens has a focal length of 6 cm and that an object is moving toward the lens at the rate of 2 cm/sec. How fast is the image distance changing at the instant when the object is 10 cm from the lens? Is the image moving away from the lens or toward the lens?
315
Student Activity Sheets
Name:____________________________________
Activity 1: Rates of Change
November 15, 2005
Problem 1: Suppose we have a plane that is flying over a RADAR tower, TA, and is on course to pass over a second RADAR tower, TB. Let u be the distance between the plane and TA, and let v be the distance between the plane and TB. What is the rate of change of u in relation to v? Problem 2: Suppose that you have a snowball that is 20 cm in diameter. It is a warm day, and the snowball is melting. The rate at which the diameter decreases is a function of the temperature. Let d represent the diameter of the snowball, let T represent the temperature, and let V represent the volume of the snowball. How fast does d change with respect to time?
316
Name:______________________________________
Activity 2: The Chain Rule
November 16, 2005
Problem 1: A car can drive 6 times as fast as a man can run. A plane can fly 12 times as fast as the car can drive. How much faster than the man can the plane fly? Problem 2: Suppose you are hiking up a mountain at the rate of 0.5 kilometers per hour. As the elevation increases, the temperature decreases at a rate of °6 per kilometer. At what rate is the temperature changing with respect to time? Problem 3: A ball is thrown into a lake, creating a circular ripple that travels outward at a speed of 3 cm per second. How fast is the area of the circular ripple growing with respect to time? Question: How does this relate to the snowball problem from yesterday?
317
Name:______________________________________
Activity 3: Related Rates
November 17, 2005
Each problem was printed on a separate page to allow ample room to work.
Problem 1: A spherical snowball is melting in such a way that its volume is decreasing at a rate of 1 cm3/min. Let d represent the diameter of the snowball in cm. At what rate is the diameter decreasing with respect to time?
Problem 2: A plane flying horizontally at an altitude of 3 miles and a speed of 600 mi/hr passes directly over a radar station. Let the distance from the plane to the radar station be represented by z. What is the rate of change of the distance from the plane to the radar station with respect to time? Problem 3: A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 1 ft. Assume that the trough fills at a constant rate. Let V represent the volume of the trough, h represent the height of the water trough, and b represent the length of the base of the water in the trough. What is the rate of change of the height of the water, h, in the trough with respect to time?
318
Name:______________________________________
Activity 4: More Related Rates
November 21, 2005 Each problem was printed on a separate page to allow ample room to work.
Problem: Coffee is poured at a uniform rate of 20 cm3/sec into a cup whose inside is shaped like a truncated cone. If the upper and lower radii of the cup are 4 cm and 2 cm, respectively, and the height of the cup is 6 cm, how fast will the coffee level be rising when the coffee is halfway up the cup? Problem: Two boats are racing with constant speed toward a finish marker, boat A sailing from the south at 13 mph and boat B approaching from the east. When equidistant from the marker the boats are 16 miles apart and the distance between them is decreasing at the rate 17 mph. Which boat will win the race?
Problem: The thin lens equation in physics is fSs
111=+ where s is the object distance
from the lens, S is the image distance from the lens, and f is the focal length of the lens. Suppose that a certain lens has a focal length of 6 cm and that an object is moving toward the lens at the rate of 2 cm/sec. How fast is the image distance changing at the instant when the object is 10 cm from the lens? Is the image moving away from the lens or toward the lens?