Utilizing Mathematica for Higher Level Thinking in Calculus

Grace McClurkin, University of Tennessee, Knoxville
Joshua Mike,  University of Tennessee, Knoxville

Overview

What is an inverted classroom?

Our goals in designing the course.

Designing the course with Mathematica.

Student feedback and future ideas.

What is an Inverted Classroom?

Traditional Classroom

Basic material is introduced in class during lecture.

Students have the opportunity to ask questions about the basics as it is introduced.

Higher level materials are sparse:

It may be a part of lecture, otherwise this is often left for the students on their own.

Inverted Classroom

Basic material is introduced via resources before class.

Higher level materials are emphasized during in-class problem solving.

Time is often allotted for homework during class.

Students have the opportunity to ask questions about more challenging concepts and problems

Bloom’s Taxonomy

A hierarchy of intellectual behavior important for learning

Accomplishing an Inverted Classroom

Before Class

Learn introductory material: Notes and some videos are provided to cover the basic and central ideas of each topic

Complete Pre-Class Practice: A few problems are given to give the students a feel for the kinds of problems that go with the material. Students are encouraged to do these problems, but they are not graded.

In Class

Learn higher-level material: Once per week, students read through and complete activities in groups.

Practice problem solving: Once per week, students ask questions and work on homework in class.

Primary Goals for Course Design

1. Clear student expectations.

2. Create manageable notes.

a) Not too long for students

b) A complete core of material

c) Simple for new instructors to use and adapt.

3. Engage students in higher thinking during class

4. Focus on conceptual understanding as opposed to computation.

a) Steep the class in a geometric viewpoint; enable students to see and create lots of pictorial examples. 

b) Focus on the ideas of linearization, limits, additivity, and coordinates.

Course Design with Mathematica

Notes

Introduce the core material to the students.

Define concepts with intuitive explanation.

Present core examples (often via Mathematica), with Mathematica input alongside output.

Sample from Notes

Pre-Class Practice

Notebook file with introductory questions and problems.

Commands are presented in bold; students are asked to use the documentation center.

Activities

Groups work to complete tasks within a Mathematica notebook file

The tasks heavily rely on Mathematica for timeliness and content.

Sample from Multivariable Limits Activity:

Task 1: Consider the function f(x,y)=(((x^2)-(y^4))/((x^2)+(y^4))).
	a) Find the limit as (x,y)->(0,0) of f along the line y = x
	b) Find the limit as (x,y)->(0,0) of f along a general line (x(t),y(t)) = t(m*t,n*t).
	c) Find the limit as (x,y)->(0,0) of f along the parabola x=(y^2). 
	d) Find the limit as (x,y)->(0,0) of f along a general parabola x=a (y^2).

f=(x^2-y^4)/(x^2+y^4);
Plot3D[f,{x,-5,5},{y,-5,5}]

a=f/.y->x;
Limit[a,x->0]

b=f/.{x->m t, y->n t};
Limit[b,t->0]

c=f/.x->y^2;
Limit[c,y->0]

d=f/.x->m y^2;
Limit[d,y->0]

Sample from Vector Field Activity:

Task 1: The electric field of a dipole can be described using the inverse square law, so that if there are point charges (Subscript[Q, 1]) and (Subscript[Q, 2]) at points (Subscript[X, 1])=((Subscript[x, 1]),(Subscript[y, 1])) and (Subscript[X, 2])=((Subscript[x, 2]),(Subscript[y, 2])), we can find the electric field in terms of X=(x,y) as EF(X)=(((Subscript[Q, 1])(X-(Subscript[X, 1])))/(|X-(Subscript[X, 1])((|)^3)))+(((Subscript[Q, 2])(X-(Subscript[X, 2])))/(|X-(Subscript[X, 2])((|)^3))). 
a) Define the electric field EF generated by a point charge of +3 at (3,3) and another point charge of -2 at (-3,-3). Then, plot your electric field over the square -5≤x,y≤5.
[Hint: You will want this quip of code to avoid the point charges themselves, TH is the threshold which you need to define to be small (Define TH to be a positive value less than 3]
	EFnew=If[Norm[EF]>=TH,EF*TH/Norm[EF],EF]

X={x,y};Q1=3;Q2=-2;X1={3,3};X2={-3,-3};
EF=Q1*(X-X1)/Sqrt[(X-X1).(X-X1)]^3 + Q2*(X-X2)/Sqrt[(X-X2).(X-X2)]^3;

TH=1/2;
EFnew=If[Norm[EF]>=TH,EF*TH/Sqrt[EF.EF],EF];
VectorPlot[EFnew, {x,-5,5},{y,-5,5}]

Sample from Hessian/Second Derivative Test Activity

Task 1: Consider the function f(x,y) =x (cos(π y)+y) - y( cos(π x)+2). 
a) Show that f has a critical point at (a,b) = (1,1) via the first derivative test.

f=x ( Cos[Pi*y]+y)-y*(Cos[Pi*x]+2);
gradf=Grad[f,{x,y}];
gradf/.x->1/.y->1;

b) Find (Subscript[V, (1,1)]) and VectorPlot V and ∇f on the same set of axes:
	i) on -1≤x,y≤3
	ii) on 0.8≤x,y≤1.2

hess=Grad[gradf,{x,y}]/.x->1/.y->1;
V=hess.{x-1,y-1};

{VectorPlot[{V,gradf},{x,-1,3},{y,-1,3}],VectorPlot[{V,gradf},{x,0.8,1.2},{y,0.8,1.2}]}

c) Use the Eigenvalues of the Hessian to determine what kind of critical point occurs at (1,1).

Eigenvalues[hess]//N

Saddle point.

d) Use the pictures from part (b) to confirm and explain your answer to part (c).

Homework and Exams

Work is completed within a notebook file, which is then submitted and graded

Students are allowed to use any Mathematica commands

Students must conceptually explain their answers.

Sample from Exam 1

1.) Find the distance between the circle (((x-1))^2)+(((y+5))^2)=9 and the line 5y-2x=3.

Grade: Numerical Answer:  1/1 Explanation Answer:  5/5

Explanation: 
How:
Find the distance between the center of the circle and the line:
-The center of the circle is c=(1,-5), which is taken from the equation of the circle.
-Define two points on the line, p1 and p2, by plugging in any value for x and solving for y.
-Define a vector between the center of the circle and any point on the line, c-p1, and a vector that runs along the line, p2-p1.
-Find the projection of the vector pc onto the vector along the line.
-Find the orthogonal component of pc by using the equation orthu v + proju v= v.
-The distance between the center of the circle and the line is the magnitude of the orthogonal.
Now subtract the radius, which is found from the equation of the circle, from the distance between the center and the line.
Why:
-The shortest distance between a line and an object is perpendicular to the line. 
-The perpendicular path between the center of the circle and the line is the orthogonal of a vector connecting the two.
-The distance of the radius is simply subtracted from this distance to find the shortest distance between the outer edge of the circle and the line.

r=3; c={1,-5};p1={1,1};p2={0,3/5};
pc=c-p1; pv=p2-p1;
proj=Projection[pc,pv];
orth=pc-proj;
distc=Sqrt[orth.orth];
dist=N[distc-r]

The shortest distance between the circle and the line is 2.57.

Student Feedback and Future Goals

Based on mid-semester feedback from students currently enrolled in the course, we’ve decided to focus on the following areas for imporvement:

Create a smoother start to the course, in particular by helping students to use Mathematica and adjust to the inverted course structure.

Rework the materials

Add more examples to notes, specifically Mathematica based examples in the first unit.

Transition the pre-class to a more guided practice in an effort to more directly introduce Mathematica commands and syntax.

Use Mathematica Course Tools to create videos to accompany the notes