tom teaching portfolio 20150506

TEACHING PORTFOLIO (SAMPLE)

PORAMATE (TOM) PRANAYANUNTANA

“For the past half-dozen years, Tom has consistently gotten amongst the very best studentevaluations in the Mathematics Department, at NYU-Poly.”

Erwin Lutwak – Chair of Mathematics

The abilities to view, think and make connections mathematically from several points of view– algebraically, numerically, graphically and verbally, are some of the most valuable skills Iteach my students. They are the skills to read and draw two and three-dimensional graphs,contour diagrams and think graphically, to read tables and think numerically. I teach mystudents to apply these skills, along with algebraic skills, to model practical problems ofthe real world. My students are from different majors – Applied Physics, Biomolecular Sci-ence, Business and Technology Management, Chemical and Biomolecular Engineering, CivilEngineering, Computer Engineering, Construction Management, Electrical Engineering, In-tegrated Digital Media, Mathematics, Mechanical Engineering, Physics and Mathematics,Science and Technology Studies, and Sustainable Urban Environments – and they all usethis subject differently. Unless my students engage themselves, they cannot view, think, ormake mathematical connections; thus, my primary goal is to encourage them to participateactively in the learning process. Since my students come from diverse backgrounds, theypossess different skills and may have different viewpoints of the same theory. They alsoexperience different practical applications of the same theory. Therefore, I choose to teachmany of same concepts in different ways: algebraically, numerically, graphically, and verballyin order for the students to have more chances to make connections with their previous orbackground knowledge. Simultaneously, I emphasize the properties and insights of the topicunder consideration. This will help the students make connections and retain the knowledge.

Using various methods I ensure participation, including discussions outside of class–recitationsand office hours, leading undergraduate research, and even assigning students to teach therecitation for the day. These techniques have worked well for the upper level classes however,for the lower levels I found that providing the students with time to pause and contemplatethe answer before advancing with the lesson is most effective. Even with these approaches,mathematics can be difficult to comprehend, therefore I use tangible examples that em-phasize real world examples. Additionally, incorporating technology–such as the graphingcalculator with CAS (computer algebraic system)–as a motivator, instead of as a tool, isuseful for keeping the students interested. The following are some examples of calculatorusage. In my Linear Algebra class, I teach students how linear independence of columnvectors of a matrix relates to its unique reduced row echelon form. This enables the studentsto inspect a matrix and find its reduced row echelon form without completely doing rowreduction steps. CAS calculators are being used for practicing this inspection technique and

Teaching Portfolio (Sample) Poramate (Tom) Pranayanuntana

verifying answers/patterns. In my Differential Equations class, CAS calculators were usedto verify the solution to the differential equation or the system of differential equations, plotslope field and solution trajectory. In my Multivariable Calculus class once we found thetangent plane, or quadratic approximation surface at some particular point, the graphs ofthe original function, its tangent plane and its quadratic approximation surface would bedrawn. So graphing calculators with CAS, if used correctly, help students visualize the mathconcepts they currently study. However, above all, when I show genuine concern for theirlearning the students respond well. A friendly comportment, a comfortable classroom at-mosphere where questions and comments are welcome, and flexible office hours have workedwell to this end. My students appreciate this time and dedication and I, not surprisingly,have found their appreciation very rewarding.

I believe that teaching applied mathematics is not just a science, but also an art. In myopinion, lecturing should not be a one-sided communication. The teachers should not onlylecture and present slides, but also ask the students for questions and opinions. They shouldalways keep observing the students’ responses to the topic taught, and should try to finddifferent strategies to get the students to understand and be able to apply the conceptsrelated to the topic taught. For example, when my daughter was in second grade, shewas mastering the definition of fractions. She had already been introduced to the idea offunctions; this opened the opportunity for me to teach her the definitions of the cosine andsine functions by first asking her to divide the length along the circumference of a unit circle(i.e. circumference is 2π) into fractions of 2π and locate the point. From that point, theline parallel to the y-axis intersects the x-axis at the cosine value of that fraction of 2π.Similarly, the line parallel to the x-axis intersects the y-axis at the sine value of that fractionof 2π. This way, I believe she not only knew the definitions, but was also able to find thesine and cosine values of some of the basic angles like the integer multiples of π/6 or π/4,for example. Of course, before she could do this, she had to have a solid foundation of thebasic operations such as the multiplication and division of fractions, adding and subtractingintegers as well as fractions. Also, she had to be familiar with the projection of a point in theCartesian coordinate system, that is the xy-plane, onto the x-axis and the y-axis. Moreover,she had to have a solid understanding of powers and radicals.

When my two kids, my daughter and my son, were in second grade, I taught them powers andradicals of numbers. Simultaneously, I would also teach them the cosine and sine functions.To demonstrate, once they knew that 25 = 2×2×2×2×2 = 32, I would tell them that

√64

is 641/2. I would ask them to factor 64, and they would get 64 = 8×8. I would then enlightenthem that the square root of 64 is nothing but 641/2 = (8×8)1/2 = 8. Using the definition ofthe power function with fraction powers, (8×8)1/2 is one of the two multiplied 8’s. This way,they could also find out that 645/6 = (2× 2× 2× 2× 2× 2)5/6 = 2× 2× 2× 2× 2 = 32, sincethey now understood that they had to take five out of the six 2’s in (2× 2× 2× 2× 2× 2).

When teaching my Linear Algebra class, I emphasize that the theory of basic linear al-gebra revolves around the concept that Linear Combination Relationship/Linear Depen-dence/Linear Independence of column vectors of same corresponding positions of a matrixand of its row equivalents are not changed through the process of row reduction. This can beused to inspect a matrix to obtain its reduced row echelon form without actually completingthe row reduction steps as seen in the following example. Ex1. Let’s consider how to find

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the reduced row echelon form of the following matrix A.

A =

0 3 −6 6 4 33 −7 8 −5 8 253︸︷︷︸~v1

−9︸︷︷︸~v2

12︸︷︷︸~v3

−9︸︷︷︸~v4

6︸︷︷︸~v5

27︸︷︷︸~v6

rref

−−−−→

1 0 −2 3 0 −240 1 −2 2 0 −70︸︷︷︸v1

0︸︷︷︸v2

0︸︷︷︸v3

0︸︷︷︸v4

1︸︷︷︸v5

6︸︷︷︸v6

︸︷︷︸

RREF (A)

.

We can observe that each of the non-pivot columns of A can be represented as a linearcombination of the pivot columns of A and the corresponding relation between each of thenon-leading columns and the leading columns of RREF (A) as follows:

−2~v1 − 2~v2 = ~v3 if and only if −2v1 − 2v2 = v3,

3~v1 + 2~v2 = ~v4 if and only if 3v1 + 2v2 = v4,

−24~v1 − 7~v2 + 6~v5 = ~v6 if and only if −24v1 − 7v2 + 6v5 = v6.

In the above Gauss-Jordan Elimination result:

A =

−20

−23 −6 6 4 3

3 −7 8 −5 8 253︸︷︷︸~v1

−9︸︷︷︸~v2

12︸︷︷︸~v3

−9︸︷︷︸~v4

6︸︷︷︸~v5

27︸︷︷︸~v6

rref

−−−−→

−21

−20 −2 3 0 −24

0 1 −2 2 0 −70︸︷︷︸v1

0︸︷︷︸v2

0︸︷︷︸v3

0︸︷︷︸v4

1︸︷︷︸v5

6︸︷︷︸v6

,it is clear that −2~v1 − 2~v2 = ~v3 if and only if −2v1 − 2v2 = v3,

A =

3

02

3 −6 6 4 33 −7 8 −5 8 253︸︷︷︸~v1

−9︸︷︷︸~v2

12︸︷︷︸~v3

−9︸︷︷︸~v4

6︸︷︷︸~v5

27︸︷︷︸~v6

rref

−−−−→

3

12

0 −2 3 0 −240 1 −2 2 0 −70︸︷︷︸v1

0︸︷︷︸v2

0︸︷︷︸v3

0︸︷︷︸v4

1︸︷︷︸v5

6︸︷︷︸v6

,it is clear that 3~v1 + 2~v2 = ~v4 if and only if 3v1 + 2v2 = v4,

A =

−240

−73 −6 6

6

4 33 −7 8 −5 8 253︸︷︷︸~v1

−9︸︷︷︸~v2

12︸︷︷︸~v3

−9︸︷︷︸~v4

6︸︷︷︸~v5

27︸︷︷︸~v6

rref

−−−−→

−241

−70 −2 3

6

0 −240 1 −2 2 0 −70︸︷︷︸v1

0︸︷︷︸v2

0︸︷︷︸v3

0︸︷︷︸v4

1︸︷︷︸v5

6︸︷︷︸v6

,it is also clear that −24~v1 − 7~v2 + 6~v5 = ~v6 if and only if −24v1 − 7v2 + 6v5 = v6.

We observe also that the pivot columns ~v1, ~v2 and ~v5 are all the linearly independent columnsfrom left to right of A, and these are corresponding to the leading columns v1, v2 and v5which are all the linearly independent columns from left to right of RREF (A).

Now, in order to inspect the matrix A and obtain RREF (A) we test column vectors of Afor linear independence and linear dependence from left to right as follows:

First we start with the first (left-most) column of A which is ~v1, if ~v1 is nonzero, then byitself, it is linearly independent that is ~v1 is the first pivot column of A, so the correspondingv1 in RREF (A) is the first leading column (the first column of identity matrix Im = I3). Ifnot, then the corresponding v1 in RREF (A) would be a zero vector.

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Then each step i of the next n − 1 steps we will test if the ith column vector of A andthe previously found linearly independent columns (pivot columns ~vp1 , . . . , ~vpi) are linearlyindependent or not. If they are linearly independent then the corresponding ith columnvector of RREF (A) is the (pi+1)th column of Im (which is I3 in our case for this example).If the ith column vector of A is linearly dependent to all the previously found linearlyindependent (or pivot) columns ~vp1 , . . . , ~vpi of A, then we solve for cp1 , . . . , cpi ∈ R such that

~vi = cp1~vp1 + · · ·+ cpi~vpi and the ith column vector of RREF (A) is

cp1...cpi0...0

:

Step 1 Since ~v1 6= ~0 (by itself, ~v1 is linearly independent) then 033︸︷︷︸~v1

rref

−−−−→

100︸︷︷︸v1

Step 2 Since ~v2 is linearly independent to ~v1 then

0 33 −73︸︷︷︸~v1

−9︸︷︷︸~v2

rref

−−−−→

1 00 10︸︷︷︸v1

0︸︷︷︸v2

Step 3 Since ~v3 is linearly dependent to ~v1, ~v2 and −2~v1 − 2~v2 = ~v3 then −2v1 − 2v2 = v3,

therefore −20

−23 −6

3 −7 83︸︷︷︸~v1

−9︸︷︷︸~v2

12︸︷︷︸~v3

rref

−−−−→

−21

−20 −2

0 1 −20︸︷︷︸v1

0︸︷︷︸v2

0︸︷︷︸v3

Step 4 Since ~v4 is linearly dependent to ~v1, ~v2 and 3~v1 + 2~v2 = ~v4 then 3v1 + 2v2 = v4,

therefore 3

02

3 63 −7 −53︸︷︷︸~v1

−9︸︷︷︸~v2

−9︸︷︷︸~v4

rref

−−−−→

3

12

0 30 1 20︸︷︷︸v1

0︸︷︷︸v2

0︸︷︷︸v4

Step 5 Since ~v5 is linearly independent to ~v1, ~v2, therefore

0 3 43 −7 83︸︷︷︸~v1

−9︸︷︷︸~v2

6︸︷︷︸~v5

rref

−−−−→

1 0 00 1 00︸︷︷︸v1

0︸︷︷︸v2

1︸︷︷︸v5

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Step 6 Since ~v6 is linearly dependent to ~v1, ~v2, ~v5 and −24~v1 − 7~v2 + 6~v5 = ~v6 then −24v1 −7v2 + 6v5 = v6, therefore

−240

−73

6

4 33 −7 8 253︸︷︷︸~v1

−9︸︷︷︸~v2

6︸︷︷︸~v5

27︸︷︷︸~v6

rref

−−−−→

−241

−70

6

0 −240 1 0 −70︸︷︷︸v1

0︸︷︷︸v2

1︸︷︷︸v5

6︸︷︷︸v6

Then we put all the corresponding columns of A and RREF (A) back together toobtain

A =

0 3 −6 6 4 33 −7 8 −5 8 253︸︷︷︸~v1

−9︸︷︷︸~v2

12︸︷︷︸~v3

−9︸︷︷︸~v4

6︸︷︷︸~v5

27︸︷︷︸~v6

rref

−−−−→

1 0 −2 3 0 −240 1 −2 2 0 −70︸︷︷︸v1

0︸︷︷︸v2

0︸︷︷︸v3

0︸︷︷︸v4

1︸︷︷︸v5

6︸︷︷︸v6

︸︷︷︸

RREF (A)

.

A CAS calculator can be used to check the work of each step and the final answer.The following figure is the screen shot of a CAS calculator used for checking theRREF (A).

Properties of eigenvalues and corresponding eigenvectors are also taught in my Linear Alge-bra class and are used for Eigenvalue Inspection of a matrix. A CAS calculator is used forchecking the answer.

Ex2. Find all eigenvalues and corresponding eigenvectors of A =

4 −5 11 0 −10 1 −1

. Since

A~v = λ~v and ~v 6= ~0 then ~v 6= ~0 ∈ Null space of (A− λI) or (A− λI)~v = ~0 with ~v 6= ~0, this

means a nonzero linear combination of column vectors of A = ~0 therefore column vectors of(A − λI) are linearly dependent if and only if λ is an eigenvalue of A. Using this fact, wecan inspect the following matrix

A− 0I =

−14 −1−5 1

1 0 −10 1 −1

we see that λ1 = 0 is an eigenvalue of A, and

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A− 1I =

−33 −2−5 1

1 −1 −10 1 −2

we also see that λ2 = 1 is an eigenvalue of A. Since tr(A) = 3 = λ1︸︷︷︸0

+ λ2︸︷︷︸1

+λ3, therefore

λ3 = 2. Therefore,

Eigenvalue A Corresponding Eigenvector

λ1 = 0 ~v1 =

111

λ2 = 1 ~v1 =

321

λ3 = 2 ~v1 =

731

Where the last eigenvector can be easily seen from

A− 2I =

−72 −3−5 1

1 −2 −10 1 −3

.

A CAS calculator can be used to check each of the answers, reducing the big amount oftedious calculation, and let students focus only on theory. The following are screen shots ofa CAS calculator used for checking each eigenvalue and its corresponding eigenvector of A.

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When teaching my Differential Equations class, I use a graphing calculator with CAS tohelp students understand theory since it reduces the calculation including differentiationsand integrations tremendously.

When teaching my Multivariable Calculus class, one very big problem that students faceis how to sketch or to figure out the shape of the surface given from its equation withoutusing a graphing calculator, especially hyperboloid surfaces. I then created a technique thatworks for any equation that has the terms of u and v only in the form of u2 + v2, for (u, v)that could be any of (x, y), (x, z), (y, z). The technique is to use polar coordinates to helpchange an equation of three variables that contain u2 + v2 into an equation of two variablesby substituting r2 = u2 + v2. Then from the graph–a curve–of the corresponding equationof two variables, we can rotate the curve around some appropriate axis to obtain the shapeof the surface of the original three-variable equation. For example, if the question is askingfor students to sketch the graph of the surface x2 + y2 − z2 = 1. Treating z as the outputand (x, y) as input, and each time we see x2 + y2, by applying polar coordinates, we canreplace x2 +y2 by r2, the equation then becomes r2−z2 = 1. Graphing this hyperbola curver2 − z2 = 1 in the rz-plane by hand, students should note the following

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(1) r cannot be zero,(2) if z = 0 then r = 1 (r 6= −1 because in polar coordinates that we use in Multivariable

Calculus, r ≥ 0), that is we know that (1, 0) is on the curve r2−z2 = 1 in the rz-plane,(3) as r →∞, z → ±r.

Since r =√x2 + y2 represents the distance r = d

xyz

, 0

0z

, especially if x = 0 then

r =√y2 = ±y, that is the r-axis can represent the positive y-axis or it can represent the

negative y-axis or if y = 0 then r =√x2 = ±x, that is the r-axis can represent the positive

x-axis or it can represent the negative x-axis or even along the straight line y = kx, we would

have r =√x2 + (kx)2 = d

xkxz

, 0

0z

, so we can rotate the curve of r2 − z2 = 1

around the z-axis to obtain the surface x2 + y2 − z2 = 1.

Linearization of function of two or more variables can be combined – added, subtracted,multiplied, divided; and composed. When I ask my Multivariable Calculus students to finda tangent plane to a surface at some particular point or to linearize a function around someparticular point, I usually ask them if they can find even easier ways to get the answeror somehow use inspection to come up with the answer or use inspection to check if thealgebraic calculation makes sense or has any mistakes. For example, Ex1. Find a linearapproximation L(x, y) of f(x, y) =

√x+ cos4(y) for (x, y) near (0, 0). Since f(x, y) ≈

f(a, b) + fx(a, b)(x − a) + fy(a, b)(y − b) with (a, b) = (0, 0), the linear approximation off(x, y), L(x, y), will have a term that contains a factor of (x− a) = (x− 0) = x and a termthat contains a factor (y − b) = (y − 0) = y. Now, cos(y) ≈ 1 − y2/2 so by multiplyingcos2(y) ≈ 1 − y2 and cos4(y) ≈ 1 − 2y2 ≈ 1 for all y near 0 (this linear approximationof cos4(y) ≈ 1 can be even easier obtained with graphing cos4(y) near y = 0). So our

f(x, y) =√x+ cos4(y) ≈ (1+x)1/2 ≈ 1+x/2 = L(x, y). This kind of question can be easily

done with inspection.

Quadratic approximation can also be done in a similar way. Ex2. Find a quadratic approx-imation Q(x, y) of f(x, y) =

√x+ cos4(y) for (x, y) near (0, 0). Since f(x, y) ≈ f(a, b) +

fx(a, b)(x−a)+fy(a, b)(y−b)+fxx(a, b)

2(x−a)2+fxy(a, b)(x−a)(y−b)+

fyy(a, b)

2(y−b)2 with

(a, b) = (0, 0), the quadratic approximation of f(x, y), Q(x, y), will have a term that contains

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a factor of (x− a) = (x− 0) = x, a term that contains a factor of (x− a)2 = (x− 0)2 = x2,a term that contains a factor of (y − b) = (y − 0) = y, a term that contains a factor(y−b)2 = (y−0)2 = y2 and a term that contains a factor of (x−a)(y−b) = (x−0)(y−0) = xy.Now, cos(y) ≈ 1−y2/2 so by multiplying cos2(y) ≈ 1−y2 and cos4(y) ≈ 1−2y2 for all y near 0.

So our f(x, y) =√x+ cos4(y) ≈ (1−2y2+x)1/2 ≈ 1+(1/2)(−2y2+x)+

(1/2)(1/2− 1)

2!(−2y2+

x)2 ≈ 1 + x/2 − x2/8 − y2 = Q(x, y). Then I would draw and show the surface z =√x+ cos4(y), its tangent plane z = L(x, y) and its quadratic approximation z = Q(x, y),

for all (x, y) near (0, 0).

Ex3. Find a quadratic approximation Q1(x, y) of f(x, y) = cos(x) cos(y) near (x, y) = (0, 0)and Q2(x, y) near (x, y) = (π/2, π/2). Since cos(u) ≈ 1 − u2/2, for all u near 0 andsin(u) ≈ u, for all u near 0. We then have f(x, y) = cos(x) cos(y) ≈ (1− x2/2)(1− y2/2) ≈1− x2/2− y2/2 = Q1(x, y), for all (x, y) near (0, 0). Also we have

f(x, y) = cos(x) cos(y)

= cos[π

2+ (x− π

2)]

cos[π

2+ (y − π

2)]

=(

cosπ

2cos(x− π

2)− sin

π

2sin(x− π

2))(

cosπ

2cos(y − π

2)− sin

π

2sin(y − π

2))

= sin(x− π

2) sin(y − π

2)

≈ (x− π

2)(y − π

2) = Q2(x, y), for all (x, y) near (π/2, π/2)

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The graphs of the surface f(x, y) and its quadratic approximations Q1(x, y) around (0, 0)and Q2(x, y) around (π/2, π/2) would emphasize the local behavior of f(x, y) near (0, 0) andnear (π/2, π/2) and this would be a good leading example toward the topic of local maxima,local minima and saddle points.

Emphasizing understandingnot memorizing formulas isthe key of my teaching thatwould lead to knowledgeretention. I have a commonmistake that many studentswould always make whenmemorizing formulas forchanging of coordinates inthe following flux integralproblem.

Ex4. Find the flux of vector field ~F =

4x4y0

through the surface S: part of a paraboloid

surface z = 16− x2 − y2 with x2 + y2 ≤ 16 and oriented upward.

Flux of ~F through S =

∫S:~r(x,y),(x,y)∈R

(~F � d ~AS

)=

∫R

(~F � (~rx × ~ry)

)dxdy︸︷︷︸dAR

=

x=4∫x=−4

y=√16−x2∫

y=−√16−x2︸︷︷︸

R

4x

4y0

�

2x2y1

︸︷︷︸

|J1| hidden here

dydx︸︷︷︸dAR

=

x=4∫x=−4

y=√16−x2∫

y=−√16−x2︸︷︷︸

R

8(x2 + y2) dydx︸︷︷︸dAR

=

θ=2π∫θ=0

r=4∫r=0︸︷︷︸T

8r2 r︸︷︷︸|J2|

drdθ︸︷︷︸dAT

= 8

r=4∫r=0

r3 dr

· θ=2π∫θ=0

1 dθ

= 1024π,

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or if students parameterize directly with variables (r, θ) then they would get


∫S:~r(r,θ),(r,θ)∈T

(~F � d ~AS

)=

∫R

(~F � (~rr × ~rθ)

)drdθ︸︷︷︸dAT

=

θ=2π∫θ=0

r=4∫r=0︸︷︷︸T

4r cos θ

4r sin θ0

�

2r cos θ2r sin θ

1

r︸︷︷︸

|J |=|J1||J2| hidden here


=

θ=2π∫θ=0

r=4∫r=0︸︷︷︸T

8r2 r︸︷︷︸|J2|


= 8

r=4∫r=0

r3 dr

· θ=2π∫θ=0

1 dθ

= 1024π.

The common mistake that students would make is plugging in an extra r with drdθ, becausethey memorize that they also need an r for the jacobian without knowing that the jacobian

which including that r was already in the magnitude of ~rr × ~rθ =

2r cos θ2r sin θ

1

r as follows:


θ=2π∫θ=0

r=4∫r=0︸︷︷︸T

4r cos θ

4r sin θ0

�

2r cos θ2r sin θ

1

r︸︷︷︸

|J |=|J1||J2| hidden here

r︸︷︷︸extra r

because of misunderstanding


=

θ=2π∫θ=0

r=4∫r=0︸︷︷︸T

8r3 r︸︷︷︸|J2|


= 8

r=4∫r=0

r4 dr

· θ=2π∫θ=0

1 dθ

=16384

5π,

which is incorrect. This is why encouraging students to try to understand the material oftopics studied is better than memorizing formulas and plug and play.

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tom teaching portfolio 20150506

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dierent skills

computer engineering

biomolecular engineering

algebraic skills

electrical engineering

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civil engineering

mathematics department