My Top Ten Technology Teachable

MomentsDan Kennedy

Baylor School

Chattanooga, TN

T^3 International ConferenceChicago, IL

March 10, 2007

Moment #1:

The Magic Numbers for Retail Wine Pricing

Take the wholesale cost of a case of wine.

Add the excise tax ($1.21/gallon today).

Divide by 12 to get the bottle cost.

Add 30% retail markup.

Discount the price 10% (a store policy).

Oh, what he would only give for the ability to accomplish all this bothersome math in a single step!

(x + 1.21 * 2.37755)

= 0.0975 x + 0.28049

÷ 12

* (1 + .30) * (1 - .10)

So, I told him sagely, all you need to do is multiply your wholesale case cost by 0.0975 and add 0.28049.

I thought the man was going to cry.

Praying I would not make this look too simple, I did a little Algebra I with the help of my calculator…

Moment #2:

The Polar Graph of Sin 666θ

1994.

My student had just unwrapped his new TI-82 graphing calculator and wanted to graph “something cool.”

I told him to put it in POLAR mode and graph r = sin 6 θ.

He liked that so much that he graphed r = sin 66 θ.

Then, of course, he decided to graph r = sin 666 θ.

The graph was the same as the graph of r = sin 6 θ !

What was going on?

The explanation is not so simple. It starts with the default θ-step in the POLAR window, which is π/24.

This would explain why the graph of r = sin 6 θ would be the same as (for example) r = sin 486 θ, since sin 486 θ = sin (6 θ + 480 θ).

θ sin 6θ cos 480θ cos 6θ sin 480θ sin 486θ

0.1309 0.7071 1 0.7071 0 0.7071

0.261799 1.0000 1 0.0000 0 1.0000

0.392699 0.7071 1 -0.7071 0 0.7071

0.523599 0.0000 1 -1.0000 0 0.0000

0.654498 -0.7071 1 -0.7071 0 -0.7071

0.785398 -1.0000 1 0.0000 0 -1.0000

0.916298 -0.7071 1 0.7071 0 -0.7071

1.047198 0.0000 1 1.0000 0 0.0000

1.178097 0.7071 1 0.7071 0 0.7071

1.308997 1.0000 1 0.0000 0 1.0000

But 666 – 6 = 660, which is not an even multiple of 24.

In fact, the calculator does not produce the same values for sin 666 θ as it does for sin 6 θ.

It actually produces the opposite values, which give the same graph in the end.

But why it produces the opposite values is still far from obvious!

θ sin 6θ cos 660θ cos 6θ sin 660θ sin 666θ

0.1309 0.7071 0 0.7071 -1 -0.7071

0.261799 1.0000 -1 0.0000 0 -1.0000

0.392699 0.7071 0 -0.7071 1 -0.7071

0.523599 0.0000 1 -1.0000 0 0.0000

0.654498 -0.7071 0 -0.7071 -1 0.7071

0.785398 -1.0000 -1 0.0000 0 1.0000

0.916298 -0.7071 0 0.7071 1 0.7071

1.047198 0.0000 1 1.0000 0 0.0000

1.178097 0.7071 0 0.7071 -1 -0.7071

1.308997 1.0000 -1 0.0000 0 -1.0000

Moment #3:

Charles Mayfield’s Trig Quiz

1992.

My students and I are still adapting to graphing calculators, but my inclination is to let them use them all the time.

As a final review of trig identities, I give them my annual “matching quiz from Hell.”

Here it is…

Match each expression on the left with the expression on the left for which it is an identity. Two letters will not be used.

1. sec2

A. sin

2. sin tan cos B. sin

3. 2 2tan seccos

C. cos

4. 2 2sin cos

csc

D. –cos

5. 2sin cos2 2

E. tan

6. (tan sec )(1 sin ) F. –tan

7. seccsc

G. cot

8. 2sec 1

tan H. –cot

9. 21 2cos2

I. sec

10. cos4 cos3 sin 4 sin3sin 4 cos3 cos4 sin3

J. –sec

K. csc L. –csc

We graded it in class, and the grades were abysmal.

One student, Charles Mayfield, had a perfect 10.

When the other students looked at him in skeptical amazement, he simply grinned and held up his TI-81.

Charles had graphed every expression.

The class gave him a round of applause.

Now older and wiser, I have learned to put the following logo on some of my quizzes:

Sometimes, though, I deliberately leave the calculator back door open, just to reward the Charles Mayfields in my classes.

His classmates never doubted that he deserved his 10, and neither did I.

Moment #4:

Baylor School’s Graduated GPA

The Challenge:

Design a sliding scale that our school could use to convert our numerical (percentage) grades to grade-point averages on a 4-point scale.

The assumptions I made:

1.Our lowest D (65) should get 1.0.

2.An average A (95) should get 4.0.

3.The GPA curve should be steeper at the low end than at the high end

I decided to use a power function of the form

1/65( ) 3 1.

30

pxG x

(65, 1.0)

p-value 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0

Grade  

65 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0

70 1.5 1.6 1.7 1.8 1.8 1.9 2.0 2.0 2.1 2.2 2.2

75 2.0 2.1 2.2 2.3 2.4 2.4 2.5 2.6 2.6 2.7 2.7

80 2.5 2.6 2.7 2.8 2.8 2.9 2.9 3.0 3.0 3.1 3.1

85 3.0 3.1 3.1 3.2 3.2 3.3 3.3 3.4 3.4 3.4 3.4

90 3.5 3.5 3.6 3.6 3.6 3.7 3.7 3.7 3.7 3.7 3.7

95 4.0 4.0 4.0 4.0 4.0 4.0 4.0 4.0 4.0 4.0 4.0

100 4.5 4.5 4.4 4.4 4.3 4.3 4.3 4.3 4.3 4.3 4.2

104 4.9 4.8 4.7 4.7 4.6 4.6 4.5 4.5 4.5 4.4 4.4

I gave the faculty a choice of curves for various p-values, and the runaway winner was p = 1.7.

Baylor continues to use this sliding conversion today, nearly 25 years later. It continues to provide teachable moments for educating students, faculty, administrators, and, of course, parents!

1/1.765( ) 3 1

30x

G x

Moment #5:

“Magic” Math E-mails from the Clueless

How many of us have received this e-mail from friends wondering what sorcery is behind this trick?

1. Pick the number of times a week that you would like to have

dinner out (try more than once and less than 10). 2. Multiply this number by 2. 3. Add 5. 4. Multiply this number by 50. (You might need a calculator). 5. If you have already celebrated your birthday in 2007, add

1757. Otherwise add 1756. 6. Now subtract the four digit year that you were born.

You should have a three digit number left! The first digit is your original number of how many times you want to eat out each week. The second two numbers are: YOUR AGE. (Oh Yes it is!!!!!). Amazingly, 2007 is the ONLY YEAR that this incredible trick will work!

This is a wonderful Teachable Moment for algebra teachers!

1. Let d be the number of days a week I want to have

dinner out. 2. Double it: 2d. 3. Add 5: 2d + 5. 4. Multiply by 50: 100d + 250.

(Who needs a calculator?) 5. Add (for me) 1756: 100d + 2006. 6. Subtract (for me) 1946: 100d + 60. (Don’t try this if you’re older than 99!)

Moment #6:

The sine of π/6 degrees

I was showing my Precalculus students how they could leave the calculator in radian mode and use the degree symbol in the ANGLE menu if they needed a temporary mode switch:

Then a student decided to compute the sine of π/6 degrees:

“Hey, look,” he cried, “the sine of π/6 degrees is the same as the sine of 30 radians. Isn’t that neat?”

Declaring the calculator to be mistaken is insufficient at a time like this. To do so is to waste a Teachable Moment.

Eventually, we figured it out!

Moment #7:

Multiple-Choice #36 from the 1988 BC Exam

I gave my Calculus class a quiz that included Multiple-Choice item #36 from the 1988 BC Calculus examination.

36. Let R be the region between the graphs of y = 1 and y = sin x from x = 0 to x = 2

.

The volume of the solid obtained by revolving R about the x-axis is given by

(A) 2

02 sinx x dx

(B) 2

02 cosx x dx

(C) 22

01 sin x dx

(D) 22

0sin x dx

(E) 22

01 sin x dx

1

0 1

The Region R

π

2

1

0 1

Area of Washer = 21- sin ( )π x

36. Let R be the region between the graphs of y = 1 and y = sin x from x = 0 to x = 2

.

The volume of the solid obtained by revolving R about the x-axis is given by

(A) 2

02 sinx x dx

(B) 2

02 cosx x dx

(C) 22

01 sin x dx

(D) 22

0sin x dx

(E) 22

01 sin x dx

So, the correct response is E.

In 1988, this was the distribution of responses:

A 6% B 1% C 18% D 21% E 48% 0 6%

The difference from the 1988 scenario was that my students had calculators.

One girl, who had incorrectly chosen D, computed her integral and compared it to the value of the correct integral in E.

She pointed out that they were the same!

This was not even difficult to explain! Consider the graphs of the two integrands on the interval [0, π/2]:

Needless to say, I gave the girl credit for D. My friends at ETS were totally unaware that there had been a double key, so the 21% who chose D in 1988 were apparently not so lucky.

Moment #8:

Numerical Integration on the TI Calculators

One of the hardest calculus topics to teach in the old days was Riemann sums.

They were hard to draw, hard to compute, and (many felt) totally unnecessary.

That was why most of us quickly moved on to antiderivatives, which is how we wanted students to do integrals.

Needless to say, when we came to the Fundamental Theorem, students found it to be the greatest anticlimax in the course.

Integration and differentiation are reverse operations? Well, duh.

Then along came the TI graphing calculators. Using the integral utility in the CALC menu, students could actually see an integral accumulating value from left to right along the x-axis, just as a limit of Riemann sums would do:

Moment #9:

Seeing Power Series Converge

One of the most powerful visualizations in mathematics is the spectacle of the convergence of Taylor series. Here are the Taylor polynomials for sin x about x = 0:

3

3 5

3 5 7

3!

3! 5!

3! 5! 7!.

x

x x

x x x

etc

y x

y x

y x

y x

Here are their graphs, superimposed on the graph of y = sin x:

Moment #10:

We All Use Math Every Day

NUMB3RS Activity: A Party of Six

Episode: “Protest”

Topic: Graph Theory and Ramsey NumbersGrade Level: 8 - 12 Objective: To see how a complete graph with edges of two colors can be used to model acquaintances and non-acquaintances at a party.Time: About 30 minutesMaterials: Red and blue pencils or markers, paper

The first six complete graphs:

If two people (A and B) are at a party, there are only two possibilities: either A and B know each other, or A and B do not know each other. Draw the two possible graphs below.

• A • A

• B • B

Draw all of the possible 3-person party graphs for A, B, and C below.

There are 64 possible 4-person party graphs for guests A, B, C, and D (Why?), but you will not be asked to draw them all. Instead, draw the 8 possible 4-person party graphs in which A, B, and C all know each other. We say A, B, and C are mutual acquaintances.

It is actually possible to color the edges of a 5-person party graph in such a way that there are neither three people that are mutual acquaintances nor three people that are mutual non-acquaintances. Can you do it?

It is an interesting fact that every party of 6 people must contain either three mutual acquaintances or three mutual non-acquaintances.

Start with guest A.

Among the remaining 5 guests, A has either at least three acquaintances or at least three non-acquaintances.

Case 1: Suppose A has three acquaintances: B, C, D.

A •

D • • C

• B A •

D • • C

• B

If any two of these are acquainted, we have three mutual acquaintances.

If no two of these are acquainted, we have three mutual non-acquaintances!

Case 2: Suppose A has three non-acquaintances: B, C, D.

A •

D • • C

• B A •

D • • C

• B

If any two of these are non-acquainted, we have three mutual non-acquaintances.

If no two of these are non-acquainted, we have three mutual acquaintances!

The Ramsey Number R(m, n) gives the minimum number of people at a party that will guarantee the existence of either m mutual acquaintances or n mutual non-acquaintances.

We just constructed a proof that R(3, 3) = 6.

Ramsey’s Theorem guarantees that R(m, n) exists for any m and n.

Intriguingly, there is still no known procedure for finding Ramsey numbers!

It has actually been known since 1955 that R(4, 4) = 18.

We do not know R(5, 5), but we do know that it lies somewhere between 43 and 49.

All we really know about R(6, 6) is that it lies somewhere between 102 and 165.

There is a cash prize for finding either one.

The great mathematician Paul Erdös was fascinated by the difficulty of finding Ramsey numbers. Here’s what he had to say:

“Imagine an alien force vastly more powerful than us landing on Earth and demanding the value of R(5, 5) or they will destroy our planet.

In that case, we should marshal all our computers and all our mathematicians and attempt to find the value.

But suppose, instead, that they ask for R(6, 6).

Then we should attempt to destroy the aliens.”