Transcript My Top Ten Technology Teachable Moments

My Top Ten
Technology
Teachable Moments
Dan Kennedy
Baylor School
Chattanooga, TN
T^3
International
Conference
Chicago, 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!
Praying I would not make this look
too simple, I did a little Algebra I with
the help of my calculator…
(x + 1.21 * 2.37755) ÷ 12 * (1 + .30) * (1 - .10)
= 0.0975 x + 0.28049
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.
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. sec   
2

2. sin  tan   cos
tan 2   sec 2 
3.
cos 
2
sin   cos 2 
4.
csc


 
 cos  
2
2
5.
2sin  
6.
(tan   sec )(1  sin  )
sec

csc
sec2   1
tan 
7.
8.

 

2
9. 1  2cos2 
10.
cos 4 cos3  sin 4 sin 3
sin 4 cos3  cos 4 sin 3
A. sin 
B.
 sin 
C. cos 
D. –cos 
E. tan 
F. –tan 
G. cot 
H. –cot 
I. sec 
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/ p
 x  65 
G( x )  3 
  1.
 30 
(65, 1.0)
I gave the faculty a choice of curves
for various p-values, and the runaway
winner was p = 1.7.
p-value
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
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
Grade
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/ 1.7
 x  65 
G( x )  3 

 30 
1
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:
MultipleChoice #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 =
The volume of the solid obtained by revolving R about the x-axis is given by

(A) 2  2 x sin x dx
0

(D)   sin x dx
2
0
2

(B) 2  2 x cos x dx
0

(E)   2 1  sin 2 x  dx
0

(C)   2 1  sin x  dx
0
2

2
.
1
The Region R
0
1
π
2
1
0
1
2
π
1sin
( x )
Area of Washer = 
36. Let R be the region between the graphs of y = 1 and y = sin x from x = 0 to x =
The volume of the solid obtained by revolving R about the x-axis is given by

(A) 2  2 x sin x dx
0

(D)   2 sin x dx
2
0

(B) 2  2 x cos x dx
0

(C)   2 1  sin x  dx
2
0

(E)   2 1  sin 2 x  dx
0
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%

2
.
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:
yx
x3
y  x
3!
3
5
x x
y  x 
3! 5!
3
5
7
x x x
y  x  
3! 5! 7!
etc.
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
Numbers
Grade Level: 8 - 12
Objective: To see how a complete graph
with edges of two colors can be used to
model acquaintances and nonacquaintances at a party.
Time: About 30 minutes
Materials: 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.
•
•
B
•
A
•
B
A
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 4person 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 nonacquaintances. 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•
•B
A•
•B
D•
•C
D•
•C
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 nonacquaintances: B, C, D.
A•
•B
A•
•B
D•
•C
D•
•C
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.”