High School Mathematics-Physics SMILE Meeting
02 December 2003
Notes Prepared by Porter Johnson

Fred J Schaal [Lane Tech HS, mathematics]        Graphing Parabolas: a Challenge
drew a graph of a parabola (concave upward) on the board. He then wrote down the formula for a parabola, y = ax2 + bx + c, for  a > 0. [The parabola is concave upward in this case.]  He then asked these two questions:

Q1:  For what value(s) of x does y = 0?
Q2For what value(s) of x does y take on its minimum value?  Why?
He then drew a horizontal line through the parabola, so that there were two zeros, x±. Since the parabola is (apparently) symmetric about its minimum (apex), we expect that xmin  = (x+ + x- )/2.  That is, the parabola is left-right symmetric about its apex. We may employ the quadratic formula to calculate the values of x at which y = 0:
x± = [ -b ± Ö(b2 -4ac) ] /(2a)

Note: some browsers will incorrectly render the square root sign (Ö).

For the case drawn, Fred pointed out that the average value of these two (real) zeros, x± . is xmin = -b/(2a).  Actually, this answer for the location of the minimum is correct, as Fred showed by differentiation:   Since y = ax2 + bx + c   ...
... the derivative is y' = 2ax + b = 0
at a minimum. Thus, the result xmin = -b/(2a) is always correct.  However, it definitely constitutes poor pedagogy as well as poor sportsmanship (what about moral turpitude?) to be forced to use calculus to solve simple algebra problems  Worse yet, it smacks of inelegance!  Fred thus posed the following question:

How do you show that xmin = -b/(2a) using simple algebra?
Porter Johnson calculated the minimum value of y:
ymin   =   y(xmin)   =   a [-b/(2a)]2 + b [-b/(2a)] + c   =   c -b2/(4a)
He then restated the problem by asking how to show that
y ³    ymin   =   c -b2/(4a) .

Who can answer either or both of these questions? More about this in the future!

Fred, you have laid down the gauntlet.  Thanks!

Walter McDonald  [CPS Substitute Teacher; X-ray technician -- Veterans Administration Hospital, North Chicago]       Fibonacci Number Programs on HP 48G Programmable Calculator
passed around information on two programs to calculate the Fibonacci numbers on the calculator, called FIB1 and FIB2.  The listings and other information from these programs (presumably) came from the HP 48G Series Advanced User's Reference Manualhttp://www.hpcalc.org/hp48/docs/books/aur.html. This is an extension of Walter's explanation of Fibonacci numbers in the Math-Phys SMILE meeting on 23 September 2003 [mp092303.html].  As explained there, the sequence of  Fibonacci numbers, F(n), is generated by iteration, starting from the seeds F(0)=0 and F(1) = 1, using the recursive formula F(n) = F(n-1) + F(n-2).  In particular, F(6) = 8, whereas F(10) = 55 and F(13) = 233.  The first program, FIB1 (Recursive Version), calculated and stored each Fibonacci number in an array, and then went on to the next one.  It took about 25.17 seconds to obtain F(13).  By contrast, the second program, FIB2 (Loop Version), stored only the last two Fibonacci numbers, and took just 0.0897 seconds to calculate F(13). This remarkable time difference was shown in a third program, FIBT (Comparing Program-Execution Time) which determines the execution time for each of the two programs.

Why are the execution times so dramatically different?  According to the handout, the Program FIB1 calculates intermediate values F(i) more than once, while Program FIB2 calculates each intermediate value F(i) only once.  Thus, the time required to calculate F(n) grows exponentially with n in FIB1, and linearly with n in FIB2FIB2 is obviously the "way to go", whereas FIB1 bogs down, even with a relatively fast CPU [central processing unit].

Porter Johnson pointed out that most currently available programmable computers are at least as sophisticated as the first commercially available mainframe computers, e.g. IBM 650. This means that (1) they can perform absolutely marvelous tasks for you if you know how to program them but (2) when a particular program crashes, you generally have to reboot the system before it will run again.  Note:  Programs that crash the computer will cause it to end up in computer limbo land, producing meaningless results!  In fact, no hand-held calculator is ever of any value, until and unless you know how to use it, and use it properly!  The more sophisticated the calculator, the easier it is to crash it and get nonsensical answers!

Very interesting, Walter!

John Bozovsky [Chicago Discovery Academy at Bowen  HS, physics]     Physics in the Film Clips
first posed the following questions:

Q1:  What happens when two bullets collide in mid-air with equal and opposite speeds?
Q2Can you pick yourself up by pulling on your bootstraps (or shoe tops, or "whatever")?  If so, how?

Ever alert to the presentation and display of concepts of physics in entertainment media, John showed us some film snippets that provide "Hollywood" answers to these basic questions.

Why is it that life seems more "real" in the movies than in everyday life?  We enjoyed this novel approach to teaching physics, John!

Professor Eduardo De Santiago  [IIT: Civil and Architectural Engineering]        Building Lighter, Stronger Bridges
This was without doubt Eduardo's most beautiful presentation yet! In preparation for the 28th Annual Chicago Regional Bridge Building Contests, teachers and their students joined our SMILE meeting for Eduardo's fifth annual presentation on building a strong yet light bridge. Eduardo's previous lectures were given in 1999 [ph120799.htm], 2000 [mp112100.htm], 2001 [mp112001.htm],and  2002 [mp111902.html].--- they dealt with the same ideas

He began by making a sketch on the board showing how a cave person would use a fallen tree as a primitive bridge to walk across a stream; a bridge takes a load from one point to another without breaking. Then he pointed out that a truss is a simple and efficient construction, relatively easy to analyze. He introduced the concepts of bending moment, tension, and shear as three important forces internal to a bridge, and he illustrated each concept with sketches and by using whiteboard erasers that he bent and otherwise stressed. From then on, one set of ideas led to another. Eduardo connected them with lucid sketches, eraser-bending, and articulate discussion. Simple ideas led to increasingly complicated combinations of ideas, and when we finally arrived at a typical truss bridge, we understood the physical ideas behind it. He did not write down any equations! He pointed out the need for symmetry (to deal with forces from any direction) and the need to minimize the number of joints. And to have fun!

When he had finished, he and some members of the Bridge Building Committee spent time with students and teachers one-on-one, answering questions. What a wonderful, phenomenological presentation, Eduardo! Thanks!  

Notes taken by Porter Johnson