High School Mathematics-Physics SMILE Meeting
Mathematics: Calculators

09 December 1997: Bill Lilly [Kenwood HS]
He showed a new calculator-based-ranger [TI-CBR], which is used with TI-82 calculators to detect motion and measure distance. It can be incorporated into programs on the calculator for automatic data-taking. The device represents an improvement on the older CBL devices. He talked about an application of measuring a motion of a pendulum. The minimum distance for detection is about 0.5 meter.  We ran out of time during Bill's presentation, and we hope he can return to show it next year.

07 April 1998  Bill Lilly [Kenwood High School; Math Dept]
He demonstrated the capabilities of the TI-CBR Calculator-Based Ranger, which is a sonic motion detector that provides information in digital and easily convertible form. The device is made by Texas Instruments for use with their calculators TI-82, 83, 85, 86, or 92. The price is \$90, and seems to be dropping with time. The "innards" of the device are the same as a Polaroid auto-focus system. The apparatus was demonstrated by hanging a bottle on a swing and swinging it back and forth as a pendulum. He was able to obtain [interpolated] numerical values of x, v, and a at time intervals of about 0.1 seconds, so that motions could be studied quantitatively, and graphs easily generated.

02 March 1999: Fred Schaal [Lane Tech HS Math]
He investigated determinants and the magic squares. Using the TI graphing calculator's internal ability to solve determinants he put in a 5x5 array and showed some to the abilities of the TI graphing calculator. Fred tried the numbers forming a "golden square" as well as a Fibonacci sequence, and in neither case did the determinant vanish.

Porter commented that the result was not what Fred expected because the elements did not have the proper relationship where an element or the ray wasn't a factor of the other numbers. Determinants can be thought of in geometric terms, where the points can be identified as vectors, and the vectors are added. there can be a solution when they reside in the same plane. You can write two equations in two unknowns in the form

A1 · X = B1
A2 · X = B2

The object is to determine the two dimensional vector X when its scalar products with the vectors A1 and A2 are given. There is no solution when A1 and A2 are parallel to one another, or when the determinant of coefficients vanishes. So it goes...

20 April 1999: FJ Schaal [Hi Tech Low Tech Lane Tech High School]
He first placed a grid on the board by using the chart that has the holes to allow chalk dust to produce a grid. He placed points (2,13) and (16,3) on the grid and used the formula

Dist = Sqrt[(x1-x2)2+(y1-y2)2]

He explained the steps and keys to press on the calculator: ( ( 2 -16 ) ( 2 + ( 13 -3 ) ( 2 ) ( = distance. Now, adding an additional point of (-12,-6), using the above formula we get distances:

Sqrt(356) = 18.868
Sqrt(981) = 31.320
Sqrt(557) = 23.601

Hero [HERON] formula for the area of a triangle of sides a, b, c:

Area = Ö(s(s-a)(s-b)(s-c))
where s = (a+b+c)/2

This gives 224 [exactly!!], whereas by graphing and counting the number of fully enclosed squares plus 1/2 the number of partially enclosed squares we get 219.

04 May 1999: Porter Johnson [IIT]
Porter took off on the use of the calculator and commented about not rounding off! By using integers and keeping the calculation to as many places as possible we may get an answer that will result in integers. If you round off the answer will be an decimal, and hide the fact that the answer from a problem starting with integers will be integers. You may not want to round the working numbers in a calculators to a few significant places and thus hide some of significance of the answer. The example consisted in calculating the area of a triangle with vertices at integer-valued x- and y-coordinates. The area of the triangle, which will come out as either an integer or half-integer, can be computed geometrically by enclosing a rectangle around the triangle, and calculating the area of the rectangle and the other right triangular pieces. Simple, non??

28 September 1999: Fred Schaal (Lane Tech)
Fred drew our attention to inequalities. For example, for a straight line y = m x + b, we might have y = (1/2) x + 2. On a graph of y vs x, below the line y < (1/2) x + 2, and above the line y > (1/2) x + 2. Our hand-held calculators seem not to enable us to display a band of (x, y) pairs that lie between a minimum and maximum set of values. Is there any way to do such a thing on any hand-held calculator? Interesting question!

12 October 1999: Walter Macdonald [CPS Substitute; Medical Technician]
He demonstrated the versatility of the HP48GX Programmable Calculator , with 128K of internal RAM [random access memory] connection with the software package SPICE48, which allows one to program circuit diagrams [although somewhat tediously, it seems] and to specify the voltage source as DC, sinusoidal AC, sawtooth, or "whatever". The calculator permits display of output currents and such in graphical form, in order to gain insight into the effects of changing resistances, capacitances, and inductances, or switching in diodes. The device is better suited to providing its gentle user with graphical insight, rather than mere numbers.

Walter also commented that the new x-ray machine in his hospital has a digital interface, so that familiarity with PCs/Macs is required in routine usage. Porter observed that the world is "going digital", giving these instances:

• General Motors has described the modern automobile as essentially an electronic device, rather than a mechanical one, as it has always been.
• The hospitals of today seem like, say, the high energy physics labs a few years back, with sophisticated electronics, computer-driven operations, schemes for fast digital tomography [MRI, PET, and CT; see http://www.islandscene.com/Article.aspx?id=56], and automated chemical analyses.
• You can buy a car, order merchandise, do banking, find a mate, and pay taxes electronically.

09 November 1999: Bill Colson (Morgan Park HS - Math)
drew 2 lines that intersect to form an angle. In his geometry class Bill uses a book titled Line Design. Some students have difficulty understanding what an angle is, and by getting them involved in such constructions using straight lines, they develop a feeling for the idea of angle. Example: start with two straight lines intersecting at 90o, and mark off equal divisions beginning at the intersection. Draw two additional lines connecting the divisions (1 & 2) and (2 & 1). The two lines intersect at one point. Now repeat this on a new drawing, making 4 lines connecting (4 & 1), (3 & 2), (2 & 3) and (1 & 4). These 4 lines intersect at 3 points. By repeating this process next for divisions 6, 5, 4, 3, 2,1 - it becomes quite clear that the intersecting straight lines come to more and more closely define a curved line. And the idea of approaching something as the limit of a process repeated an "infinite" number of times becomes an experience that is useful in more advanced math. Bill passed around examples of beautiful art work his students made using these ideas. Very nice!

29 February 2000: Fred Schaal (Lane Tech HS)
showed us classroom use of plotting calculators, and he used one to project to the screen so we could see what was going on. Betty Roombos followed Fred's directions to do this, and it worked well. Fred had her (and others with calculators he had passed out) plot the equations of straight lines and find their intercepts with the x and y axes. He used number pairs to determine a set of lines:

 x-intercept y-intercept line 10 1 y = -(10/1)x+10 9 2 y = -(9/2)x+9 8 3 y = -(8/3)x+8 7 4 y = -(7/4)x+7 6 5 y = -(6/5)x+6 5 6 y = -(5/6)x+5 4 7 y = -(4/7)x+4 3 8 y = -(3/8)x+3 2 9 y = -(2/9)x+2 1 10 y = -(1/10)x+1

The intersections of successive pairs of lines were determined by using the plotting computers, and one could even zoom up close to the intersections to see more detail and better determine the actual number pair locating each intersection. He then used the intersection points as data, and did a linear, quadratic, and cubic fit to those points on a calculator. He compared the results with the "tangent curve" described below by Porter Johnson. A fine phenomenological math lesson!

29 February 2000: Comments by Porter Johnson (IIT)
The formula for line segment lying in first quadrant and intercepting the x-axis at x = a and intercepting the y-axis at y = b - a, where 0 < a < b, is

x/a + y/(b-a) = 1

Let us regard the parameter b as being fixed, while a varies continuously between the values 0 and b. A solid region in the first quadrant is filled by these lines. To determine the boundary of that region, we solve the above relation for y, to obtain

y(a) = (b - a) (1 - x/a) = b + x - a - (bx)/a

In this relation, keep x fixed, and vary a. The maximum value of y under such variation can be calculated by setting the derivative dy/da to zero:

dy/da = - 1 + (bx)/a2 = 0 ,
so that
a = Ö(bx)

We substitute this value of a into the expression for y(a) to obtain

ymax = b + x - 2 Öbx = (Ö b - Ö x)2.

To show that this is indeed the maximum value of y(a), calculate the quantity ymax - y(a):

ymax - y(a) = a + bx/a - 2 Ö(bx) = ( a - Ö(bx) )2 /a .

Clearly, the right side is non-negative, and it is zero when a is chosen as above. The formula for ymax gives the greatest value that y can have for a given value of x, and thus lies at the top of the region traced out by the straight lines described above. Thus, the top of the region is given by

y = (Ö b - Ö x )2.
or
Ö x + Ö y = Öb.

This curve, which represents the envelope of all the straight lines, can also be written as

(x - y)2 + b2 = 2 b (x + y)

The curve is a parabola with the line of symmetry [axis] lying along the line x=y. It is symmetric under interchange of the variables x and y. The point closest to the origin has coordinates

x = b/4 = y .

This is the "symmetry point of the parabola", and is the "tangent point" for the curve with a = b/2, namely

2x/b + 2y/b = 1 .

Every other point on this curve is a "tangent point" for one and only one of the straight lines described above, which have intercepts a and b-a, respectively.

Here is an Excel-generated image of the lines and the asymptotic curve:

Incidentally, these "envelope curves" occur frequently in geometrical optics, in which light rays move in straight lines in a uniform medium. Clearly, the "bundle" formed by all light rays can have a nontrivial structure. The boundary of that "bundle" is called a caustic in geometrical optics. As an example, see http://www.math.harvard.edu/archive/21a_spring_06/exhibits/coffeecup/index.html The Coffeecup Caustic. Here is a brief description of the effect, taken from that reference.

You are drinking form a cylindrical cup in the sunshine. Sometimes, when the sun shines into the cup, you can see a crescent of light as the sunshine reflects from the inside of the cup onto the surface of the drink. A picture of a real cup is shown, and you can do your own on-line computer simulations of the effect. Check it out!

14 March 2000: Porter Johnson (IIT Physics)
explained Coffee Cup Caustics to us with some mathematical detail, but nicely presented.

Coffee Cup Caustics

The Coffee Cup Caustic is shown on the website http://www.math.harvard.edu/archive/21a_spring_06/exhibits/coffeecup/index.html, on which the reflected image from a real coffee cup is shown, as well as a "Monte Carlo" simulation of the event.

Let us suppose that rays parallel to the y-axis strike an upper semi-circle of radius R from the inside, and are reflected. If we let the angle between the reflection point be q, the coordinates of the point of reflection are x = R cos q and y = R sin q. The angle of incidence of the ray, relative to the normal, is p/2 - q, and the angle of reflection has that same value, as well.

The reflected ray travels at an angle p/2+2 q relative to the positive x-axis, and strikes the circle again at a point an angle 3 q from the horizontal, at the point (x = R cos3 q, y = R sin 3 q). The equation for the reflected ray is

y - R sin q = (x - R cos q) tan (p/2 + 2q).
or
y = R sin q - (x - R cos q) / tan 2q.

As q varies, we generate a series of straight lines. To find the envelope of those straight lines, we must determine the maximum value that y can have for a given value of x, and the appropriate choice of q, by setting the derivative dy/dq equal to zero. Thus we obtain

dy/dq = R cos q - R sin q/ tan 2q + 2 (x - R cos q)/ sin2 2q = 0.

Thus, we obtain the relation

x = R (3 cos q + cos 3 q ) / 4 = R cos3 q.
and
y = R sin q - (x - R cos q)/ tan 2q = R sin q (1/2 + cos2 q ) ,
or,
y(x) = R ( 1/2 + (x/R)2/3 ) Ö (1 - (x/R)2/3 ).

The reflected straight path runs between two points on the circular rim, corresponding to angles q and 3 q with respect to the x-axis. These two points on the boundary have coordinates (x, y) = (R cos q, R sin q) and (R cos 3 q, R sin 3 q), respectively. Each such straight line is tangent to the "caustic curve" at one point, which lies precisely one fourth of the way along the path, with coordinates
x = R (3 cos q + cos 3 q )/4
and
y = R (3 sin q + sin 3 q )/4

We cycle through the various striking points by letting q vary between 0o and 180o, or p radians. For each value of q, we obtain a point on the envelope, or caustic curve.

A template of a 360o Protractor was handed out, and participants made their own caustic by drawing lines from the positive x-axis [right on the middle] from angles q to 3 q. Here is a set of angles relative to the horizontal to use:

 10o --- > 30o 20o --- > 60o 30o --- > 90o 40o --- > 120o 50o --- > 150o 60o --- > 180o 70o --- > 210o 80o --- > 240o 90o --- > 270o

The right side of the caustic will arise, as if per magic!

The full curve, which was produced using the software package EXCEL, is given below. An excellent reference on using EXCEL in graphics is the book EXCEL for Scientists and Engineers by William J Orvis, Second Edition [ISBN 0-7821-1761-9].

01 February 2000: Fred Schaal (Lane Tech HS)
invited a number of us up front, gave us each a piece of polar graph paper (paper with a set of concentric circles on it, and radii drawn in every 10o). He had each of us cut out a segment of angle, one of us 10o, next 20o, then 30o, etc. We then taped the edges together to form a set of cones. Fred proceeded to lay out a set of equations for the lateral area (LA) of a cone, and ended up with a result that the altitude of a constructed cone lying on the table is:

h = R( k/180 - k2/3602 )0.5
where k is the angle (in degrees) of the cut-out segment.

We used sticks of spaghetti to poke down through our cones' tops to the table below, marked the stick, and got an experimental value of h. And then we calculated h from the above expression. Both observed and calculated values matched surprisingly well! A beautiful phenomenological math lesson! Thanks, Fred!

28 March 2000: Fred Schaal (Lane Tech HS)
Fred brought us back to something he had done with us earlier: The truncated cone. Start with a circle of radius R cut from paper. Draw two radii on the circle from its center, and so define an angle of 360o - ko between them. Cut out the sector between them, as shown:

`                     s = 2 p r = p [k/180] R`

Tape the radii together, to form a truncated cone surface with the paper. Let r be the radius of base of the cone, and h its altitude. Then

r = R * [ko/360o]

and

h = (R2 - r2)1/2 = R * [ 1 - (ko/360o)2]1/2.

The conical volume is given by
V = pr2h/3

= p R3/3 * [ko/360o]2 * [ 1 - (ko/360o)2 ]1/2

"What value should the angle ko have in order for volume V to be a maximum?" was the question Fred asked us. But lacking for any volunteers, Fred left us with the unanswered question. Maybe next time somebody will have an a solution?

02 May 2000: Porter Johnson

### Using EXCEL to Study Volume of Cut-out Cone See Notes for 28 March 2000 HS Math-Phys Class

If you take a sheet of paper with a circle of radius R, and cut out a sector of opening angle ko, the volume of the cone formed by that sheet is

V = [p R3/3] * [ko/360o]2 * [ 1 - (ko/360o)2 ]1/2.

The formula may be simplifying by defining the ratio x = ko/360o, so that for radius R = 1

V(x) = [p x2 /3] Ö (1 - x2)

One may show by elementary calculus that the volume is maximized by the choice

x = Ö (2/3) = .861497,

corresponding to angle k = 293.9388o. The maximum volume is

Vmax = 0.403067.
We may use the graphics program EXCEL [a standard component of the Microsoft Office program package] to create a a data table, as well as a graph. An EXCEL data table is arranged in cells, which look somewhat like the following:
 *** A B C D 1 _____ _____ _____ _____ 2 _____ _____ _____ _____ 3 _____ _____ _____ _____ 4 _____ _____ _____ _____ 5 _____ _____ _____ _____
Let us identify these cells by giving the Column and Row; eg, C3 or A5. You may enter data into a given cell by moving the mouse to the cell in question, and clicking.

Here is a set of steps that will result in the data table and the graph: First, type these descriptive words in lines #2 and #3

• In Cell A2 type cone vol
• In Cell A3 type angle
• In Cell B3 type x
• In Cell C3 type angle
• In Cell D4 type volume
Now, type in these numbers in line #4
• In Cell A4 type 290
• In Cell B4 type = A4/360
• In Cell C4 type = A4
• In Cell D4 type = PI() * B4^2 / 3 * SQRT(1 - B4^2)
Now, type this number into line #5:
• In Cell A5 type 291
Your EXCEL table should look like this:
 *** A B C D 1 _____ _____ _____ _____ 2 cone vol _____ _____ _____ 3 angle x angle volume 4 290 .805556 290 .402645 5 291 _____ _____ _____
Note that the formulas typed on line #4 do not appear, but only numerical answers.

At this point you should save the EXCEL table [after all, you've worked hard to create it!].

• Take the mouse and "enclose" [i.e. "highlight"] both cells A4 and A5, which contain the numbers 290 and 291, respectively. If you pull downward on the lower right corner of the rectangle with a + sign there, you will get evenly spaced points as far down as you pull; say, 290 to 310.
• Now, enclose the cell B4, and pull down on the + sign at the lower right corner, so as to fill cells in column B as far down as column A is filled.
• Similarly, enclose cell C4, and pull it down.
• Finally, enclose cell D4 and pull it down.
Your table should look like this:
 *** A B C D 1 _____ _____ _____ _____ 2 cone vol _____ _____ _____ 3 angle x angle volume 4 290 .805556 290 .402645 5 291 .808333 291 .402830 6 292 .811111 292 .402963 7 293 .813889 293 .403042 8 294 .816667 294 .403066 9 295 .819444 295 .403035 10 296 .822222 296 .402946 11 297 .825000 297 .402798 12 298 .827778 298 .402589 13 299 .830556 299 .402320 14 300 .833333 300 .401986 15 301 .836111 301 .401588 16 302 .838889 302 .401123 17 303 .841667 303 .400590 18 304 .844444 304 .399987 19 305 .847222 305 .399313 20 306 .850000 306 .398564 21 307 .852778 307 .397739 22 308 .855556 308 .396837 23 309 .858333 309 .395855 24 310 .861111 310 .394791

The next objective is to make a graph of the volume versus angle using EXCEL. To do so, complete the following steps: EXCEL

• Using the mouse, capture the data in column A, and then in column D.
• From the toolbar menu, click on the "chart wizard" icon.
• Pick the "xy scatter" option
• Then pick the "smooth curve" option
• Include the "major grid lines" and "minor grid line" options for x and y
• Finish drawing the graph.
The result should look like the image shown here:
Here is the graph of cone volume over the full angular range 0o to 360o:o

02 May 2000: Fred Schaal (Lane Tech HS)
did the Law of Sines with us as an exploration by hand-held calculator. He drew a triangle on the board and identified its three angles as A, B, C, and the lengths of the sides opposite as a, b, c, as shown:

`          `
Then
a/sinA = b/sinB = c/sinC.

Choosing values for A, and keeping c and B fixed but letting the side a increase with A, he obtained the following table of values, which he wrote on the board:

A a / sin A
6o 16.3
19o 14.4
46o 17.8
83o 15.4
99o 20.2
153o 16.3

Triangles were actually drawn for the various values of A, and a was then measured, and the above ratios were calculated based on these experimental measurements. Fred noted that the ratio does not change much, and wondered at this.

Fred set up the calculation of the curve for x(T), y(T), where

x = T[1+1/(2p) ] cos T
y = T[1+1/(2p) ] sin T
for 0 < T < 20p.

He displayed the result as a set of points with the aid of a TI graphing calculator, a transparency projector and LCD display. The graph looked like a roll of carpeting viewed end-on, a kind of spiral around the origin. Most interesting! A good example of how to connect the abstract to the concrete so as to make math "real" for students!

02 May 2000: Bill Lilly (Kenwood HS)
gave us a handout titled "Pendulum," which was a lab exercise. Using a TI-CBR Sonic Ranger, high frequency sound from the Ranger was reflected from a swinging balloon [held on the end of string by Pearline Scott (Franklin School)]. The output from Ranger (for 10 s) was stored in the calculator and shown on the projector screen as a graph with the aid of Fred Schaal's LCD display. It looked like a sine wave, and Bill described how students could analyze the data for frequency, effect of shortening string, lengthening string, amplitude effects, velocity minima and maxima with position, etc. Another beautiful example of concrete/abstract connections!

26 September 2000 Fred Schaal (Lane Tech HS)
presented "92s to the Rescue." He gave one hand-held TI 92 to Betty Roombos (Lane Tech HS) to enter commands and data. The LCD display of the TI 92 was projected onto the screen for all to see. Fred placed on the board: a/sinA = ? - referring to the Law of Sines. He instructed Betty how to draw a triangle and label its vertices with angles A, B, C, and the sides opposite with a, b, c. Betty was a whiz, and with Fred's further direction she soon had the angle A labeled with its value of 41.99o, and the side a = 2.28 cm, taken as a "measured" value. The ratio R = a/sinA was calculated as 3.41 cm. Then, as Betty dragged the top vertex, C, around the screen of the TI 92 (as one does with a mouse on a PC), the triangle changed its linear and angular dimensions, and the changing value of R was seen to range from 3.10 to 4.77 and higher. When the sizes of angles were measured to be in radians instead of degrees, the range of values of R continued to be the same. Beautiful! What a great way for students to develop an intuition for the Law of Sines and what it tells us about the geometry of plane triangles. Thanks, Fred, for phenomenological math!

26 September 2000 Fred Farnell (Lane Tech HS)
walked carefully across the front of the room in front of us and asked,

"What would a graph of this motion look like?"

Using the same projection setup as Fred Schaal, Fred F. connected a sonic ranger (CBR) to a TI 83, and to the projector. He explained that a high frequency sound wave was sent out by the CBR, which would then be reflected from Fred's body back to the CBR where the reflected wave would be detected. Next, he walked carefully, as before, away from the CBR which put data about Fred's position (D) and speed (v) at various times (t) into the TI 83. He then caused the D vs t data to be displayed on the screen, and data points appeared to lie roughly in a straight line with positive slope. Using
y = mx + b

and a program in the TI 83, he found the best-fit values of m and b to the data and plotted it. It seemed like it was not a very good fit to us.

Next, Fred repeated the experiment, but this time ran away from the CBR. The D vs t graph of his motion appeared parabolic, typical of constant acceleration motion. When he found the best fit to the parabolic equation,

y = ax2 + bx + c,

and plotted it on the same graph as the experimental points, the result did not look like a best fit to many of us. He also tried fitting to an exponential equation, with the same kind of result. This leaves something for us to explain, but Fred certainly showed us what wonderful things can be done experimentally right in the classroom these days. Good phenomenological physics, Fred!

10 October 2000 Fred Farnell (Lane Tech HS)
explained why the fit of a straight line equation to experimental constant velocity points (and its subsequent v vs t graph) at the last meeting was so poor. It turned out that the constant (non zero) velocity occurred over an 8 second period, but the data taking ran for 15 seconds, and the velocity was zero for the last 7 seconds! When Fred ran new data and kept only the portion for non-zero velocity, the fit turned out great! He did this "live". Same story for motion of constant acceleration. Thanks for restoring our faith, Fred!

24 October 2000 Fred Schaal (Lane Tech HS)
set up the projector to display the output of a TI 92 calculator so we all could see it, and gave  the TI 92 to Betty Roombos to manipulate. Fred asked Betty to draw a triangle and label its vertices with  A,   B,   C and we saw it as it happened. Then he advised Betty what to do to construct the bisectors of the angles at the vertices A and B. The angle bisectors intersected at a point which was labeled F. Next, he had Betty construct a ray connecting vertex C with the point of intersection, F. The angles ACF and BCF were "measured" by manipulation of the TI 92 and labeled with their values, which both happened to be 37.05 o! This showed that the constructed ray was a bisector of the angle at vertex C, and was convincing evidence that the bisectors of all three angles intersect in a common point! Wow!

Then Fred asked Betty to "grab" the vertex point C and move it around. As this was done, we could see the shape of the triangle ABC change, and the values of the two angles also changed, but remained equal. Very nice! Similar exercises may be done to show that medians meet at a point; perpendicular bisectors of sides meet at a point. Porter Johnson asked: What about the "Nine Point Circle Theorem?" Anyone know? Check the website http://mathworld.wolfram.com/Nine-PointCircle.html.

A perfect complement to what you did last time, Fred! (...the actual construction on the white board, which did not display it well.)

07 November 2000 Fred Schaal (Lane Tech HS)
sketched out on the board a kind of geometry problem he will do next time using the TI 92 calculator. He drew two straight lines (representing mirrors), intersecting at some angle. Then he drew a triangle with vertex points A, B, C and sketched its reflection across the nearest of the mirrors. He pointed out that the order of the points, A, B, C was ccw (counter clockwise) for the triangle, but cw (clockwise) for the vertex points - A' ,  B' , C' - of its reflection. Sketching the reflection of the those points across the second mirror (straight line) located vertex points A", B", C" - which followed a ccw order. It could be interesting to see how Fred will do this with the TI 92. But this provoked discussion about reflections in mirrors, and brought out that it takes two mirrors to see your reflection as others see you!

21 November 2000 Fred Schaal [Lane Tech HS]
displayed the output of his TI 92 Calculator on the screen. He chose the "geometry" option and drew a triangle.  Then he added a line, and reflected the triangle about the line.  He noted that the second triangle could not be superposed on the first one, because it was "left-handed".  Next, he added another [non-parallel] line, and made reflection of the second triangle.  He showed that the third triangle could be superposed by rotation onto the first one.  Thus, two reflections correspond to a spatial rotation.

10 December 2002: Fred Schaal [Lane Tech HS, Mathematics]    TI-92 Graphics Calculator Attacks Triangle!
Fred
used the TI-92 calculator to illustrate basic geometrical ideas; specifically, general properties of triangles.  First he hooked a calculator to a translucent imaging screen placed on the overhead projector, so that the entire class (we) could see the calculator display.  Then he gave the calculator in question to one of the participants, who used the menu to bring up "geometry", and then "insert triangle".  The participant then clicked on three vertex points in succession, and the triangle appeared.  Fred then showed her how to shift one of the vertices of the triangle by using click and drag. Then she clicked on an icon to construct the altitudes of the triangle, which were seen to intersect at a point, the ortho-center of the triangle. When she moved a vertex, the altitude lines and ortho-center changed continually.  Then she clicked on another icon to obtain perpendicular bisectors of each side, which intersect at a point, the circum-center of  the triangle.  Again, as a vertex was moved, the circum-center also changed.  Interestingly, neither the ortho-center nor the circum-center must always lie inside the triangle.  These calculators make geometry fun, Fred!

25 February 2003: Walter McDonald [CPS Substitute -- VA Hospital Technician]      Approximation of Functions
Walter
showed us how to do Taylor eries polynomial expansions of trigonometric functions on his HP 48GX Programmable Calculator.  The idea is to truncate the expansion of, say, sin x in powers of x after a few terms, and to plot the results on the calculator.  The expansion:

sin x = x - x3 / (1 ´ 2 ´ 3) + x5 / (1 ´ 2 ´ 3 ´ 4 ´ 5) - ...
requires about 10 terms for accurate representation of the graph of sin x vs x over one period -p < x < p. He obtained similar results for the series involving cos x:
cos x = 1 - x2 / (1 ´ 2) + x4 / (1 ´ 2 ´ 3 ´ 4) - ...
He found that the series for the functions [ tan x,  sec x, tan-1x, sin-1x, cos1x ] converged only at small x, and that the functions cot x and csc x did not possess series expansions.  Porter Johnson suggested expanding the functions (x cot x and x csc x)  in the latter cases, and then dividing by x at the end.  This feature of  automated Taylor series expansions is not present on the TI 83 or TI 86 Programmable Calculators, but does occur on the HP 48.

Fascinating, Walter!

11 March 2003: Fred Schaal [Lane Tech HS, Mathematics]      TI Interactive Software
Fred
had intended to transfer images from his laptop computer to the screen in order to display TI Interactive Software, but discovered that, without a projector, there was no hope of doing so.  Such transfers of images from one type of system to another are always problematical, and the safest step is to use your own system for the entire process.  Thanks for trying, Fred!

25 March 2003: Fred J Schaal [Lane Tech HS, Mathematics]     Laptop Capers:  TI Interactive
Fred
showed the TI Interactive Software, which emulates a TI-83 Calculator on a Windows-based PC.  With the assistance of Monica Seelman, Fred went into the graphics mode, and first displayed a graph of the function y = x2 - 8, using the ranges (-10 £ x , y £ +10). A very nice upward parabolic curve appeared, which left the region of graphing at x » ± 4.2. Next Fred plotted the graph of y = 8 - x2, which corresponded to the same parabolic curve, but inverted. The two curves intersected at the line y = 0, for x = ± Ö» ± 2.8. Finally, Fred asked for the area between the two curves, which can be formally written as the integral

4 ò0Ö 8 dx [ 8 - x2] = 4 ´ [8 x - x3 / 3] (x = Ö 8) = 4 ´ 8 ´ 2/3 ´ Ö 8 » 60.33

Very visual and user-friendly, Fred!

08 April 2003: Fred Schaal [Lane Tech HS, Mathematics]        Green Line '92-ing!
Fred
had been playing around with his calculator while riding the el, and began studying the following problem:  For a given triangle with vertices (A, B, C), draw the following lines:

• The bisector of interior vertex angle A, which we assume to be acute.
• The median from the center of side AB to vertex C
• The altitude from side AC to vertex B
These three lines intersect to form a smaller triangle within the original triangle ABC, in general. . What is the area of that triangle, in relation to the area of the original triangle?  By trying this for different triangles, Fred found that it was difficult to make the area of the smaller triangle more than about 18% of the area of the original triangle. Is there a general result? If so, what is it, and how is it proved?

Fred, you have constructed an unusual problem! Very Good!

07 October 2003: Fred Schaal [Lane Tech HS, mathematics]        Lubbock*** [TI-83] Helper
Fred
used his calculator and plotter to extend his lesson of 29 February 2000 [mp022900.htm] concerning tangent lines and asymptotic curves.  This lesson had, in turn, been an extension of a lesson by Bill Colson on 11 November 1999 [ph110999.htm] on producing intricate geometrical patterns. These comments by Porter Johnson are excerpted from the first lesson:

The formula for line segment lying in first quadrant and intercepting the x-axis at x = a and intercepting the y-axis at y = b - a, where 0 < a < b, is

x/a + y/(b-a) = 1
The asymptotic curve, which represents the envelope of all the straight lines, can also be written as
y = (Ö b - Ö x )2     or      Ö x + Ö y = Ö b      or     (x - y)2 + b2 = 2 b (x + y)
The curve is a parabola with the line of symmetry [axis] lying along the line x = y. Bill used the graphing calculator to extend the problem to an extended region:   -a < x < a and -a < y < a.  What did it look like? Amazingly, the original asymptotic curve (caustic) was symmetrically extended to all four quadrants:
Ö |x| + Ö| y| = Öb

How come?

Fred's trusty programmer, Bill Colson, developed and displayed the graphs for xp, where the real, non-integer variable p increases from 1 to 2.  The curves all passed through x = 1, but became steeper below that point for x < 1 as p was increased.  For x > 1, it was the other way around.  Fred showed the graphs for negative x as well.  Surprisingly, for p = 1.1, 1.3, 1.5, 1.7, and 1.9 the curves were negative, whereas for p = 1.2, 1.4, 1.6, and 1.8 they were positive.  How come?

Comments by Porter Johnson:  For negative x, there is a real, positive value of  p whenever p is a rational fraction of the form a / b, where the integer b is even.  Furthermore, there is a real, negative value of  xwhen p=a / b is rational, when b is odd. The gremlins and lubbockians*** of calculator programming were definitely on your side for this one, Fred.

Q: How many Texas Tech football supporters does it take to change a light bulb?
A: Don't be silly; there's no electricity in Lubbock*** [home of Texas Instruments™]!
Bigger plots are definitely better!  Good, Fred!

21 October 2003: Babatunde Taiwo [Dunbar Vocational HS, science]        Graphics Calculator and Motion Sensor
Babatunde
showed that his TI-83 graphing calculator, CBL Ranger motion sensor, and Vernier Logger Pro [http://www.vernier.com/soft/lp.html] graphical interface could be used to display graphs of idealized motion, as well as record the motion of moving objects. First he displayed a distance-time graph that looked like the one shown here, with distance marks in meters and time marks in seconds.

` `
He set up the motion sensor and software so that it would record the position of an person walking across the room. The objective was to reproduce the graph given above with one's own motion --- by being at the right place at the right time. After three trials, Babatunde was able to reproduce the curve pretty well. Walter MacDonald then tried it with this curve, as well as the curve given below. He was able to reproduce the curves on his first attempts.
`      `
Babatunde then showed the curves of velocity and acceleration versus time, as determined numerically from the positions. He then pointed the motion sensor downwards, dropped a ball from a height of about a meter, and used the motion sensor to record its vertical position versus time. He was able to follow the ball through several bounces. Then he showed the graphs of velocity and acceleration versus time. The acceleration was negative, and of modest relative size, when the ball was in flight, whereas when the ball struck the floor it became quite large and positive. [Because of the intrinsic time resolution of the apparatus, we were seeing some time averaged acceleration of the ball, rather than its instantaneous acceleration.] These are very visual exercises in the kinematics of motion, which allow us to experience the direct connection of motion with distance-time graphs.  Thank you for showing this to us, Babatunde!

For additional information see the websites: Activity Investigating Motionhttp://www.math.mtu.edu/gk-12/investigatingmotion.html and Utilizing the Graphing Calculator in the Secondary Mathematics and Science Classroomhttp://www.esc4.net/math/

Babatunde also showed a spinning top toy that he had recently obtained at American Science & Surplus [ http://www.sciplus.com/category.cfm?category=13]:

Here, Kitty Kitty ... A top that’s fun! A top that doesn’t need string or practice or coordination. Twist the winder, push the button, and the colorful plastic top bounces around on its spring bottom, lights up brightly (batteries included), and spins for a very long time. A nice distraction for a jaded child or a bored tabby cat

18 November 2003: Fred J Schaal [Lane Tech HS, mathematics]        STO FRM Not That Bad!
Fred
asked if anybody knows the standard "slope-intercept" equation for a straight line.  Somebody suggested ...

y = m x + b .
Another very familiar form is the linear relation
A x + B y = C
Dividing the latter relation by C, and adopting the notation a = C/A and b = C / B, we may write this relation as
x / a + y / b = 1
With this latter form, the x-intercept is a and the y-intercept of the curve is b. Neato!

Bill Shanks mentioned that an elliptical curve with semi-major axes (a,b) can be written in a similar form:

(x / a)2  +  ( y / b)2 = 1
Earl Zwicker extended the curve by shifting the center of the ellipse to the point (x0, y0):
[ (x  - x0)/ a ]2  +  [ ( y - y0) / b ]2 = 1
Porter Johnson pointed out that the full diamond [rhomboid] shape can be written as
| x / a|  +  | y / b | = 1
Fred also mentioned that the planet Mercury will be visible in the evening sky in about a month, just South of the brighter planet Venus.

Thanks for showing us the neat Algebra, Fred!

04 May 2004: Fred Schaal   [Lane Tech HS, Mathematics]           Fraction Action
Fred reminded us of a universal feature of programming algorithms on digital computers, in that they perform arithmetic exactly with integers, whereas with real numbers the calculations may be inexact because of round-off error.  Consequently, arithmetic with fractions can be done exactly -- provided the numbers that arise are not too large. On his trusty TI-81 programmable calculator, Fred went into fraction mode, and was able to verify the relation:

1 / 19 + 1 / 87 = 106 / 1653
On the other hand, the computer was not able to obtain the identity
4 / 123 + 5 / 841 = 3979 / 103443
giving some result like 0.0384656, since it encountered a denominator that was too large for storage, and had to retreat to real numbers.  Fred also pointed out that the TI-81 can be used for mixed fractions, in several different ways:
2 + 2 / 3 + 3 + 3 / 8 (frac)= 151/24
... or ...
2 + 2 * 3^(-1) + 3 + 5 * 8^(-1) (frac)= 151/24
Neato, Fred!