My summer begins Friday. One of my goals over the summer months is get my thoughts down about the mathematics behind the applet below. It is stunningly simple, but it is the details I want to make clear. So, I present it to you as a sort of summer inquiry activity for teachers. As you move the green points around and as you drag the purple-ish circle around and change its radius, what happens?
Follow Up: After sharing this with some of my students after one of their exams, they expressed a desire to be be able to place points and draw lines, so I updated the applet to reflect their wishes.
I was pleasantly surprised over how much time my students spent with this applet and the quality of their observations.
Wednesday, May 26, 2010
Friday, May 21, 2010
Can't Quite Get My Head Around The Numbers
A lot of BIG numbers is this story.
- A month's worth of leaking oil could fill enough gallon milk jugs to stretch more than 11,300 miles.
- At worst, it's enough to fill 102 school gymnasiums to the ceiling with oil.
- The Gulf has five billion drops of water for every drop of oil.
Thursday, May 20, 2010
Random or Real?
A WCYDWT moment from Mind Your Decisions centering on the online game ARORA. Can you tell the difference between random data an real data?
Sunday, May 16, 2010
This Changes My Game
I have been trying to create moodle questions that get my students to construct something in GeoGebra, get some sort of information from the construction, and answer the question. I have been struggling with the idea that any calculated moodle questions I come up with is ultimately reduced to an algorithm. Nevertheless, if my students are unaware of the algebra behind the algorithm, and need to think of what to construct, then perhaps that is O.K.
One question I have used recently is this:
find a good window setting (which is essentially scrolling out until you can see the entire circle...do they know what they are looking for? Goal #2),
graph the line x = 33 (which is difficult to do on my calculator of choice, the Nspire, or other graphing calculators),
constructing the two intersection points, and taking the larger of the two y-values (recognizing that this is the desired answer to the question),
To answer the question correctly, there is still a lot of intuitive mathematical understanding that is being displayed, especially when you ask them why the answer must be 71.79.
I have been encouraging my students to find different web-based resources, lately, and one of my students found a web-based calculator web2.0calc.Imagine my surprise when one of my students did this to solve this problem:
producing this result
Not a CAS, but a pretty nifty numerical solver. And intuitive! When I asked the student what made him try this, he told me that he just wrote the equation, and substituted 33 in for x like the question asked him to do. There were two answers for y, so he used the larger of the two. Intuitive.
THIS changes my game.
Go ahead and try this for yourself on the calculator below.
One question I have used recently is this:
A circle has a center (17,15) and a radius 59. Write the equation of the circle. Then, find the larger of the two y-values when x=33.By algorithm, the better algebra students will easily be able to solve this circle equation for the two y-values, but the students in this particular class are not my better algebra students. What I hope would happen is for them to graph the circle in GeoGebra (meaning they must write the equation correctly...#1 goal),
find a good window setting (which is essentially scrolling out until you can see the entire circle...do they know what they are looking for? Goal #2),
graph the line x = 33 (which is difficult to do on my calculator of choice, the Nspire, or other graphing calculators),
constructing the two intersection points, and taking the larger of the two y-values (recognizing that this is the desired answer to the question),
To answer the question correctly, there is still a lot of intuitive mathematical understanding that is being displayed, especially when you ask them why the answer must be 71.79.
I have been encouraging my students to find different web-based resources, lately, and one of my students found a web-based calculator web2.0calc.Imagine my surprise when one of my students did this to solve this problem:
producing this result
Not a CAS, but a pretty nifty numerical solver. And intuitive! When I asked the student what made him try this, he told me that he just wrote the equation, and substituted 33 in for x like the question asked him to do. There were two answers for y, so he used the larger of the two. Intuitive.
THIS changes my game.
Go ahead and try this for yourself on the calculator below.
Friday, May 7, 2010
Monday, May 3, 2010
What I Wish I Did First
Though things have worked out fine, I wish I had given this Magic Point first.
Sunday, May 2, 2010
Function Family Guy
I discussed my approach to function families in an earlier post. The instructions for each of these graphing activities were the same:
For example, to enter the equation of a parabola from vertex A to B in the graph below, it would look like this on the Nspire
But like this in GeoGebra
I think this might be one area where the Nspire is superior, but even as I write this sentence, I think of how graphing circles and and other equations is so much easier in GeoGebra. You do not need to isolate y in order to graph a function in GeoGebra. You can graph non-functions like x = 4. Ellipses and Hyperbolas are peices of cake (or of cones...you choose).
For example, graphing circles in GeoGebra look like this
but look like this in the Nspire
where you must graph the circle in two parts, top and bottom. Of course, I hope that by now, my students will graph a circle on the Nspire like this,
using function notation and the algebraic representation of geometric transformations, reflecting the top half of the circle in f1 over the x-axis, then translating the reflected circle up six units.
Write an equation for each of the following exactly as you would enter them into your Nspire, sketching the graph of each as you go. Only when you have checked your equations and graph with me should you graph them on the Nspire.Nothing fancy, but it was fun. Kids whined about the numbers (too many decimals), and argued about the finished result (there are two famous football-shaped headed cartoon characters). We did all of these on the Nspire (hmm...might be my first Nspire-related post), but we also used GeoGebra. There are some differences in how you enter the equations which make the Nspire a little better for this activity.
For example, to enter the equation of a parabola from vertex A to B in the graph below, it would look like this on the Nspire
But like this in GeoGebra
I think this might be one area where the Nspire is superior, but even as I write this sentence, I think of how graphing circles and and other equations is so much easier in GeoGebra. You do not need to isolate y in order to graph a function in GeoGebra. You can graph non-functions like x = 4. Ellipses and Hyperbolas are peices of cake (or of cones...you choose).
For example, graphing circles in GeoGebra look like this
but look like this in the Nspire
where you must graph the circle in two parts, top and bottom. Of course, I hope that by now, my students will graph a circle on the Nspire like this,
using function notation and the algebraic representation of geometric transformations, reflecting the top half of the circle in f1 over the x-axis, then translating the reflected circle up six units.
- A parabola with vertex A, from A to B.
- A parabola with vertex A, from A to C.
- A parabola with vertex D, from D to F.
- A CUBIC with inflection point D, from D to E.
- The LEFT HALF of a circle from B to E.
- The RIGHT HALF of a circle from C to F.
- A line segment from (1,1) to (1.5, 0.5).
- A line segment from (1.5, 0.5) to (1,0).
- The LEFT HALF of a circle, centered at (1,-1) and radius of 0.5.
- A circle centered at (-3,1) with a radius of 1.25.
- A circle centered at (4, 0.5) with a radius of 1.25.
- A parabola from the vertex (3,2) and stopping at (4, 2.25).
- A parabola from the vertex (-2, 2.5) and stopping at (-3, 2.75).
- A parabola with vertex (4,0), opening downward, from (3, -0.25) to (5, -0.25).
- A parabola with vertex (-3,0.5), opening downward, from (-4, 0.25) to (-2, 0.25).
- Circle centered at (1,3) with radius 0.5.
- Square root from vertex (1.5, 3) to (0.5, 3.25).
- Square root from vertex (0.5, 3) to (-0.5, 3.25).
- Square root from vertex (0.5, 3) to (-2, 2.75).
- A cubic from inflection point (-2, 2.75) to (-3,2).
- A parabola with vertex (1, 3.75) from (0, 3.5) to (2, 3.5).
- A circle with center (-3,1) and radius 1.
- A cubic from inflection point (-2, -0.75) to (-3,0).
- A square root from vertex (0.25, -1) to (-2, -0.75).
- A parabola from vertex (0,-1.25) to (-1, -0.75).
- A parabola with vertex (2.25, 3.75) from (1, 3.5) to (3, 2).
- The bottom half of a circle centered at (2.25, 2) with radius 0.75.
- A parabola with vertex (-0.25, 3.75) from (-1, 2.75) to (0, 3.5).
- A square root from vertex (2.75, -1) to (2.5, 1.25).
- A parabola with vertex (1, -1.5) from (-1,-1) to (3, -1).
Saturday, May 1, 2010
The Magic Point
I provided this applet to my students a couple of weeks ago. It fit well with where we were at the time, exploring matrix transformations and all. I left it on my class Moodle for a week before we began talking about it in depth. My only question to the students was "What does point D have to do with anything?"
I pose the same question to you.
Or perhaps you will share a different question to ask.
I pose the same question to you.
Or perhaps you will share a different question to ask.
Wednesday, April 14, 2010
What If Not - Circle as Directrix
As another application of What If Not problem posing, let's take a look at the parabola construction. What if the directrix was not a line, but instead, what if the directrix were a circle?
As illustrated in the mathlet below, begin with a circle and a point (Focus) different from the center. Place a point on the circle (Drag Me). As we did with the parabola construction, we are looking for those points equidistant from the Focus and the point Drag Me. This equidistant point will lie on the perpendicular bisector of the Focus and Drag Me, so we construct that next. Following the lead of the parabola construction, we construct a perpendicular to the circle from the point Drag Me; this is simply a radius. The intersection of this radius and the perpendicular bisector is equidistant from the focus and the circle. The locus of this point as Drag Me moves along the circle is an ellipse.
What is the next logical (at least in my mind) What If Not question to ask? What if the circle was not the directrix, but the ellipse was the directrix? What would happen then?
As illustrated in the mathlet below, begin with a circle and a point (Focus) different from the center. Place a point on the circle (Drag Me). As we did with the parabola construction, we are looking for those points equidistant from the Focus and the point Drag Me. This equidistant point will lie on the perpendicular bisector of the Focus and Drag Me, so we construct that next. Following the lead of the parabola construction, we construct a perpendicular to the circle from the point Drag Me; this is simply a radius. The intersection of this radius and the perpendicular bisector is equidistant from the focus and the circle. The locus of this point as Drag Me moves along the circle is an ellipse.
What is the next logical (at least in my mind) What If Not question to ask? What if the circle was not the directrix, but the ellipse was the directrix? What would happen then?
Monday, April 12, 2010
What If Not?
I imagine (or I hope) most teachers of Geometry, Algebra 2 and above are familiar with the construction of a parabola. As illustrated in the mathlet below, we begin with a line which serves as the directrix and a point not on the line which is our focus. Place a point on the directrix (labeled as Drag Me!). Since each point on the desired parabola is equidistant from the focus and the directrix, we next construct the perpendicular bisector of the "Drag Me" point and the focus, knowing the equidistant point in question must lie on this line. This bisector happens to be tangent to the parabola (which provides an interesting way to find the equation of a tangent line). We next construct a perpendicular to the directrix at the "Drag Me" point. The point where this line intersects the perpendicular bisector is on the parabola. In fact, the locus of this point as you drag the "Drag Me" point along the directrix is the parabola. If you right-click (or ctrl-click) on this point, you can choose trace the point.
This is where the Art of Problem Posing and the What If Not strategy comes into play.
I played around with the following idea a couple of years ago in one of my classes. I did this as a lark, not quite certain where it would lead. I am glad I posed the questions!
Applying the What If Not strategy to the parabola construction, I changed the construction question. What if the directrix used in the construction were not the directrix, but instead, the parabola? In other words, what if we mimicked the parabola construction using the parabola as our directrix? What would happen?
Let's begin with a parabola, its focus, and its directrix (we still need this line so we can change the direction the parabola opens). Place a point (Drag Me! of course) on the parabola. To mimic the construction, we are looking for a point that is equidistant from the point Drag Me and the focus. This point must lie on the perpendicular bisector of these two points, so we construct it. Continuing to mimic the construction, we next construct a perpendicular to the parabola at the point Drag Me. To do this, we construct a tangent at the point Drag Me, and then a perpendicular to this. The intersection of the perpendicular bisector and this perpendicular is the desired point equidistant from the parabola and the focus. The locus of this point is known as Tschirnhausen's Cubic.
What else can you do with this? It is a nice exercise to derive a parametric equation of this cubic. What if you used some other shape, say, a circle for the directrix in this construction? Do you recognize this change in the construction as leading to an ellipse or a hyperbola? What if you repeated this construction using Tschirnhausen's Cubic as the directrix? It's never ending! Applying the What If Not strategy to a common problem is a great way to find something new hiding beneath the surface!
This is where the Art of Problem Posing and the What If Not strategy comes into play.
I played around with the following idea a couple of years ago in one of my classes. I did this as a lark, not quite certain where it would lead. I am glad I posed the questions!
Applying the What If Not strategy to the parabola construction, I changed the construction question. What if the directrix used in the construction were not the directrix, but instead, the parabola? In other words, what if we mimicked the parabola construction using the parabola as our directrix? What would happen?
Let's begin with a parabola, its focus, and its directrix (we still need this line so we can change the direction the parabola opens). Place a point (Drag Me! of course) on the parabola. To mimic the construction, we are looking for a point that is equidistant from the point Drag Me and the focus. This point must lie on the perpendicular bisector of these two points, so we construct it. Continuing to mimic the construction, we next construct a perpendicular to the parabola at the point Drag Me. To do this, we construct a tangent at the point Drag Me, and then a perpendicular to this. The intersection of the perpendicular bisector and this perpendicular is the desired point equidistant from the parabola and the focus. The locus of this point is known as Tschirnhausen's Cubic.
What else can you do with this? It is a nice exercise to derive a parametric equation of this cubic. What if you used some other shape, say, a circle for the directrix in this construction? Do you recognize this change in the construction as leading to an ellipse or a hyperbola? What if you repeated this construction using Tschirnhausen's Cubic as the directrix? It's never ending! Applying the What If Not strategy to a common problem is a great way to find something new hiding beneath the surface!
Thursday, April 8, 2010
I think I am finally starting to recover...
I guess I shouldn't complain. I know there are quite a few things worse than spending 11 days in Italy for spring break. I was lucky enough to chaperon my son's Latin Club trip over there. I can't imagine a better trip.
- Three days in Venezia and the surrounding Islands
- Three days in Firenze with a side trip to Siena
- One day in Assisi
- Three days in Rome, including Holy Thursday, Good Friday, and Easter Sunday. Actually ate dinner at a place where I could look out the window from my table and see the Colosseum!
- One day on the Island of Capri (which has to be the most beautiful place I have been, not including HOME!)
Wednesday, March 24, 2010
The Complete Quadrilateral: Still MORE Points on The Orthocentric Line
Reflect the Focal Point over each side of the quadrilateral.
Yep...you guessed it.
Yep...you guessed it.
Tuesday, March 23, 2010
The Complete Quadrilateral: The Hervey Point
Recalling the Four Triangles, the circumcenters of these triangles are concyclic, lying on the Circumcentric Circle, whereas the orthocenters are collinear, lying on the Orthocentric Line. Construct the perpendicular bisector of each circumcenter/orthocenter pair for each triangle. These perpendicular bisectors are concurrent at the Hervey Point.
Monday, March 22, 2010
The Complete Quadrilateral: Even More Points on The Orthocentric Line
Are you familiar with the Orthopole? I first ran across this at a presentation by my friend Ray Klein, a T3 National Instructor from the Chicago area. If you are not familiar with it, the following applet will illustrate the construction.
In the Complete Quadrilateral, construct the Orthopoles of the Four Triangles. The Orthopoles lie on the Orthocentric Line.
In the Complete Quadrilateral, construct the Orthopoles of the Four Triangles. The Orthopoles lie on the Orthocentric Line.
Sunday, March 21, 2010
The Complete Quadrilateral: More Points on the Orthocentric Line
Using pairs of opposite sides, drop perpendiculars from the midpoint of one to the other. These pairs of perpendiculars intersect in six points total, which lie on the Orthocentric Line.
Saturday, March 20, 2010
The Complete Quadrilateral: The Pedal Line, The Orthocentric Line, and The Focal Point
This makes a pretty picture. Drag the slider slowly to the left to reveal each line or point one at a time. If you do anything with conics and/or their constructions, this is an unexpected conic, which are the best kind!
Draw the Pedal Line. Recall the Pedal Line is the line through the feet of the perpendiculars dropped from the Focal Point to each side. Recall the Focal Point is the intersection point of the four circumcircles of the Four Triangles.
Draw lines through the midpoints of pairs of opposite sides.
These seven lines, along with the four lines of the quadrilateral, are tangent to the parabola whose focus is the Focal Point and whose directix is the Orthocentric Line.
Draw the Pedal Line. Recall the Pedal Line is the line through the feet of the perpendiculars dropped from the Focal Point to each side. Recall the Focal Point is the intersection point of the four circumcircles of the Four Triangles.
Draw lines through the midpoints of pairs of opposite sides.
These seven lines, along with the four lines of the quadrilateral, are tangent to the parabola whose focus is the Focal Point and whose directix is the Orthocentric Line.
What Does This Say About Grades and Grading?
Granted, it is a game design class, but could the same thing be applied to other education settings? Followed a link in the above article to this, conaining the following quote
"The elements of the class are couched in terms they understand, terms that are associated with fun rather than education," [Sheldon] told iTnews. "There will always be a portion of the class who will not be motivated to learn no matter what an instructor may try. Those that are not as involved, one or two out of a class of forty, are pretty much drifting through life anyway thanks to factors the classroom can't really address."
I buy that!
Friday, March 19, 2010
The Complete Quadrilateral: Circles on Diagonals
Construct the diagonals of the quadrilateral, and then construct circles with the diagonals as diameters. The centers of the circles are the midpoints of the diagonals, and we know they lie on the Midline. The three circles intersect in two distinct points on the Orthocentric Line. Therefore, the Midline and the Orthocentric Line are perpendicular. In circle-talk, these three circles are coaxal, and the Orthocentric Line is the radical axis of the three circles.
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