The four dog problem from Steve Strogatz:

N dogs start in the corners of a regular N-gon. Each runs directly toward the dog on its left. How far do they run before colliding?

— Steven Strogatz (@stevenstrogatz) October 27, 2015

Here is how it was used in my class.

If the dogs blindly travel half way towards the next dog before re-aiming, then you get this picture, where each vertex represents the dog’s position:

If you raise each square up in the z-dimension and print the resulting 3d shape then you can get this:

If the dogs travel 12% of the way before re-aiming then you get this picture, and this 3d shape:

The generalized interactive 2d version of this concept is a lot of fun to play with. In this demo, your mouse horizontal controls how much to spin each square, and your mouse vertical controls how much to scale each square. Matt Enlow was the driving force behind this super fun demo. Some samples:

Here’s the code for both of these shapes in madeup:

thickness = 3 size = 10.0 theta = 0 z = 0 repeat 15 t = 0 while t <= 360 x = size * cos (t + theta) y = size * sin (t + theta) t = t + 90 moveto x,y,z end extrude 0,0,1,thickness theta = theta + 45 thickness = thickness * (0.5)^0.5 z = z + thickness size = size * (0.5)^0.5 end

thickness = 1 size = 10.0 theta = 0 z = 0 repeat 35 t = 0 while t <= 360 x = size * cos (t + theta) y = size * sin (t + theta) t = t + 90 moveto x,y,z end extrude 0,0,1,thickness theta = theta + 8.775 thickness = thickness * 0.84 z = z + thickness size = size * 0.8775 end]]>

@stevenstrogatz great @Mathologer video on the cardioid and its family: https://t.co/ElDxW67N3X

— Anna Lukács (@launknaacs) November 6, 2015

What a great video, linking linear functions, cardiods, and the family of Mandelbrot sets. Rang my bell.

Here’s some of my previous lessons and experiments that involved these things:

- The Mandelbrot Set viewed through Pre-Calculus. Presentation, Lesson Plans, and Interactives. Links to the various blog posts are at the bottom of the page.
- Processing Experiments:
- Waning Moon – Steps through linear functions with slope 1 to 4.

- Interactive Waning Moon – Uses the mouse to control the slope and the y-intercept of the linear function.

- Linear Waning Moon – Interactive. Same process, but instead of a circle, inputs are in a line at the top and outputs are in a line at the bottom.

- Linear Mod Art – Interactive, similar setup as above, but useful for teaching how it works.

- Sine Mod Art – Same as above, but using a sine function instead of a linear function.

- Waning Sine Moon – Interactive circular version using sine.

- Waning Exponential Moon – Interactive circular version using an exponential function.

- Waning Moon – Steps through linear functions with slope 1 to 4.

The nice thing about the processing versions is you can pretty easily create large image files that you can print:

]]>

RAY: You’re one of a hundred people standing in line to get onto an airplane that has 100 seats. There’s a seat for every person who’s in line, and each of you has a boarding pass for your assigned a seat. The first person to walk onto the plane drops his boarding pass and, instead of picking it up, decides, “I’m just going to sit anyplace.” He takes a seat at random.

Now, every other passenger will take either his assigned seat or, if that seat is taken, that passenger will take any seat at random.

TOM: I’ve been on that flight!

RAY: You are the last passenger to walk onto the plane. Obviously, there’s going to be one seat left, because everyone else is sitting in his correct seat, or not.

The question is: What are the chances that you get to sit in your assigned seat? I’m going make this multiple choice.

A: 1 out of 2.

B: 1 out of 10.

C: 2 out of 50.

D: 1 out of 100

E: Zero.

What strategies could you give students to help them reason their way through this puzzle?

]]>Take this function: f(x) = x^2 + a*cos(ax). When you take values of **a** between 0 and 8 and back to 0, you get the following:

I’d like to make an image that represents the entire family of this function. Problem one: the function values get pretty large in this window, so find the remainder of the function output after dividing by 5. Here’s the gif of that output:

Problem two: How do we represent this in one static picture? If you take the first image (where a = 0)

and take the function values from 0 to 5 and make the pixels for row 1 colored based on the function value (0 being black, and 5 being white), then you get this row of pixels (expanded to 20 pixels high for ease of viewing):

So this row of pixels represents **a** = 0. If you set **a** to be 0 at the top of the picture, **a** to be 8 at the middle of the picture, and **a** to be back to 0 at the bottom of the picture, stitch all these lines together you get:

Click on image for full 4k by 4k resolution.

Here’s another with the function y=(10a)/(1+x^2) with a going from 8 to 0 to 8.

Click on image for full 4k by 4k resolution.

Here’s another with the function y=ceil(ax) – ax + floor(ax) with a going from 0 to 5 to 0.

Click on image for full 4k by 4k resolution.

And lastly, here’s the last one, but colorized. The 0 to 255 now controls the hue of the color (with full saturation and balance, using HSB color).

Click on image for full 4k by 4k resolution.

Here’s the code for those who are interested (or view it live on openprocessing)!

int max_a; float mod(float a, float b) { //desmos mod and processing mod act differently. //desmos mod always returns a positive value return (abs(a % b)); } float func(float x, float a) { //type 1 max_a = 8; return (mod(x*x - a * cos(a*x),5)); //type 2 this counts from a= 8 to 0 to 8 //max_a = 8; //a = max_a - a; //return (mod((10*a)/(1+x*x) ,5)); //type 3 //max_a = 5; //return(mod(ceil(a*x)-a*x+floor(a*x) ,5)); } void setup() { size(1000,1000); colorMode(HSB,255); int w,h; float x,a,f,hue; max_a = 5; loadPixels(); for (int i = 0; i < pixels.length; i++) { w = i % width; h = int(i / width); x = map(w,0,width,-6,6); if (h < height/2) { a = map(h,0,height/2,0,max_a); } else { a = map(h,height/2,height,max_a,0); } f = func(x,a); hue = map(f,0,5,0,255); pixels[i] = color(hue,255,255); } updatePixels(); //save("type3c.png"); }

edit 10/16/15:

Here are a couple more pictures for you (the code on openprocessing is up to date with these new types):

Here is the code to sketch out this 2d shape on the x-y plane and rotate it around the x-axis:

to func x out = -1 * (x - 1)^2 + 5 out end moveto 0,0,0 x = 0 xmax = 3.0 numPoints = 100 while (x < xmax) out = func(x) moveto x,out,0 x = x + xmax//numPoints end moveto xmax,0,0 moveto 0,0,0 nsides = 100 revolve 1,0,0,270

When you click the solidify you get this 3d shape (only rotated 3/4 of the way around for sake of visulization):

Here’s the magic step. Click **Download **and open the .obj file with your 3d printer software and hit print:

Want to rotate around the y-axis instead and practice shells? No problem.

And print!

The AP (but not the IB) curriculum has students find the volume of a solid created by extruding out a known cross-section from a given area. For instance, on the 2010 AP Calculus AB exam, they asked the following question, zone in on part (c):

This type of question is difficult for students to visualize. Madeup can make some great models (that can be printed) for the students. While I think the programming is tricky enough that I wouldn’t encourage you to bring the code directly to novices; it is pretty clean and easy to modify. For example: take the same area as above and find the volume by taking cross sectional squares with one side on the xy plane and perpendicular to the x-axis.

In the madeup world, the code looks like this:

to func x out = x ^ 0.5 + 1 out end moveto 0,0,0 x = 0 xmax = 3 numRect = 14 while (x < xmax) out = func(x) moveto x,out,0 moveto x + xmax//numRect,out,0 moveto x + xmax//numRect,0,0 moveto x,0,0 x = x + xmax//numRect extrude 0,0,1,out end moveto xmax,0,0 moveto 0,0,0

When you hit the extrude button, it takes the 14 rectangle slices and brings it up the z-axis to make a square. The result is the 3d shape:

Here’s the shape (approximated) with 4 rectangles:

And here’s the shape with 100 rectangles:

Print!

Enjoy! As always, please fire away with questions/comments/etc.

]]>Also check out Dan Meyer’s fall contest (ends 10/6/15) that asks the students to be creative in making their own Loop-de-loop mathematical art.

Programming connection: have the students make their loop-de-loop with python and turtles, here is a 3-5-2 loop-de-loop:

or with Scratch, here is a 2-4-5:

Enjoy! ]]>

#1TMCthing Improve my questioning in class, have students interpret each other’s answers before I barrel in and ruin the learning. #tmc15

— Dan Anderson (@dandersod) July 27, 2015

Thankfully this message found much traction in the #mtbos. Rachel and Sam Shah started up the BetterQs blog where many guests have written posts on questioning. I wrote a post today, Hints and One Helpful Question. If you’d like, go over and read it, and considering adding BetterQs to your blog aggregator of preference.

Thanks,

Dan

]]>RAY: There’s a rare disease that’s sweeping through your town. Of all the people who are exposed to it, 0.1 percent of the people actually contract the disease. There are no symptoms until the disease actually occurs. However, there’s a diagnostic test that can detect the presence of the disease up to a year before it strikes. You go to your doctor, and he administers the test. It comes out positive. You say, “I’m done for!” Then you get a little bit encouraged. You say, “Wait a minute, doc, is this test 100 percent accurate?” Your doctor responds, “Well, not really. It’s 95 percent accurate.” In other words, 5 percent of the people who take the test will test positive but they don’t really have the disease. Here’s the question: What are the chances that you actually have the disease?

]]>NEW BARBER MATH

RAY: A barber had his first customer of the day, who happened to be a friend. When he was done, the barber refused to take the money from the customer. The fellow said, “Look, I know we’re friends, but, business is business. I want to pay for my haircut.”The barber said, “Here’s what we’ll do. You open the cash register. I don’t have any idea how much money is in there. But, you match whatever is in there, and then take out 20 bucks.”

The customer says, “Okay,” and he does that.

The barber says, “Gee, I kind of like this.” So, the next customer comes in, he gets his haircut, and the barber says, “You can do the same thing my first customer did. Open the cash register, match what’s in there, and then take out 20 bucks.”

The second customer does that, and he leaves. The third customer does the same. The fourth customer, after receiving his haircut, opens the cash register, and says, “I can’t do it. ”

The barber says, “Why not?”

“There’s no money in here. Not a cent.”

The question is, how much money was in there to start?

Code is found here.

Hi definition images are found here.

Update: Here’s what happens if you zoom in on one of these graphs.