math – Montessori Muddle

Area of a Triangle

The area of a triangle is one half of the length of the base times the height:

$A = \frac{1}{2} \cdot b \cdot h$

For my Geometry class, I made this set of six triangles to show that as long as the base and height are the same, all these triangles will have the same area.

Each student measured a triangle and found its area, which is a useful exercise in itself to get them to transfer the ideas and equations out of the book, and then the all compared their results. Their calculated areas were all within 5% of the actual value, which was not unexpected given that some small measurement error was inevitable.

Since you can use any side as the base, not everyone measured the equivalent side and height, so I had to demonstrate that similarity as I summed up the exercise.

For the next time I use this set, I’ve marked the one side that is 10 cm on each triangle for students to use as the base.

Basic 3d Printing with OpenScad

In Geometry we’re using OpenScad to create basic, 3d geometric solids and printing them to get to know the 3d printers.

A basic example is creating a cylinder with a radius of 10 mm and a height of 5 mm. The code in OpenScad looks like this (OpenScad Cheat Sheet):

cylinder(r=10, h=5);

and it looks like:

After we Render the object, we export it as a 3d .stl file. This we can load into the 3d printer software.

For our smaller, Prusa, printer, the program is PrusaSlicer. You’ll need to slice the object and then export the G-code. The program that actually prints the object is Pronterface.
For the larger Lulzbot printer, we use the Cura program and can print directly from the software once we’ve sliced the model.

The end result, on the Prusa printer looks like this:

Math Flowcharts

Flowchart in progress. Showing topics being covered in basic statistical graphing. The topics they are working on at the moment are highlighted in yellow. The worksheets attached to each topic are linked on the left side in the highlighted area.

To help students track their progress in math we’ve started requiring them to map what they’re doing on flowcharts. Right now, they’re doing it on paper, but we’re working on getting it to be all electronic.

Tracking with the charts helps them see how what they’ve done fits into the bigger picture. It allows them to be able to go back up the chart to previous topics if they need to review, and look forward to what comes next (and to work ahead if they would like).

The image above shows a sample what a student’s flow chart would look like while they are working on a subject (statistics in this case). The topics they are working on at the moment are highlighted in yellow. The worksheets attached to each topic are linked on the left side in the highlighted area. Links to references (Khan Academy for example) are linked on the right–there are only a few on right now (see the Mean topic on the upper right of the flowchart).

The topics on the flowchart can be expanded (using the green button on the top right of each topic) to show more detail.

Expanded window for the topic "Mean". — Expanded window for the topic “Mean”. Detail on the topics is a little sparse at the moment as we’re focusing on setting up all of the flowcharts first.

At the moment, I’m uploading the flowcharts that we’re currently using up on the website myself. Students can use the website to get worksheets and links to references, but if they mark what they’re doing on the webpage it’s not saved. We’re currently working on making it possible to create and edit flowcharts on the website itself. After that, we’ll be setting it up so students can log in with their school accounts and track their progress electronically. One ultimate goal is to map the entire upper school curriculum with these flow charts so a student might be able to track their work all the way from 7th to 12th grade.

Projectile Paths

Paths of a projectile. The half oval represents a line connecting the maximum heights of each projectile. — Paths of a projectile.

I had my Numerical Methods student calculate the angle that would give a ballistic projectile its maximum range, then I had them write a program that did the the same by just trying a bunch of different angles. The diagram above is what they came up with.

It made an interesting pattern that I converted into a face-plate cover for a light switch that I made using the laser at the TechShop.

Maximum Range of a Potato Gun

One of the middle schoolers built a potato gun for his math class. He was looking a the mathematical relationship between the amount of fuel (hair spray) and the hang-time of the potato. To augment this work, I had my Numerical Methods class do the math and create analytical and numerical models of the projectile motion.

One of the things my students had to figure out was what angle would give the maximum range of the projectile? You can figure this out analytically by finding the function for how the horizontal distance (x) changes as the angle (theta) changes (i.e. x(theta)) and then finding the maximum of the function.

Initial velocity vector (v) and its component vectors in the x and y directions. — Initial velocity vector (v) and its component vectors in the x and y directions for a given angle.

Distance as a function of the angle

In a nutshell, to find the distance traveled by the potato we break its initial velocity into its x and y components (v_x and v_y), use the y component to find the flight time of the projectile (t_f), and then use the v_x component to find the distance traveled over the flight time.

Starting with the diagram above we can separate the initial velocity of the potato into its two components using basic trigonometry:

$\cos{\theta} = \frac{v_x}{v}$
$\sin{\theta} = \frac{v_y}{v}$ ,

so,

$v_x = v \cos{\theta}$ ,
$v_y = v \sin{\theta}$

Now we know that the height of a projectile (y) is given by the function:

$! y(t) = \frac{a t^2}{2} + v_0 t + y_0$

(you can figure this out by assuming that the acceleration due to gravity (a) is constant and acceleration is the second differential of position with respect to time.)

To find the flight time we assume we’re starting with an initial height of zero (y₀ = 0), and that the flight ends when the potato hits the ground which is also at zero ((y_t = 0), so:

$! 0 = \frac{a t^2}{2} + v_0 t + 0$

$! 0 = \frac{a t^2}{2} + v_0 t$

Factoring out t gives:

$! 0 = t ( \frac{a t}{2} + v_0)$

Looking at the two factors, we can now see that there are two solutions to this problem, which should not be too much of a surprise since the height equation is parabolic (a second order polynomial). The solutions are when:

$! t = 0$

$! \frac{a t}{2} + v_0 = 0$

The first solution is obviously the initial launch time, while the second is going to be the flight time (t_f).

$! \frac{a t_f}{2} + v_0 = 0$

$! t_f = - \frac{2 v_0}{a}$

You might think it’s odd to have a negative in the equation, but remember, the acceleration is negative so it’ll cancel out.

Now since we’re working with the y component of the velocity vector, the initial velocity in this equation (v₀) is really just v_y:

$! v_0 = v_y$

so we can substitute in the trig function for v_y to get:

$! t_f = - \frac{2 v \sin{\theta}}{a}$

Our horizontal distance is simply given by the velocity in the x direction (v_x) times the flight time:

$! x = v_x t_f$

which becomes:

$! x = v_x \left(- \frac{2 v \sin{\theta}}{a}\right)$

and substituting in the trig function for v_x (just to make things look more complicated):

$! x = \left( v \cos{\theta} \right) \left(- \frac{2 v \sin{\theta}}{a}\right)$

and factoring out some of the constants gives:

$! x = -\frac{v^2}{a} 2 \sin{\theta}\cos{theta}$

Now we have distance as a function of the launch angle.

We can simplify this a little by using the double-angle formula:

$! \sin{2\theta} = 2 \sin{\theta}\cos{theta}$

to get:

$! x = -\frac{v^2}{a} \sin{2\theta}$

Finding the maximum distance

How do we find the maxima for this function. Sketching the curve should be easy enough, but because we know a little calculus we know that the maximum will occur when the first differential is equal to zero. So we differentiate with respect to the angle to get:

$! \frac{dx}{d\theta} = -\frac{v^2}{a} 2 \cos{2\theta}$

and set the differential equal to zero:

$! 0 = -\frac{v^2}{a} 2 \cos{2\theta}$

and solve to get:

$! \cos{2\theta} = 0$

$! 2\theta = \cos^{-1}{(0)}$

Since we remember that the arccosine of 0 is 90 degrees:

$! 2\theta = 90^{\circ}$

$! \theta = 45^{\circ}$

And thus we’ve found the angle that gives the maximum launch distance for a potato gun.

The Joy of Math

Regardless, if learning is to be as efficient and deep as possible, it’s essential that it be done freely. That means giving children a voice in which activities to participate, for how long, and also the level of mastery they want to achieve. (“This is the biggest clash with traditional curriculum development,” Droujkova notes.)
— Vangelova (2014): 5-Year-Olds Can Learn Calculus

This article provides a lot of evidence to support the notion that the conceptual aspects of calculus and other “higher level” forms of math should be taught at all age levels, not just at the end of high school or in college.

The presentation below elaborates:

Dilation

A quick program that animates scaling (dilation) of shapes by scaling the coordinates. You type in the dilation factor.

dilation.py

from visual import *

#axes
xmin = -10.
xmax = 10.
ymin = -10.
ymax = 10.
xaxis = curve(pos=[(xmin,0),(xmax,0)])
yaxis = curve(pos=[(0,ymin),(0,ymax)])

#tick marks
tic_dx = 1.0
tic_h = .5
for i in arange(xmin,xmax+tic_dx,tic_dx):
    tic = curve(pos=[(i,-0.5*tic_h),(i,0.5*tic_h)])
for i in arange(ymin,ymax+tic_dx,tic_dx):
    tic = curve(pos=[(-0.5*tic_h,i),(0.5*tic_h,i)])

#stop scene from zooming out too far when the curve is drawn
scene.autoscale = False

# define curve here
shape = curve(pos=[(-1,2), (5,3), (4,-1), (-1,-1)])

shape.append(pos=shape.pos[0])
shape.color = color.yellow
shape.radius = 0.1
shape.visible = True

#dilated shape
dshape = curve(color=color.green, radius=shape.radius*0.9)
for i in shape.pos:
    dshape.append(pos=i)

#label
note = label(pos=(5,-8),text="Dilation: 1.0", box=False)
intext = label(pos=(5,-9),text="> x", box=False)

#scaling lines
l_scaling = False
slines = []
for i in range(len(shape.pos)):
    slines.append(curve(radius=shape.radius*.5,color=color.red, pos=[shape.pos[i],shape.pos[i],shape.pos[i]]))


#animation parameters
animation_time = 1. #seconds
animation_smootheness = 30
animation_rate = animation_smootheness / animation_time

x = ""
while 1:
    #x = raw_input("Enter Dilation: ")
    if scene.kb.keys: # event waiting to be processed?
        s = scene.kb.getkey() # get keyboard info

        #print s
        if s <> '\n':
            x += s
            intext.text = "> x "+x
        else:
            try:
                xfloat = float(x)
                note.text = "Dilation: " + x

                endpoints = []
                dp = []
                for i in shape.pos:
                    endpoints.append(float(x) * i)
                    dp.append((endpoints[-1]-i)/animation_smootheness)
                #print "endpoints: ", endpoints
                #print "dp:        ", dp
                for i in range(animation_smootheness):
                    for j in range(len(dshape.pos)):
                        dshape.pos[j] = i*dp[j]+shape.pos[j]
                    rate(animation_smootheness)
                    if slines:
                        for i in range(len(shape.pos)):
                            slines[i].pos[1] = vector(0,0)
                            slines[i].pos[-1] = dshape.pos[i]

                for i in range(len(shape.pos)):
                    dshape.pos[i] = endpoints[i]
                    slines[i].pos[-1] = dshape.pos[i]

                for i in range(len(shape.pos)-1):
                    print shape.pos[i], "--->", dshape.pos[i]
                    
            except:
                #print "FAIL"
                failed = True
                intext.text = "> x "
            x = ""

Toothpick Shapes’ Sequences

Using a sequence of connected shapes to introduce algebra and graphing to pre-Algebra students.

Make a geometric shape–a square perhaps–out of toothpicks. Count the sides–4 for a square. Now add another square, attached to the first. You should now have 7 toothpicks. Keep adding shapes in a line and counting toothpicks. Now you can:

make a table of shapes versus toothpicks,
write the sequence as an algebraic expression
graph the number of shapes versus the number of toothpicks (it should be a straight line),
figure out that the increment of the sequence–3 for a square–is the slope of the line.
show that the intercept of the line is when there are zero shapes.

Then I had my students set up a spreadsheet where they could enter the number of shapes and it would give the number of toothpicks needed. Writing a small program to do the same works is the next step.