Math Challenge: Can You Draw This?

 

Please, do not visit her post just yet

Fawn Nguyen wrote a post on Friday that caught my attention.

She divided her math class into pairs, making one person in each pair a Describer and the other person a Drawer.

She then gave the Describers a figure the Drawers were supposed to draw at 1:1 scale and three rules designed to prevent the Drawers from seeing the figure, the Describers from seeing the Drawers’ drawings, and either from using gestures and body language to signal information.

They were allowed to talk all they wanted.

I thought today I would be your Desciber. We can use the comments to this post to talk to each other.

Exercise in visualization and communication

This Challenge is an exercise of visualization, communication and knowledge. Visualizing and knowing what one is seeing (recognition: visualization, knowledge); describing this efficiently and effectively (oral communication); visualizing what is described (listening, visualization); connecting that visual to known shapes and images (knowledge); and efficiently and effectively drawing the visualized figure to scale (visual and haptic communication, knowledge) are foundational skills our students need to use and communicate mathematics both within and beyond the classroom.

I participated in a similar exercise Dr. David Pimm conducted in a Math ed graduate class. The Describer picked a part of a much larger and more complex figure than Nguyen’s figure and described it and its location to the Drawer. The Drawer visualized, located and identified, rather than drew, what the Describer was seeing. We were required to sit on our hands, a behaviour that would have made Nguyen’s Drawers’ jobs much harder. But the Describer and Drawer in Pimm’s class had to confirm orally that the visualizations match (secret messages, discrete mathematics), a slightly different skill than Nguyen’s Drawers were learning.

Both exercises are simultaneously frustrating and engaging, like an addictive game one struggles in, cannot win and yet cannot put down. By design, the exercises target these emotions; one’s communication and visualization skills; and one’s knowledge. These are emotions and skills that a majority of their time the mathematician and the teacher — indeed all people — encounter and need to cope with. (For this reason, Pimm’s exercise was perfect for preservice teachers.)

 

 

Good luck. Enjoy this math Challenge.

Ready, set, …

There is only one rule here: do not visit Nguyen’s post no matter how tempted you are to do so until you have finished drawing your figure.

Feel welcome to comment me with queries, comments, your final drawing, instructions for a figure you find or design (please provide a URL to this figure, so others and I can find it, and yet not see it here), and most importantly your reflections (both experiential and critical) after taking up this challenge.

Draw a figure following these instructions.

  1. Use a ruler and a compass to draw this figure. Blank paper; a sharp pencil (2H or harder); eraser; and either coloured pencils (green, light blue and dark blue), another sharp pencil (HB or softer) or a pen (or three of different colours) may also help.
  2.  

  3. With a sharp hard pencil (2H or harder), lightly
  1. construct three concentric circles with radii of 32 mm, 59 mm, and 78 mm.
  2. draw a diameter across the smaller, inner (32 mm) circle that also intersects the larger, outer (78 mm) circle twice, with ticks.
  3.  

  4. poke your compass into one of the intersections of the inner circle and its diameter;
  5. without lifting the poked end of your compass, stretch your compass to the opposite arc of the middle (59 mm) circle; and
  6. scratch an arc above and below (orthogonal to the diameter) the common centre of the concentric circles.
  7. repeat the last two instructions after poking the other intersection of the inner circle and its diameter.
  8. use your ruler to “connect” each intersection of the arcs and the common centre of the circles.
  9. with your ruler set as just instructed, draw a diameter across the inner circle and two ticks intersecting the outer circle.
  10.  

    Notice, the two diagonals now frame four 90° angles.

     

  11. poke your compass back into one of the two holes that you just made in the inner circle.
  12. scratch an arc across the middle of the two 90° angles adjacent to the diameter you are “in”.
  13. repeat the last two instructions for each of the remaining three intersections of the inner circle and its two diameters.
  14. use your ruler to “connect” the intersections of opposite arcs and the common centre of the circles.
  15. with your ruler set as just instructed, draw a diameter across the inner circle and two ticks intersecting the outer circle.
  16. repeat for the other pair of opposite arcs.
  17.  

    There are now four diagonals that frame eight 45° angles.

     

  18. once again create arcs to bisect these angles following instructions 2i — 2l.
  19. for each pair of opposite angles, with your ruler set as instructed, scratch a tick where the ruler intersects the middle circle (eight ticks total).
  20.  

  21. starting with one tick intersecting the outer circle, connect that point of intersection with an adjacent point of intersection in the middle circle.
  22. connect that middle-circle intersection with the inner-circle intersection along the same radius as the already connected outer-circle intersection.
  23. complete the square by connecting the inner-circle intersection and outer-circle intersection with the intersection in the middle circle that the unfinished square opens toward.
  24. repeat the last three instructions to complete the ring of eight squares.
  25.  

    The construction is complete; however, you might want to emphasize the figure by colouring or darkening its components.

     

  1. With a sharp soft pencil (HB or softer) or three pencil crayons of different colours, or with a pen or three pens of different colours, darken
  1. the outer circle in one colour; Nguyen used green.
  2. the horizontal and vertical line-segments (as you look on the figure) in another colour; Nguyen’s was dark-blue.
  3. the diagonal line-segments in a third colour; Nguyen used light-blue.
  4.  

    I stippled the squares and scumbled or hatched alternate rhombi to emphasize the three dimensional effect.

  1. Check Nguyen’s post to compare your drawing with the original figure.

So did you do it? Did you get it right? What did you learn? I would love to know.

Send me your queries, your comments, your final drawings, your instructions for a new figure, and your reflections (both experiential and critical).

Advertisements

The Algorithm in the Code: Building an Inquiry Tool

A couple of days ago, I posted a Math Challenge posed by David Wees some weeks ago. The code emulated Euclid’s Algorithm of Coprimes and GCFs.

 

 

First analysis

Analysis of the code reveals that, when a=0, b=b and, when b=0, a=a. However, a reaches zero at the code’s onset, while b does the same after the code runs through scenarios when b≠0. This implies that one of the two values reaching zero is key to the code and the quantity of the other value when this happens is informative.

Tabulating the difference between possible values of a and b within an arbitrary range of integers might illustrate how b=0 is reached. This process falls in the Planning and Implementation steps of David Coffey’s thinking-stage charts. Here is my table for mapping the “moves” of the code within the range of -9<a<9 and -9<b<9.

 

 

Playing with b≠0

Notice that a>b below the a=b or 0 diagonal. So, for instance, the difference between a=6 and b=4 is 2, found in the bottom-left triangle of the table. In this triangle, according to the code, the difference a-b equals a. So, now a=2 and b=4.

Repeating the process using a=2 and b=4 produces a difference of 2, this time in the top-right triangle. The new difference b-a equals b. Now a=2 and b=2, which produces a difference b-a of 0. Since b-a = b, the code ends with b=0 and a=2.

 

 

Cases

There are six distinct cases where the code returns unique case results.

Case 1: a = b

When a=b, the code returns their common value. Why? As shown in the example above, the step after a=b is b=0 and the value of a is returned. This value is that when a=b.

Case 2: either a = 0 or b = 0

A starting value of a=0 returns b. A value of b=0, returns a. This is a rule built into the code. But what would happen if the rule were not followed?

Let’s take our b=0 and a=2 example beyond termination. Continuing the while-loop produces a difference a-b of 2, in the bottom-right triangle. This difference returns a=2 and b=0, exactly where we started.

What if a=0 and b=2? The difference b-a returns b=2 and a=0, another recursive repeat.

So, a=0 returns b and b=0 returns a. If a=b=0, zero is returned (in agreement with Case 1).

Case 3: a < 0, b < 0 or both < 0

When either a or b or both are negative, the code never resolves to termination (except when a=b, Case 4). In fact, the greater value iterates to infinity in steps of the lesser negative value.

Let us try a=3 and b=-2 (we could easily have tried a=-2 and b=3). The difference a-b returns a=5 and b=-2, which in turn returns a=7 and b=-2, then a=9 and b=-2, ad infinitum.

a=-3 and b=-2, on the other hand, returns (a=-3,b=1), (a=-3,b=4), (a=-3,b=7), again ad infinitum.

 

 

Case 4: a = b < 0

Contrary to Case 3, when a=b<0, the common value of a and b is returned, in agreement with Case 1. Ignoring this, Case 3 is followed; however, there is no condition that rectifies the ambiguity of which direction, toward infinite a or infinite b, the map should follow.

Case 5: +a and +b share a common factor

When a and b share a set of common factors, the greatest of these factors is returned, as per the a=6 and b=4 example which returned 2, the greatest common factor of 6 and 4.

Case 6: +a and +b are coprime, or relatively prime

When a and b do not share a common factor, 1 is returned, since 1 is the only natural number that is a divisor of both.

Let’s map a=3 and b=8. As you can see from the table below, 1 is returned.

 

 

The analysis of cases weaves over and through David Coffey’s thinking-stage charts’ Analysis through Verification stages.

Interpretation and Pedagogy

I was introduced to the formal Extended Euclid’s Algorithm via induction within a discrete mathematics university course. It was taught to me as a means to learn modular mathematics, so not much emphasis was placed on explaining the Extended Algorithm nor the induction. In fact, given this challenge posed by David Wees, or perhaps more so the table derived from it, the manner in which I learned the Extended Algorithm was probably the worst possible.

David’s challenge and the table offer great entry tasks into the study of GFCs, coprimes, Euclid’s Algorithm and several branches of mathematics that build from them. Before the Algorithm is even named and formalized, students get to explore its mechanisms and formalize their own rules based on their mapping activities. Once they master the code and table, they can learn the corresponding Algorithm schema with emphasis on matching the items of the schema to the mapping on the table and the methods in the code. Then the Algorithm can be named and its uses illustrated.

For those students who do not know code, the teacher can interpret the code with them and offer scaffolding afterward. The code is probably easier to understand than an instruction list, if instead of treating it as code, the teacher treats it as an outline of process. Notice the subtle difference here between instruction (do this) and process (this is how this works).

The table doesn’t just determine GFCs and coprimes, it illustrates how greatest common factors and relatively prime, or coprime, numbers are calculated. It also illustrates why negative integers do not produce finite results, except where a=b, and why a=0 returns b and b=0 returns a.

One question that might remain is what the table and code return. In the case of positive integers, the returns are obviously GCFs or 1. Interpretation can determine whether the initial values are coprime or related by common factor. But what does the return of a when b=0 and the return of b when a=0 mean?

Quite simply the returns are the divisors of the numbers being analyzed. So, if one of those numbers is zero, it stands to reason that the other number is a viable divisor of zero. For instance, when b=0 and a=3, the return of 3 signifies that zero is divisible by three. Arithmetically, when a=b=0, infinity or undefinable should be returned, since conventionally no number “can” be divided by zero. This is the one flaw in this code and table.

In order for the constructed table to be a viable tool for learning Euclid’s Algorithm, it should be printed out or created with non-erasable ink and the mapping should be done with pencil and eraser. The table can be used several times then to build literacy, mastery and fluency of Euclid’s Algorithm.

Do you have any tasks that engage students in active learning of the outcomes, content, skills and concepts you are teaching?

Math Challenge: Do you know what algorithm this is?

David Wees came up with this challenge. Determine what algorithm this code emulates. You will find the answer more informative if you create a table to see the pattern of moves the code makes. The table can then be used to introduce the algorithm formally to your students. Or better yet, get them to build their own tables from the code. Try a range of integers to test the code. What patterns exist?

 

 

I will post my answer in a couple of days.

Math: My Solutions to mathhombre’s Problems

Part of the job of a teacher is to model her or his Subject-specific thinking to help students understand the subject techniques and the Subject processes, patterns and metacontent. To this end, I dedicate this post to modelling how I tackled the math problems, created by John Golden as part of a math final for preservice teachers, that I mentioned in my last post.

mathhombre's Twitter avatar I enjoy these types of problems. They are chuck full of math — and Math — from geometry and trigonometry to algebra and arithmetic. But even more so, they require mathematical thought, imagination, creativity, logic and perseverance to solve them. Had John expanded the questions to ask for metacognitive reflection and recording, at least the trigonometry question could easily pass as a Math lab.

Once again I invite you to visit his post and take a crack at his problems before reading on.

The remainder of this post contains my solutions to these problems. (I went overboard with the trigonometric problem. I blame it on play and slightly improved health.)

Trig Problem 2

The first problem John offers is a circle geometry / trigonometry problem. He presents a diagram and simply asks the test taker to “Figure out some of the missing information in the diagram”. The only information the student is sure of from this task is that some information — more than one piece — is missing. Interpretation, logic and problem solving are all the tools the student has to continue with the question.

As I mentioned in my last post, what I really like about this question — and the next one — is its openness. The student isn’t asked for all possible missing information. He can choose what she wants to explore and discover.

I began my activity to this task by stating two assumptions on which the rest of my work rests. I assumed that point C is the center of the big circle. I also assumed that point A is the center of the small one. You might think this obvious given the diagram, but I like stating these assumptions right away. That way everyone knows where I stand and the rest of my work can not be faulted based on ambiguity on these points.

I then stated the explicit givens, that angle ADE is 30°, angles AED and ACD are 90°, and segment AC is 3 units long. (All of this and the rest of the original problem diagram is in light blue in the answer diagram below.)

Here is a diagram of my answer.

 

 

Click to enlarge or visit the GeoGebra construct from which it derives.

Joy, Pythagoras

Based on the assumptions that C is the center of the big circle and A is the center of the small one, I calculated and labelled the radii for the big and small circles. So CD, CI, CL and CM are each 6 units long and AC, AG, AH, AI, AJ and AK are each 3 units long.

This made apparent that triangle ADC is a right triangle with legs CD of 6 units and AC of 3 units, allowing (through Pythagoras’s Theorem) hypotenuse AD to be calculated as square root 45 or 3√5, which is also the hypotenuse of triangle AED.

Making angles

Equally obvious to the use of Pythagoras’s Theorem to calculate the common hypotenuse of triangles ADC and AED, is the use of the 180° (or π) “rule” for use in triangle AED.

This concept is so well known, it has almost become pneumonic. Yet, think on how profound the notion that the internal angles of all triangles always add up to a common constant, and that this constant, 180°, is also found in: the angles that accumulate to make a line, the conversion from degrees into radians and the command that sends us back the way we came.

In triangle AED, angle DAE is 60° because angle AED is 90° and angle EDA is 30° and these add up to 180°.

Attempting a similar algorithm in triangle ADC requires some labelling. I set the measure of angle ADC to unknown x, which makes the measure of angle CAD to 90-x by the 180° rule. But the fun does not stop there. Since the angles touching line EF on one side at common point A also add up to 180°, angle FAC is 30+x (60+(90-x)+(30+x)=180).

One could circle point A applying this rule to label all angles surrounding A, but another way is to apply the Vertically Opposite Angle Rule (yet another pneumonic one can ignore since the 180° rule produces it) to determine that angle JAK is 60°, IAJ is 90-x and HAI is 30+x.

Similar rules can be used to determine that all the angle measurements about points C and E are 90°, but also to calculate that angles CFA, LFN, ABE and VBW are 60-x in measure. (Oh, by the way, I added points B, L, N, M, P, Q, R, V, W, S, T and U to the diagram. The last five are hidden off the edge, but are used solely to allow labelling of angles VBW, UDT, TDS and LFN vertically opposite of and equal in measure to angles ABE, EDA, ADC and CFA respectively.)

The value of x, and consequently of the other angles, is calculated applying the Law of Cosines in triangle ADC.

AC² = AD² + CD² – 2(AD)(CD)cos(x)
3² = (3√5)² + 6² – 2(3√5)(6)cos(x)
9 = 45 + 36 – (36√5)cos(x)
cos(x) = [(-72)/(-36)](1/√5)
cos(x) = 2/√5
x = 26.57°

From this, 90-x = 63.43°, 30+x = 56.57° and 60-x = 33.43°. Using the 180° rule to back check these values confirms them.

These calculations take care of all the obvious angles that the problem suggests are missing.

Measuring sides

Observation of the diagram to this point reveals that quadrilateral AEDC is not a kite since AC ≠ AE (AH = AC and AE = AH + EH), angle DAE (at 60°) ≠ CAD (at 63.43°) and angle EDA (at 30°) ≠ ADC (at 26.57°).

The lengths of AE and DE need to be calculated.

I used the Law of Sines to calculate the lengths of AE and DE. This is fairly easy since the sines of 30°, 60° and 90° produce easy ratios, 1/2, √3/2 and 1 respectively. The calculation looks like this.

sin(30°)/AE = sin(60°)/DE = sin(90°)/(3√5)
1/(2AE) = √3/(2DE) = 1/(3√5)
AE = 3√5/2, DE = 3√15/2

I repeated this process to calculate the lengths of AF, CF, BA and BE.

sin(33.43°)/3 = sin(90°)/(AF) = sin(56.57°)/(CF)
AF = 3sin(90°)/sin(33.43°) = 5.45 units
CF = 3sin(56.57°)/sin(33.34°) = 4.54 units

sin(33.43°)/(3√5/2) = sin(90°)/(BA) = sin(56.57°)/(BE)
BA = 3√5sin(90°)/2sin(33.43°) = 6.09 units
BE = 3√5sin(56.57°)/2sin(33.43°) = 5.08 units

It did not escape me that triangles ABE and ACF, with equivalent interior angles, are mathematically similar.

Having calculated the length of CF as 4.54 units and the length of CL as 6 units, it holds that FL = CL – CF = 1.46 units.

Unfortunately, the same can not be said for calculating the lengths of EQ, EP and FN, which remain undetermined. Perhaps there is a trick to calculating them. I was considering the power of a point using secants BD (through Q) and BM (through I) to calculate EQ, but the lengths of DQ, BQ, MI and BI are unknown. Unfortunately, the length of “chord” RQ, subtended by angle EDA and creating triangle QDR, also can not be determined because the length of DQ is again unknown. With point A and angle DAE mathematically set, the length of chord PN might be calculable, paving the way to calculating the lengths of EP and FN. I have not yet figured out how to do this. This leaves EQ, EP and FN unmeasured.

Measuring JR

The length of JR is 1.02 units. But I cheated to determine this!

Whereas all the other determined values described in this post are calculable from the initial information provided in this post and John’s original problem, calculating the length of JR required mechanical measurement of another length, mainly the length of DO. GeoGebra did the measuring, and had it not done so, the length of JR would be undetermined like those of EQ, EP and FN.

Wait! So why didn’t I measure the lengths of EQ, EP and FN? For the lengths of these segments, I have no theory to back my measurements with. For calculating the length of JR, I do.

The length of JR can be calculated using the Side-Splitter Theorem. (Funny, I used to call this the Parallel Projection Theorem. The Side-Splitter Theorem sounds like a comedy act.)

Segment AC splits triangle DOR parallel to leg OR of that triangle. The proportions of triangle DCA are already known to be CD:AC:AD = 6:3:3√5. So knowing the length of DO or OR can provide the length of DR in triangle DOR.

Here is where the mechanical measurement comes in. I used GeoGebra to measure (not calculate) the length of DO at 9.60 units. From this, using proportional scaling of similar triangles DCA and DOR, I calculate the lengths of OR and DR as 4.80 units and 10.73 units respectively.

The length of JR = DR – DA – AJ = 10.73 – 3√5 – 3 = 1.02 units.

Still if this were a real exam and I did not have access to GeoGebra, I would not be able to determine JR’s length.

Why stop there?

That essentially covers the basic values of John’s original problem that any preservice teacher would reasonably be expected to determine on a test. But why stop there?

I am under no time pressure and I saw right from the start that I can calculate values associated with those circles. I am sure John did not expect such a step, but here I am noticing patterns and recognizing opportunities. So, here John are my bonus marks.

The circumference (2π*radius) of the small circle is 6π since its radius is 3 units. That of the big circle is 12π from its radius of 6 units.

The central angles of these circles produce isosceles triangles with radial legs of either 3 units (small circle) or 6 units (big circle) and chords subtended by these angles.

In particular, the small circle’s central angles GAH and JAK at 60° suggest equilateral triangles AHG and AKJ. They also suggest that arcs GH and JK have arclengths of π. Since these triangles are equilateral, GH and JK are 3 units long just like the radius of the circle. This implies an arclength of one radian or π. Independently, 60° = (60°/180°)π = π/3, while the arclength of an arc of a circle is calculated as the measure of its central angle times the radius of the circle (angle*radius). In this case angle*radius = (π/3)*(3) = π.

Since the other central angles of the small circle are not 60°, their angles times three are used to calculate the arclengths of their arcs and the Law of Cosines is used to calculate the lengths of their chords.

The length of CG, for instance, is 3.15 units as calculated below.

(CG)² = (AC)² + (AG)² – 2(AC)(AG)cos(angle CAG)
(CG)² = 3² + 3² – 2(3)(3)cos(63.43°)
CG = 3.15 units

Arc CG, corresponding to this chord and similarly subtended by central angle CAG, has arclength 1.06π units ((63.43°/180°)*3 = 1.06).

For the big circle, there are four isosceles triangles with 6 unit-long radial legs and 6√2 unit-long chords (calculated using Pythagoras’s Theorem since the central angles around C are 90°). The corresponding arcs are all 3π in arclength.

I used the 180° rule for triangles, used in the Making Angles section, to calculate the other angles of each of these isosceles triangles.

For triangle CAG, angles AGC and GCA are each equal to 180° – angle CAG = 180° – 64.43° = 58.29° in measure.

And that exhausts the values I calculated.

What I missed

I noticed upon reviewing my answer diagram that I forgot to calculate the “exterior angles” around B, D and F, that is angles WBI, EBV, NFC, KFL, CDU and SDE. However, I did calculate these in the GeoGebra construct using the 180° rule for lines.

Angle WBI = angle EBV = 180° – 33.43° = 146.57°
Angle NFC = angle KFL = 180° – 33.43° = 146.57°

Angle SDE = angle CDU
180° – angle TDS – angle EDA
180° – angle ADC – angle UDT
180° – 26.57° – 30°
Angle SDE = angle CDU = 123.43°

Geometry Problem 1

Relative to the circle geometry / trigonometry problem above, John’s line geometry problem is trivial. There is not much to do with it except fill in the missing angles following the given information.

I started out by assuming that lines JK and LM were parallel to each other. This was a pretty good assumption, given that John was testing student-teacher knowledge of parallel lines, perpendicular lines, transversal lines and associated angles.

Here is my answer. I will leave the explanation to you.

 

 

Click to enlarge or visit the GeoGebra construct from which it derives.

Benefitting Our Students

Solving an unfamiliar problem in front of, or better yet, with our students can greatly benefit them. They get to see how an unfamiliar problem is solved. They get to see how we approach and think on a problem. They get clearer — or messier — explanations as we think aloud. They even get to see us make mistakes and how we recognize and resolve them.

Sometimes modelling mathematical activity when faced when an unfamiliar task or challenge can flop on us and backfire on our kids. But how we deal with these flops is also informative.

Most of the time, our flops are minor — and often silly (yes, 1 + 2 = 3, not 4) — and our modelling, after correcting the error, leads to a correct answer.

But what our students gain is our thinking, reasoning and problem solving expertise. They learn the difference between solving and solution, between proving and proof. They learn to be — and how to be — creative, logical, persistent and progressive in our tackling of a problem. They learn to touch Math, instead of just math.

And that is where we want our students to be.

Gone are the days where the answer to a particular problem is satisfactory. Now thinking is where it’s at. We have to model our thinking and discuss solving with our students. We have to teach them to think, so they can excel beyond our meager abilities.

Our students deserve better and modelling is one key way (another would be rich math problems and labs) to ensure they get the meta out of their math.

Math: Let Students Earn Their Weight

mathhombre's Twitter avatar In a post he published yesterday, John Golden offered two geometric problems he had included in a math final for his preservice math teachers. Please visit his post and take a crack at his problems.

 

 

 

 

What I really liked about these problems was how open they were. The first problem, giving a circle geometry diagram, challenges John’s students to “Figure out some of the missing information in the diagram”. The second problem, also giving a diagram — this one of a series of lines, segmented only by the diagram edge, asks students to “Find more angles”.

At first, I thought this was a rather vague way to offer problems, particularly on a final exam where marks are summative and presumably weighted more. Then, I gave it some thought. I am not sure what John had in mind, but having questions which ask for “some” or “more” answers leaves the onus of how much work the student does and what weight the problem has on a test on the student. A way to assign a mark to such questions is to make “Each correct answer worth one mark”. Then all one has to do is make the test worth out of so-many maximum marks (less the total possible to get), order the questions so the “big” ones come first and voila … one has a test that the student weighs.

But what about students who answer only some questions thoroughly and skip others? One could make each question worth a minimum amount of marks, or make all questions equally difficult or multi-outcomed. There are likely other ways to ensure all outcomes are covered by the student.

It is an idea, one that just occurred to me as I read John’s post. What I really like about it is that it is truly open. The student cannot prejudge how much a question is worth and allocate effort accordingly. The grade the student gets is tied to the value the student, not the teacher, applies to each question.

What do you think? Do we herd our children with our judgement of worth? Should we be teaching them to do this intrinsically?

Oh, and by the way, my next post will show my solutions to John’s questions.

Transition into School: Studying Streams to End Summer

Goodbye International Year of Youth

Today is International Youth Day, the last day of International Year of Youth. Both are a special celebration of youth, our future and childhood, all of which are deserving of endless recognition. As teachers, and some of us as parents, we are keenly aware of the value, potential and capacity of youth. We celebrate youth every day and every year.

In North America, the school year will resume in a couple of weeks. We are all busy preparing ourselves and our classes for our students. We are eager to get started and perhaps even a bit nervous to do so (stage fright is so exhilarating).

Update: Due to the fact that I am not an organization arranging a celebration of youth related events, I do not have permission to display the UN International Year of Youth logo. So, I have exchanged my image for a link to the logo. Please select the link to see the logo.

 

 
 

The International Year of Youth logo depicts a planet filled with colourful speech bubbles. The speech bubbles and the sense of community they convey symbolize the theme of the International Year of Youth: “Dialogue and Mutual Understanding”. The logo illustrates that the entire world can get involved in the International Year of Youth and can promote dialogue and mutual understanding. The words “International Year of Youth” appear below the logo together with date of the Year (August 2010 – 2011) and the slogan for the Year “Our Year. Our Voice”. [The slogan was chosen by the global youth themselves.]

Refer to UN Observances for other official international UN events.

Just because the International Year of Youth is ending, doesn’t mean we shouldn’t keep youth in mind. In fact, with the last couple of weeks of Summer still ahead of us, now is a perfect time to think about outdoor science, whether we engage in it now or in September. And what better field camp or trip to transition out of Summer and into school with than to study the local creek or river.

 

 

Multi-disciplinary science

Stream science is multi-disciplinary. It involves problem solving in hydrology, geology, geohydrology, geomorphology, chemistry, ecology, physiology, animal behaviour, lux physics, physics, mathematics, art, sonology and civil, environmental and restoration engineering in a simple, concrete, measurable, authentic and holistic setting.

Students use levels, flow meters, D-frames, drift nets, bio-samplers, measuring tape, surveying equipment, sketch supplies, still and video cameras, sound recorders and computers to conduct various experiments to measure, inventory and analyze the morphology and ecology of streams. Not only do they explore the characteristics of streams and stream ecosystems, they use cool tools in authentic situations. If the stream needs restoration or habitat maintenance, the activities the students do aid the community as well.

 

 

Streams are defined by mostly one-direction flow (of water, substrate, energy, organic matter, chemistry/water quality, biota, riparian and hyporheic character, and stream morphology) and ecosystem instability (dynamic disturbance). They are dynamic systems, and students generally only get to visit a snapshot of these ecosystems.

Some stream topics are:

 

 

Here I will discuss only some mathematical aspects of streams to illustrate the interdisciplinary nature of stream study.

The math in stream morphology

This section corresponds to the blue ‘Units’ and ‘Types’ sections of the concept web above.

Streams are periodic in shape, though the shape of the stream and length of the periodicity change with time and location. This shape or stream morphology is determined by landscape features and stream discharge, the volume of water that passes across a cross-section of a stream within a certain time. The greater the discharge, the more drastic its effect (e.g, a 100-year flood is actually a measure of volume, not frequency). The predominant discharge is bankfull discharge, a discharge that fills a channel from bank to bank. Due to its predominance, bankfull discharge literally carves the stream channel.

 

 

To understand how bankfull discharge carves the channel, one must understand the nature of unconstricted currents. Streams roll in three dimensional Euler loops as they flow. We see this helical structure in currents in the atmosphere and ocean. On land, the landscape constrains these loops. Different stream structures or types form as the Euler current is embedded onto different landscapes and substrates.

 

 

Nonetheless, there is a predictable underlying base pattern to stream morphology. I will present only the mid- or reach-scale morphology, leaving the stream-long and micro- or substrate-scale morphologies to the students to research and study. Reach morphology is determined by bankfull width, which is proportional to bankfull discharge. The following diagram illustrates this morphology, opening a whole new insight into streams that most students never before notice. Equally inspirational hydraulic patterns are waiting to be discovered at other scales of stream morphology.

 

 

The reach morphology of the stream cycles downstream with changing periodicity and structure in response to changes in the landscape and substrate the stream cuts through. In addition, bankfull width changes as the stream encounters different obstacles and constrictions. So, though the ideal stream form is predictable, the actual morphology of any stream is dynamic. The form also migrates downstream as the stream erodes and deposits bed and bank substrate.

Stream and riparian landscape and ecology are strongly influenced by stream morphology, hydraulics and hydrology. In addition, like the landscape, biota influence stream morphology.

All this makes stream science a truly interdisciplinary study and a great way to transition students out of summer and into school.

What do your community and you do to help students embrace the last two weeks of summer?