Task 2

[James Crook]

Task 2 was:

"I want ONE person to come forward and tell me that everyone in the
group now knows how to multiply 2x2 matrices and find the inverse of
2x2 matrices (and knows what a matrix is). This will test your
ability to communicate, your ability to check understanding, and your
ability to make a decision and share openly as a group. Remember, if
two people come forward it's no good. It needs to be ONE person. How
you do that - it's up to you."

So what needs to happen?

[James Crook]

3 days left on this challenge.

Hint: using e-mail to each other might be a good idea.

--James.

[James Crook]

2 days left on this challenge.

--James.

[James Crook]

1 day left on this challenge.

[Nicole Shih]

Is this all you wanted?

1.000[d  -b]

ad-bc[-c  a]

where

[a  b]

[c  d] are four numbers

and then multiplying two by twos are

[a1  b1]  [a2  b2]=

[c1  d1]  [c2  d2]

 

[a1a2+b1c2   a1b2+b1d2]

[c1a2+d1c2   c1b2+d1d2]

[James Crook]

Yes, that's exactly what I wanted and it's correct.  

However, I now know you know.  I don't know that Jonathan knows! 

He could just cut and paste your answer, and I still wouldn't know that he knows.

So as a group, how can we check that everyone understands?  

This is about moving beyond the model where there is just one teacher in a group.

It takes a little while to get used to.  I started moving beyond 'just-one-teacher' in the group by asking for a mathematical fact.  That way you were each teaching the group something by sharing a mathematical fact.  Now I am taking it a little further.

--James.

[Shri R Ganeshram]

Hmm,
I sent a message earlier asking for you guys to let me(sh...@mit.edu) know if you know; that way, I could help you catch up if needed and then let James know (as a single representative) but some of you have yet to let me know. :/
________________________________________
From: l2lea...@googlegroups.com [l2lea...@googlegroups.com] On Behalf Of James Crook [james....@gmail.com]
Sent: Thursday, June 23, 2011 6:09 PM
To: l2lea...@googlegroups.com
Subject: Re: Task 2

Yes, that's exactly what I wanted and it's correct.

However, I now know you know. I don't know that Jonathan knows!
He could just cut and paste your answer, and I still wouldn't know that he knows.

So as a group, how can we check that everyone understands?
This is about moving beyond the model where there is just one teacher in a group.

It takes a little while to get used to. I started moving beyond 'just-one-teacher' in the group by asking for a mathematical fact. That way you were each teaching the group something by sharing a mathematical fact. Now I am taking it a little further.

--James.

On 23 June 2011 22:31, Nicole Shih <nicole.s...@gmail.com<mailto:nicole.s...@gmail.com>> wrote:
Is this all you wanted?
1.000[d -b]
ad-bc[-c a]
where
[a b]
[c d] are four numbers
and then multiplying two by twos are
[a1 b1] [a2 b2]=
[c1 d1] [c2 d2]

[a1a2+b1c2 a1b2+b1d2]
[c1a2+d1c2 c1b2+d1d2]

On Thu, Jun 23, 2011 at 4:41 AM, James Crook <james....@gmail.com<mailto:james....@gmail.com>> wrote:
1 day left on this challenge.

On 22 June 2011 19:30, James Crook <james....@gmail.com<mailto:james....@gmail.com>> wrote:
2 days left on this challenge.

--James.

On 21 June 2011 09:46, James Crook <james....@gmail.com<mailto:james....@gmail.com>> wrote:
3 days left on this challenge.

Hint: using e-mail to each other might be a good idea.

--James.

[Abe Rabin]

Yeah, Shri told us that if anybody didn't know, they should e-mail him.

[Jonathan Joo]

Ok, here's my go at the pyramid problem.

Volume=Base Area*height*1/3.

So there are two cases:

The base of the pyramid is a square 10x10. Thus, the area of the base is 100.

The height he mentioned was 5, so 5*100*1/3 is 500/3. that's one possible volume for the pyramid.

The other case is if the sloped sides are each of length 10. Then, we can form a triangle using the height, (5), and the sloped sides, which are 10. The sloped side is the hypotenuse, and we know the height, so it's just a matter of 5^2 + x^2 = 100.

X^2 = 75

X is the square root of 75.

But then the square root of 75 is the distance from the center of the square base to a corner. That means that the actual side length of the base of the square is sqrt(75)*sqrt(2), so sqrt(150).

From there, we just do sqrt(150)*sqrt(150) to get the area of the base, so 150, then multiply by the height and 1/3.

So 150*5*1/3. This is 250.

I hope my math is right, kind of hard to describe my thinking process without a diagram to show it.

-Jonathan Joo

---

From: Shri R Ganeshram <sh...@MIT.EDU>
To: "l2lea...@googlegroups.com" <l2lea...@googlegroups.com>
Sent: Thursday, June 23, 2011 6:52 PM
Subject: RE: Task 2

[James Crook]

Yes, that's correct.  Thanks.

Has everyone seen a proof, using double integrals, that the volume of a pyramid is 1/3 the area of the base times the height?  It works whatever the shape of the base, so it works for a cone too.

It's not difficult.  The key idea is to slice the pyramid into lots of slices.  The slices are all the same shape.  The ones near the base are the largest, and the slices get smaller and smaller as you move towards the apex.  Would someone who know this please take over the explanation?  I'd like everyone in the group to understand how this works.  

The exact same idea works for more complex shapes.  For example we can, and will, slice a sphere into circular slices to show that the formula for the volume of a sphere follows from the formula for the area of a circle.  This uses the Pythagorean Theorem in a rather nice way.  Instead of the constants in these formulas for volumes of solids 'coming from nowhere' we are going to show how the formulas are all related to each other.  It's no accident that we see pi here again in the formula for the volume of a sphere and not some totally new constant.

--James.

[James Crook]

OK.  So this is not just about the maths, it is also about teaching.

You learn very quickly as a teacher that people are reluctant to say that they don't understand something.  People are especially reluctant to say if they think they should know something, and are not sure that they do.  

One of the ways around that is diagnostic tests.  We are NOT trying to rate people and say who is better than someone else.  A diagnostic test identifies bits of maths that people don't quite know.  For example the question about The Pythagorean Theorem was a diagnostic.  I was pleased to get Abe's answer "It depends. Is the 12cm side a hypotenuse or a leg?" - brief and to the point - and from it I know that he has not just learned Pythagorean Theorem by rote, but also knows how to apply it in 'both' directions. 

Another much more complex diagnostic test is here http://onlinemathcircle.com/wiki/index.php?title=Differential_equations - it's in the 'prerequisites' section.  This is a test to see if someone knows the right things for a university differential equations course.  It's not a test of how clever they are.  That's not what matters.  It's a test to find out what they need to know.  If someone doesn't know these things they will struggle with solving differential equations.  As a teacher and student it is far better to follow up on the diagnostics, rather than pile on more maths that won't be properly understood.  Unfortunately in University students can go through whole courses not understanding and not saying.  Then they go to another course that builds on that.  We need to do everything we can to change that.  If maths is taught well and built up in a good way it is complex, beautiful and fun.

So back to "Task 2".  Asking 'does everyone understand this' WILL NOT WORK as a way to find out if people know.  This is a matter of psychology, not maths.  I think everyone 'knows' matrices, in that they've seen them, and know basically how they work.  But lets complete Task 2 and check this out, and at the same time fix it if needed.

Here is a list of people:

James

Abe

Shri

Abe

Kelsey

Jonathan

Nicole

James

Each person is to challenge the person below them in the list with a diagnostic matrices problem.  NOT a contest problem, but something that should be not hard - if you know matrix basics.

For example, Nicole could ask Me:

What is:

1 2

3 -4

times:

5 6

-7 8

It's a good test of matrix multiplication because the matrices aren't diagonal matrices, all the numbers are different, and there are some negative numbers to check that I get the signs right.  If I give the right answer then it's a good indicator I know matrix multiplication.  But if instead I say:

5 12

-21 -32

I am sure she can guess what is going wrong and put me right.  If someone did think that was the right answer, then we would definitely have needed to sort that out before doing anything more advanced with matrices.  

So, please get to it.  You have the list.  Right now challenge the person below you in the list with a diagnostic matrix problem with an e-mail just to them.  Make up a problem yourself.  Don't wait to receive a problem to do this.  Don't use the exact same example I gave.  Check both multiplication and inverse.  Check that they get the right answers back to you.  If they do, tell Shri, and Shri can then tell me when everyone in the group has shown they know.  If they don't get the right answer either explain to them yourself or get Shri or me or Nicole to explain.  If they don't give you any answer at all after one day, tell me, and send me the problem you gave them.  Also tell me if you don't receive a problem within one day.

--James.

[Abe Rabin]

Just wanted to say, my name is on the list twice, should it be like that?

Also, isn't there a general solids volume formula for any shaped base with edges drawn to a single point above it?
I'll look it up.

[James Crook]

Correction:

James

Abe

Shri

Kelsey

Jonathan

Nicole

James

On 24 June 2011 13:00, Abe Rabin <honest....@gmail.com> wrote:

Just wanted to say, my name is on the list twice, should it be like that?

Also, isn't there a general solids volume formula for any shaped base with edges drawn to a single point above it?
I'll look it up.

Yes, one third base * height (as measured perpendicular to base).  We are going to show why it is one third, rather than just quoting the formula, and also check whether it is 1/4 for the four dimensional equivalent.

--James.