As is usually the case when an great event like this is brought to my attention I've sadly already agreed to be somewhere else. The seminar I would very much have liked to have joined is being presented by the University of Birmingham's Chris Sangwin: Automatic e-Assessment of Mathematical Word Problems (STACK). Chris has been doing some great work with computer aided assessment (CAA) systems for mathematics. The result of this work is STACK - System for Teaching and Assessment using a Computer Algebra Kernel.

And what's great about STACK is that it fully integrates into Moodle. Click here for details.

*How to make marking algebra homework a nightmare for teacher*Imagine you've completed the first question of your math homework and the answer is 3x+4. Part of the reason your math teacher wants you to write down your answer as 3x+4 (i.e. in that order) is that if all the class has written their answer with the term in x first then you don't have to confirm the math, you can just visually check that the answers are laid out correctly. But say a student has given their answer as 4+3x? It's still mathematically correct (albeit against the usual convention of writing linear expressions in the form something times x plus a constant) but, because we aren't really attempting to check the math - just the way the answer is laid out (lazy, I know) - then it's all to easy to accidentally mark an answer as incorrect when it is, in fact, right.

My example of 3x+4 is a fairly trivial linear expression. What about a quadratic? If the answer is x

^{2}+4x+4 then my student could have written that... or (x+2)(x+2), or (x+2)

^{2}... you get the idea. All are mathematically/algebraically equivalent but having to check the math for every question you're marking can become a bit of a chore.

Which is why, in exams, questions are usually very specific about how you should answer a question. For example "factor/factorize completely" or "write your answer in its simplest terms" or words along those lines.

I always get asked what the right way is of answering such-and-such a question and, to be honest, outside of an exam, I'm not that precious about it. If I'm giving one-to-one tuition in their home my usual answer to a student will be "Well, I could have come to your house by car, jet pack, helicopter. I could have come straight here, gone via Scotland (

*I live a long way from Scotland*)." In fact I could have gone round and round the world three times. As long as I get there, and as far as the student is concerned, it really doesn't matter.

So what decides which route I take? What decides which mathematical method we should use? The answer is

**. Cost (in the accounting/economics/project management sense of the word, i.e. in the "cost benefit analysis" sense) is all about taking the most effective choice within known constraints. I tell my students to chose a method (i.e. make a choice) that they feel most comfortable with. A choice that, when they are actually doing the math, they feel they are less likely to make a mistake. That, for them, will be the most cost effective route through the problem. It's a very common project management approach (effectively the one we're using for the international Moodle rollout I'm currently working on) and it can work wonders in math teaching.**

*cost*

*A worked example*In the best traditions of math textbooks, let's explore what I mean about choices having a cost with a worked example. Chris included this poser in his abstract:

**In a railway journey of 90km an increase of 5 kilometers per hour in the velocity decreases the time taken by 15 minutes. What is the velocity?**

I thought it might be interesting to work through this problem but to also explain, in terms of cost, why I personally would use particular math techniques.

I know this is probably going to sound geeky but the elegance of algebra really appeals to me. Expressing a problem in algebraic terms is going to be, for me, the most cost effective. Once I've translated the problem into algebra I'm going to continue with the algebraic approach and solve any equations I can construct to find the unknowns.

What about the less algebraically inclined? Well, if you got as far as coming up with equations you could plot them and see where the lines or curves (or whatever - I don't know yet) cross. That, of course, depends if you're more graphically motivated.

If you happen to know that speed/velocity (there's more on the difference further on in this post) is given by dividing the distance travelled by the time taken to travel that distance then there is yet another good way... but, unless you're lucky, it could take you some time. That is the way of

*trial and improvement*(basically trail and error): take a guess and work in small steps from there until you get at least a good approximation to (or at best the exact value of) the answer.

For those that are interested here are the steps I took:

**1)**

*The units aren't SI units. Convert distances into metres and time into seconds.*

90km is equivalent to 90000m

15 minutes is 900 seconds

5 kilometers per hour is 5000m in 3600 seconds. That simplifies to 25 eighteenths metres per second.

**well, basically that's the professional physicist in me. If you want to know what happens when people who really should know better don't convert their measurements to standard units then read this.**

*Reason:***2)**

*Use speed equals distance divided by time relationship...*

We rote learned this one when I was a budding young physicist in science lessons at school - so one could argue that either you know this or you don't.

**two aspects here to consider. One is that experience tells me because this is a "speed/distance/time" question so use the "speed equals distance..." relationship. Second is that I also rote learned**

*Reason:**two unknowns takes two equations*.

This is knowledge in my head. How it got there was, as I say, and rightly or wrongly, rote learning.

**3)**...

*to construct two equations*

Firstly... assign some letters: 'v' for velocity and 't' for time. My two equations are

t=90000/v

and

(v+25/18)(t-900)=90000

**That takes an understanding of rearranging equations, for example v=d/t can be written as vt=d. To do that we just need to understand that '=' means everything on one side is the same as everything on the other and, because of that, everything you do to one side you've got to do to the other. Emphasis of this last point is criminally (I think) missing from a lot of basic math teaching in the UK.**

*Reason:***3)**

*Substitute the first equation into the second*

Remember that, although I have two equations with two unknowns, what ever the values of v and t in the first equation they have exactly the same value in the second equation. They are, in the trade, simultaneous.

Now I have just one equation... with no t in it, just v:

(v+25/18)(90000/v-900)=90000

**I'm in my algebraic comfort zone.**

*Reason:***4)**

*Multiply out the brackets*

How long have you got? There are lots of different ways of multiplying out brackets. There's FOIL (first, outer, inner, last), the "box" method - where v+25/18 is the length of one side and 90000/v-900 is the length of the other and the area of the box is the area of four smaller boxes added together (if you're familiar with it you'll know what I mean). My method of choice is "happy smiley face"... I guess because *cough* I'm a happy, smiley kind of guy...

I end up with

90000-1250+125000/v-900v=90000

which, with a bit of algebraic tomfoolery, I can manipulate that into

900v

^{2}+1250v-12500=0

Why write it in that form remembering that I can write it however I like?

**now it's a quadratic I can solve it using the quadratic formula.**

*Reason:***5)**

*Put the numbers through the quadratic formula*

In fact I end up with two numbers, one positive and one negative. Can I can ignore the negative velocity? (Your author puts on his best James Bond 'Q' immitation) Now pay attention 007...

A negative speed doesn't make much sense but a negative velocity might. Speed is a scalar quantity but velocity is a vector. A negative velocity means you're going the opposite direction. However, in the context of this question the negative solution can quite happily be ignored.

So...

If you substitute a=900, b=1250 and c=-125000 into the quadratic formula you end up with a positive answer of 11 and 1/9 metres per second. That's the equivalent of 100/9 times 3600 metres every hour, or... to cut a long story short... 40km per hour.

**6)**

*Test your answer*

Don't just give the answer. If you have time (and more so if you're not feeling very confident of your result) justify your answer.

How long will it take to travel the 90km at 40km per hour? The answer is 8100 seconds.

How long at 45km per hour? That works out to be 7200 seconds.

That is a difference of 900 seconds, or 15 minutes. That gives us good justification that our answer is correct*

*Conclusions*How did you get on with Chris' problem? Which method works best for you? How do you measure the cost of a given mathematical process? I'd be interested to know what you think.

Is measuring the cost of choices before you in a project your method of managing a project? I'd be interested to hear about your experiences.

*Note that I'm not going to claim I was 100% correct. I might have made a mistake in my justification that wrongly proves an incorrect result. Again, it's the scientist in me: by showing you my working I'm hoping that any mistakes will be picked up by you - my peer reviewer, as it were. Don't hesitate to let me know if you think I'm wrong.

Just a quick follow-up to this post. I was working through this problem with one of my students last week. When it came to step 5 (where I use the quadratic formula to solve a quadratic) he completed the square - which I have to confess I didn't even entertain (even though it was me that taught him that method!).

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