kitchen table math, the sequel: Steve H on setting up problems

Thursday, September 2, 2010

Steve H on setting up problems

re: how many unknowns?
I like to use more variables than are needed because I find it easier to create correct equations. I know that I can always turn the algebra crank later without much thought.

r+s=12 is easy and I know that it's correct. I also know that the half perimeters are pi*r and pi*s.

I then look for enough equations to meet my unknowns. That is what's funny about this problem. You don't have enough information to directly solve for the answer before you look at the choices. There are not enough equations for the variables. Even if you use just r and (12-r), you have no equation, unless, that is, you plug in each answer.

I don't like problems like this, because my first reaction is that you don't have enough information. You do, however, if you look at the possible answers.

Also, why is there no variable in the answer? It's just a unique aspect of this particular problem. What if one of the semicircles is replaced by half of a square? You would have something like this:

4r + (12-r)*pi

for the perimeter. the variable does not disappear when the expression is reduced.

You can't trust what you think because problems try very hard to trick your understanding. You just have to follow the facts (equations) and see where they lead you. As I always say, let the math give you the understanding, not the other way around.


gasstationwithoutpumps said...

"You don't have enough information to directly solve for the answer before you look at the choices."

That's wrong. The entire problem can be solved without looking at the answers. You can't solve for r and s separately (insufficient information for that), but the thing is, you don't need to know r and s to solve the problem, you only need to know their sum, and that is given.

SteveH said...

Let me put it this way. How do you know beforehand that the variable will drop out? You think you need one of the answers to solve the problem. Then you see that the answers have no variable. Only afterwards do you realize that you really don't need the answer. That was why I made the comment that you have to stick to the facts and see where they lead.

SteveH said...

This reminds me of some ratio problems where you you assign variables to the numerator and denominator, but find later that you only need one variable for just the fraction.

For example:

(a/b)^2 + 2(a/b) + 1 = 0

If you set x=a/b, you can solve this, assuming you really don't need to know what 'a' or 'b' are. However, when you are setting up the problem, this might not be obvious.

Allison said...

I stuck with the facts and never had that problem in the first place.

Problem says
"Find length of arcs". I write down r, I write down s, I write down r+s is 12, and then I write down
Pi r + Pi s. I immediately factor, put in 12. All done.

I didn't know beforehand that they would drop out. And I didn't think I needed to know either one--because I just did what I was supposed to do, given what the problem asked. I never worried about r or s.

This is, I think, an important point: don't solve problems you're not asked to solve!

Now, in fact, I didn't write anything down but Pi r + Pi s, but the steps in my head were clear.

ChemProf said...

"don't solve problems you're not asked to solve!"

Speaking of novices versus experts, that's a biggie! I can't tell you how often I see my students doing just that. They can spend so much time trying to solve a problem that isn't what was asked for (and that they don't have enough information to figure out!)