kitchen table math, the sequel: solution manual - section 6, problem 5

Tuesday, August 31, 2010

solution manual - section 6, problem 5


How do you solve this?
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31 comments:

Anonymous said...

let r be the radius of the left circle, s the radius of the right circle.

Each half circumference is pi* radius,
so the total length is pi*r+pi*s,
but r+s=12, so total length is 12pi.

August 31, 2010 at 6:53 AM
Catherine Johnson said...

arrgh

I was stumped by the fact that I didn't know the relative lengths of the two radii.

sigh

August 31, 2010 at 8:39 AM
Anonymous said...

One way to walk through a problem you don't know is to ask yourself

what do I know?

how can I write what I need in terms of what I do know?

I know r+s.
I know the lengths of the indivdiul curves, too.

So, the curve around R is Pi*r, where r is the radius of circle with center R. The curve around S is Pi*s, where r is the radius of circle with center S.

The sum of the curves is Pi*r + Pi*s.

I know r+s. How can I rewroie Pi*r + Pi*s in terms of r+s?

August 31, 2010 at 8:43 AM
Catherine Johnson said...

My problem was that it didn't occur to me to assign different variables for the two radii.

It didn't occur to me I was going to end up with a nice, neat little equation I could solve via factoring.

August 31, 2010 at 12:29 PM
Anonymous said...

But it didn't occur to me, either. I didn't know that when I started writing the problem in terms of r and s. That's why it's important to start writing things down in the way the question asks for things--because you don't know what's going to factor ahead of time. You hope that as you do the problem, you'll see it.

Not assigning two different variables is different: that means you didn't notice the circles were different. Which is odd, because the picture shows they are, and more, because in mathland, unless you're told things are the same, you MUST assume they are different, so you must use different variables.

August 31, 2010 at 12:44 PM
John Warner said...

I knew there was a formula connecting the radii r + s = 12 so s = 12 - r
the length of the first semicircle is pi * r. The length of the second semicircle is pi * s = pi * (12 - r)
The whole length is pi * r + pi * (12 - r) = pi * r + pi * 12 - pi * r = 12pi
I also knew that because they were asking the question there must be a solution and I have confirmed my thought that the relative sizes of the radii do not matter.

August 31, 2010 at 2:00 PM
Catherine Johnson said...

Not assigning two different variables is different: that means you didn't notice the circles were different.

I did notice they were different; that's why I despaired, thinking that if I set up equations with two variables I would have two equations & never the twain shall meet.

August 31, 2010 at 2:16 PM
Catherine Johnson said...

you don't know what's going to factor ahead of time

I've been assuming that people who are proficient in math know this - !

August 31, 2010 at 2:17 PM
Catherine Johnson said...

my thought that the relative sizes of the radii do not matter

Here's what's frustrating to me.

First of all, I did see that the two radii were different.

Second, since I couldn't find any way to figure out the ratio of the two radii, and since this is an SAT question and thus has an answer, I, too, thought the problem must be solvable without knowing the ratio.

But that was as far as I got.

At that point, I was careering back and forth between "two equations! oh no!" & "don't know the ratio! oh no!" --- and as I was taking a timed test, I had to throw in the towel.

I've seen several people advise starting by simply working your way through tests, leaving timed practice for later on.

I should probably just work problems right now without setting the timer.

I also need to study beginner probability and functions. Am planning to work through Dolciani's chapter on quadratic functions soon.

August 31, 2010 at 2:24 PM
Anonymous said...

Maybe we're talking about different things. I'm trying to talk about my thought process as I do the problem.

I certainly knew by the time I'd written down pi r + pi s that I'd been given r+s, and so I knew I'd factor it that way.

But as I literally thought about each sentence of the problem, I didn't know that the sum of the two arcs would correspond to this sum and that I'd factor it to get the right thing.

I was just doing what I think of as thinking mathematically: using only what I was told, no more, no less, and seeing where I was trying to go.

I didn't despair about two equations or two unknowns (which is solveable, by the way) because I didn't go off worrying about a problem that wasn't there yet.

This is really the crux, I think. Mathematical thinking requires forcing yourself to be disciplined in a way which only addresses what's present, and not allow yourself to drift off into other things.

This is a lack of mastery, but it's something deeper. If you see two unknowns and it drives you to a kind of pattern-matching panic, where you are searching the mind's database for "examples of two unknowns" then you are not thinking mathematically. It may be that as a test strategy, such a method works, but I find it difficult to believe. A person thinking mathematically isn't searching their mind for any string to grasp hold of. They are step by step deriving or inferring what they can from what they've got.

This is what's wrong with math ed today. Its lack of coherence leads kids to grasp at straws instead of building a rational picture of what's known.

August 31, 2010 at 2:26 PM
Anonymous said...

Your panic is why test panic is real. Fear makes thinking impossible. It's difficult to explain to someone how it's not that one is taking a test that leads to the fear, but the actual question in front of oneself that can bring on the fear--the fear that "I don't know how to do THIS" can spiral.

Certainly, building up one's muscles by not worrying about the time helps. So does reading the test and skipping a question where you don't know what to do--go solve what you can, get over that initial wave of sickness and fear, and then calm down.

What tricks do you do to get over writer's block, or the terror of the blank page?

August 31, 2010 at 2:30 PM
Catherine Johnson said...

People who think mathematically never search their mind's database for relevant knowledge, principles, and examples?

August 31, 2010 at 2:32 PM
Anonymous said...

I must not be saying this clearly.

Yes, mathematical thinking often means solving a problem by reducing it to another already solved problem.

This is not like searching the WHOLE database at all. It's not like saying "there's a polygon in this problem; what do I know about polygons." It's about inferring what's known, what's needed to be known, and what you need to do to get from here to there. Did the problem really ask for me to know things about polygons, or did everything I needed to know about polygons appear in the problem statement? Math is a coherent whole, so seldom are you really searching for some bizarre unknown fact that sits on the side and magically solves the problem. Even if there is such a one, you should be able to get there from here in logical steps.

So the principles and examples follow from what you know, instead of doing a key word search.

Maybe I can't phrase this well at all, but here's an example of thinking mathematically:

You see a math problem on a midterm which asks you to prove a theorem. You start to solve it, and realize that to prove what you want to prove, you need a certain lemma.

But you don't know the lemma. Instead of searching the db of your mind, grasping for record of having heard the lemma, you realize: lemma must be true, so write on midterm "given lemma,...." and proceed.

The idea being: math follows from itself. There may be times when things are obscurely true in research, but as you study a subject, you aren't grasping from faraway fields as much as you are following a path stone by stone.

August 31, 2010 at 3:19 PM
Anonymous said...

What I'm trying to say is that what mathematicians know aren't obscure facts that reveal some secret connection between things--so it's not like they are desperately looking for some secret key or formula to help them. They are plodding through the definitions they have in front of them, and the definitions they know in their mind.

August 31, 2010 at 3:37 PM
lgm said...

Did c=2 π r pop into your head when you read the problem? How about when you glanced at the answers, following the test strategy of crossing out the obviously wrong?

Instead of thinking 'I must factor', I have found it more useful to think "I must use the Distributive Property" wisely.

August 31, 2010 at 6:19 PM
Catherine Johnson said...

What tricks do you do to get over writer's block, or the terror of the blank page?

I'm pretty low fear --- !

August 31, 2010 at 6:25 PM
Catherine Johnson said...

Tonight I have been boldly assigning variables right and left!

Speaking of low-fear.

August 31, 2010 at 6:26 PM
Catherine Johnson said...

I have practically no test anxiety, either.

I probably have the opposite. I work better under pressure (or at least I used to).

August 31, 2010 at 6:26 PM
Catherine Johnson said...

Here's another factor (a factor!)

A couple of years ago I read a fabulous observation, by a math professor I think, about brilliant math students versus works-hard math students. (This may have appeared in a study; I don't remember now.)

The observation was that the difference between brilliant and works-hard was that the brilliant students produced elegant solutions and proofs, while the works-hard students went on wild goose chases.

I love that!

I've spent the past 5 years energetically chasing down wild geese, catching them most of the time.

It's fun, but it's not elegant.

Lately I've been thinking I should be less sloppy; it's a deliberate practice issue, I think. If I'm going to carry on practicing problems I can already do, which I am for the time being, then I should try to produce more streamlined solutions.

This comes into play with the SATs in a major way, because SAT problems don't 'mess around.' They're direct, everything you need to solve the problem is present (I haven't seen any 'not enough information' questions yet), and nothing is extraneous.

With SAT math questions, one of the clues that I'm on a wild goose chase is that my attempted solution is way too complicated.

I'm sure both of these elements (self-knowledge re: propensity to go on wild goose chases and knowledge of the nature of SAT questions) strongly influenced me to think that I was wrong to consider two variables.

August 31, 2010 at 6:39 PM
Catherine Johnson said...

Did c=2 π r pop into your head when you read the problem?

Yes.

Instantly.

August 31, 2010 at 6:40 PM
lgm said...

When you read the part that said RS=12, did r+s=12 pop in?

(r=radius of circle w/center R, s=radius of circle w/center S)

August 31, 2010 at 6:52 PM
lgm said...

that's the missing key.

imho many kids would get that but miss the distributive property part.

August 31, 2010 at 7:00 PM
Anonymous said...

Catherine, if you haven't yet read Polya's "How to Solve It", then stop working on SAT problems and go to the nearest library or bookstore that has a copy. Read through it once before picking up the SAT problems again.

Polya explains mathematical problem solving (in the form of instructions) clearly than anyone else.

August 31, 2010 at 7:01 PM
Catherine Johnson said...

When you read the part that said RS=12, did r+s=12 pop in?

Wait!

I just re-read.

Here's what popped into my head:

* the two radii were presumably of unequal lengths; at least, nothing in the wording of the problem told me that the two radii were equal

* c = 2 pi 4

* the two radii added together = 12

What did not pop into my head was the idea of assigning the variable s (or any other variable) to the radius of circle S and the variable r to the radius of circle R and going from there.

August 31, 2010 at 7:12 PM
Anonymous said...

You need to practice symbolic transcription and interpretation.

You need to learn how to turn what you're reading into symbols. Then, when you're done, you need to learn how to turn it back into concepts to make sure you've answered the right question.

Every sentence needs to be rewritten in mathematics. "the two radii...": those are represented in a very compact notation by variables. You need not write r = s or r != (!= means not equals) s, because neither fact is specified. "the distance from ..." is easily transformed into these symbols, then, so you have RS = 12 yields r+s = 12.

This is true throughout all of the problems you post here. You'd be better off spending time practicing these translations before trying to chug through the actual solution.

August 31, 2010 at 7:24 PM
ChemProf said...

One thing that would confuse a lot of my students that may be confusing you is r = radius of left half-circle and R as the center point of the half-circle. Sometimes it can be easier to make anything you don't know x and y, as that starts to look more like something you are used to.

So x is one radius, y is the other, and x + y = 12. This is actually what saves you -- you are absolutely right that for the two semi-circles you have two equations "and never the twain shall meet". You'd really have one equation (xPi + yPi) with two unknowns. However, since x + y = 12, that gives you two equations two unknowns!

What did you write down when you started the problem, if you didn't assign variables?

August 31, 2010 at 7:46 PM
Catherine Johnson said...

However, since x + y = 12, that gives you two equations two unknowns!

hmmmm...

I will mull.

I didn't write anything down -- I assumed that path was wrong & I kept trying to see why I was wrong, what I was missing.

Then I ran out of time.

August 31, 2010 at 7:49 PM
Glen said...

I would never create two unknowns in a situation like this, where the two radii are not independent. Since the distance from R to S was given as 12, the radius of one circle made a good unknown, and the radius of the other was 12 minus that SAME unknown. Either circle would do, of course.

The length of the curve can then be expressed in terms of the one unknown for both semicircles. Using the left circle, and calling its radius r, the right has to be 12-r, so the two semicircles added together were,

= pi*r + pi(12-r)
= pi*r + pi*12 - pi*r
= 12pi

If I took part of my $100 and gave it to a friend, there would be only one unknown. Whether you made it the amount I gave him, or the amount that I kept, or the percent I gave him, or the percent I kept, or the difference in dollars or percent or fraction between what he got and what I kept, or the ratio of our money, or whatever, there is only one unknown. Everything else in such a problem can be expressed in terms of that one unknown, which usually makes the problem easier to manage.

September 1, 2010 at 4:31 AM
ChemProf said...

"I didn't write anything down" - that's the first thing I'd try changing. When you are solving new problems, you need to SEE what you are trying to figure out, and writing it down can actually help you see what the path should be.

September 1, 2010 at 11:30 AM
Catherine Johnson said...

"I didn't write anything down" - that's the first thing I'd try changing.

That's strange - apparently ktm ate one (or two) of my comments.

I left two this afternoon.

Heck.

One of them said that I am definitely taking chemprof's advice.

The other had to do with how much I appreciate all the help --- I've been thinking about the comments here all day long, rehearsing them.

THANK YOU!

Hope this comment stays put.

September 1, 2010 at 4:15 PM
Catherine Johnson said...

I just remembered the other comment, which related to Glen's comment:

I would never create two unknowns in a situation like this, where the two radii are not independent. Since the distance from R to S was given as 12, the radius of one circle made a good unknown, and the radius of the other was 12 minus that SAME unknown. Either circle would do, of course.

THAT was what I was trying to come up with ----- and couldn't.

September 1, 2010 at 4:16 PM

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