how many unknowns, part 2

gasstationwithoutpumps said:

Although Glen would never create 2 unknowns, preferring r and 12-r to r and s, I often find it easier to create multiple unknowns when initially setting up the problem, then remove the unnecessary ones. In this case, it was easier to remove (r+s) as a single unit, and never worry about manipulating 12-r.

I can't tell you all how important these threads have been to me: how much I'm learning (I hope I'm learning - !) and how rich the experience has been.

It's led me to think about the question of self-teaching a bit. Until last night, I had simply never thought about 'how many unknowns' in the way you all are talking about unknowns now. I had never thought about it because, where unknowns are concerned, the books seem to suggest that less is more.

Mind you, I don't think any math book I've used has directly stated that 12 - r is superior to r + s=12. I'm pretty sure I inferred that it was based in the fact that I don't recall any instances of r + 12 where 12 - r was a possibility.

This strikes me as the kind of thing a good math teacher would bring up in class, perhaps as an aside?

Or something that would come up in discussion?

What do you think?

how many unknowns?

re: how many unknowns in the two half-circle problem, Glen wrote:

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.

THIS is what I was trying to do.

THIS is what I always do, if possible.

I don't know what the problem was.

Inflexible knowledge?

Heat prostration?

I'm half serious about the heat. I took the test outside in 85+ temp. All summer long I've had severe performance deterioration any time I work in the heat. One day, when the temperature was close to 100, I found myself unable to solve even the simplest of problems. I sat at the picnic table working the same problems over and over again in slow motion. Five, 6, 7 times. Or more. I'd crawl through the problem, check my (wrong) answer, then go back to the beginning and crawl through it again and then again until finally the correct answer appeared.

Then I'd go on to the next problem and do that one 6 or 7 times.

I love summer. Have to soak up the sun while I can.

fair warning

If all goes as planned, I am going to begin working through the Unit 5 worksheets from the Arlington Algebra Project, as lgm suggested. Tonight.

I say 'fair warning' because there's no answer key.

Teaching Company high school courses on sale

I'm probably going to spring for Algebra II, God help me. (There's a workbook, too, for about $30 extra, but it's not listed on the web site. If you order, call the customer service rep.)

This review persuaded me:

An Oldie But A Goodie
April 30, 2009

To put this TTC course in a historical perspective, it's course #102 with a copyright date 1994. If you can remember back then, there wasn't much of an internet, computer graphics were in infancy, and DVD technology was still being developed. Does this mean this course should be re-done and "modernized"? Well, that'd be nice, but it's really not necessary. But it does mean the visual aids, such as they are, are very basic, and the teacher makes extensive use of the simple old blackboard and flip-chart easel.

Now, I kind of liked that, because it reminded me of when I first took Algebra II in high school, 50 years ago. But if you - or the kid you got this for - get bored if there are no special effects, animations or related "eye candy" at every turn, then this course will indeed be a poor choice. This course is for those who not only want to learn algebra, but are willing to do what it takes to learn algebra. For those people, this course is an excellent investment. It's going to take some work on the student's part, first and foremost being concentration. To state it another way, if you want to learn algebra and how to work algebra problems, this course is a valuable asset, and if you don't, I'm at a loss as to what to recommend.

Most TTC Science & Math course are survey courses, meaning they "sell" the subject and its high points in an entertaining manner, and take you up to, but not into, problem-solving. That's NOT true with this one. It doesn't pretend, or even try, to get you fired up about algebra. This is a pure dirty-hands work-course. Frequently, the instructor will put a problem on the board, and ask the student to stop the "tape" and solve the problem. When you restart the course, the instructor will show the solution and what steps are involved. If you skip these problems, you have no way of gauging how much you've learned - or not. To learn algebra, you have to work algebra - there's no ducking this - and this course offers plenty of that kind of opportunity.

I'll also note that, although it's not on the web page, this course still comes with a workbook, aligned with the lecture topics, with answers in the back. If you're looking forward to a formal classroom course, working these problems is certainly in your best interest.

Another reviewer mentioned "Algebra II for Dummies". This is also a good self-study approach. There's a companion workbook, which I also recommend. (The price for both is only about $30.) Between this DVD course and the Dummies book and the workbooks, you should have all you need to fully move around at this algebra level. And if you'll be taking a high-school course, these two items will prepare you about as much as is possible.

Regarding the instructor - ok, Jay Leno he's not. But I enjoyed his "blue collar" kind of instruction, and I had no problem at all understanding him or the information he was trying to get across. And no, he doesn't look at the camera all the time, but this makes it "seem" like a classroom environment, where the teacher certainly doesn't (I hope) spend the entire period glaring at you.

I really enjoyed this course and got a lot out of it. It was well worth the price. For the intended audience, I definitely recommend it."

This is annoying. The chemistry course, which I bought just a few months ago, has now been revised.

true confession

I found the wild-goose-chase problem, the one that took me many steps not to do while C. solved it at once using "logic and reasoning."

Find the value of r so the line that passes through each pair of points has the given slope.

44. (-2, 7), (r, 3), m = 4/3 ^*

C.'s solution:
7 - 3 = 4/3 (-2 - r)
4 = (-8 - 4r)/3
12 = -8 - 4r
20 = -4r
r = -5

My non-solution:
y = 4/3x + b
7 = 4/3(-2) + b
7 = -8/3 + b
21/3 + 8/3 = b
29/3 = b

I seem simply to have stopped at this point. I don't know why. Looking at my solution page now is a bit like examining a single-vehicle accident scene trying to determine how the driver flipped his car twice in broad daylight & nice weather.

Alternatively, I didn't stop with 29/3 = b (I remember having done more steps) but the rest of my perambulations are recorded on some other piece of paper, not this one.

We'll never know.

knock on wood
true confession
wild goose chase


^* source: Glencoe Algebra, p. 261

help desk, part 3 slope again

Steve H left this demonstration (proof?) of why, when calculating slope, it's OK to use either point as "Point 1":

(y1-y2)/(x1-x2)

= -(y2-y1)/[-(x2-x1)]

= -1*(y2-y1)/[-1*(x2-x1)]

= -1/-1 * (y2-y1)/(x2-x1)

= (y2-y1)/(x2-x1)

Here's my question.

What are the justifications for each step?

(y1-y2)/(x1-x2)

= -(y2-y1)/[-(x2-x1)] mutiplicative identity ??

= -1*(y2-y1)/[-1*(x2-x1)] mutiplicative identity again??

= -1/-1 * (y2-y1)/(x2-x1) commutative property of multiplication??

= (y2-y1)/(x2-x1)

I'm definitely ready for something more formal...although I'm not sure where I'm going to find it.

Think I'll check Dolciani's & Foerster's books.


update from Steve:

I think schools should spend much more time on these identities. They should show how they are used in all sorts of interesting ways, forwards and backwards. I don't think I really learned algebra until this happened. [Catherine here: I agree absolutely. I'm really feeling a need for this, and it's not really something I can "provide" for myself - at least, not without a HUGE amount of effort.]

(y1-y2)/(x1-x2)

= -(y2-y1)/[-(x2-x1)]

This is the Distributive Law, in reverse, if you will.

[Catherine: I didn't see this! Now I do!]

= -1*(y2-y1)/[-1*(x2-x1)]

I suppose you could call this the Multiplicative Identity. I mentioned before that the sign "belongs" to the term that comes after it. I always like to think of a sign as a +1 or a -1. [Catherine: me, too]

= -1/-1 * (y2-y1)/(x2-x1)

I suppose you could call this the Commutative Property of Multiplication law. In other words, it doesn't matter which way you multiply fators. That's why I liked to have students circle the factors in a rational term. Then they know that they can move the factors to any position.


= (y2-y1)/(x2-x1)

I got rid of the -1/-1 using the Multiplicative Inverse law.


It's interesting that in the list of basic identities I found online that there wasn't an identity for

a/1 = a

I suppose this would come from the Definition of Division

help desk, part 2

I need a royal road to geometry.

Tomorrow I begin Lesson 88 in Saxon Algebra 2.

I won't finish the book until the very end of the summer, which means I will have taken 12 months to get through the thing.

Next I start Saxon Advanced Mathematics, which will take me another year to get through if I'm lucky.

fyi: Saxon publishes four high school books:

• Saxon Math Algebra 1
• Saxon Math Algebra 2
• Advanced Math
• Calculus
  

Geometry and trigonometry are integrated throughout the first 3 books.

So I'll spend next year working my way through Advanced Mathematics while Christopher, who will be in 8th grade, finishes Math A: algebra and some geometry. (Math A is still a 1 1/2 year course, as far as I can tell, though people keep saying NY is going back to the old algebra 1 - geometry - algebra 2 sequence. His class began Math A in January.)

Freshman year, if he stays on the fully accelerated track which I expect he will, he'll be taking geometry - real geometry, with proofs. Or so I hear. (Is this a separate, souped-up geometry course that's neither Math A nor Math B? Don't know! I get all my info from my neighbor, whose son is a year ahead in school.)

Assuming you aren't hopelessly confused by now,^* you may see the problem.

He's catching me.

I'm 2/3 of the way through Saxon Algebra 2, and I have yet to do a proof. Unless there are a lot of proofs in Advanced Mathematics (which there may be - don't know)^** I'm going to be starting geometry-with-proofs the same time Chris does.

That's not good.

I need a royal road.

Does anyone know whether Tammi Pelli's Proofs Workbook might work?

At 64 pages, it's the right length.


(large image)




(large image)




^**update: proofs in Advanced Mathematics

Just checked the scope and sequence for Advanced Mathematics:

Proofs
Elements of Proofs
Understand basic logic and reasoning
State the contrapositives of conditional statements
State the converses and inverses of conditional
statements
Do proof outlines
Do formal proofs
Theorems
Prove the chord-tangent theorem
Prove theorems about secants and tangents
Prove theorems about chord products
Prove the Pythagorean theorem
Prove similarity of triangles
Prove the law of sines
Prove that equal angles imply proportional sides


questions about Math A
Math A Regents exams (archived)
Math A Toolkit
Home Instruction Schools Regents Exam Review
sample Saxon Math lessons
Saxon scope and sequence



^* I'm hopelessly confused, but I'm used to it