Tuesday, February 6, 2007

Death By Triangle...

Well, I thought that since we only had the morning class today that there wouldn't be too much information for me to cover on the scribe post...
I was wrong...
Mr. K started the class with quick chat about net etiquette and not posting your last names or pics of yourself on the internet because "The internets memory is infinite". You may delete it yourself but it's not really gone. He said it had something to do with RSS and gave examples of how long forgotten original webpages of big sites can still be found in the internets infinite weave. Then people had problems with the homework so we went over those.
Question 11

Find the arc length of a bicycle wheel, given a radius of 24 inches and a central angle of 48 degrees.



The problem with this seemed to be how the answer sheet shows how to get the answer. Mr K expained that it uses a formula that we should completely disregard because their is a better method that is more powerful in solving problems, porportionality. He then showed how porpotions were used to derive that formula.

After that question number 10 was next on the list which served as the perfect segway into our lesson for the day.

Question 10
In what quadrant(s) is the point P(theta) if we know that:
a)cos(theta) is negative?
b)sin(theta) <> 0?
Starting that question he described the unit circle, which is a circle with a radius of 1.
P(theta) is where the angle and the circle meet and would be in a certain quadrant.
Remember:
Cast our CAST because proportional thinking is much more powerful.
Sine is a function of y.
Cosine is a function of x.
Tan is Sine/Cosine
Mr. K then challenged us to come up with side lengths from angles in the unit circle faster than him using our calculators. Of course it was fixed and he had knowledge that we did not. So he shared...

All the angles that he had chosen were 30 degrees away from the x-axis, these are called related angles. With these angles in the unit circle the ratio of the opposite arm of the triangle to the hypotnuse was 1:2, and will always be that because dilatation has no effect on that ratio. Through the pythagorean theorem you can then find the other arm to be the square root of 3.

The ratio of opposite arm to hypotnuse to adjacent arm is 1:2:sqr(3) then.
The next triangle we learned about was the one that got a man hurled off a cliff. The right angle isosceles triangle. The pythagoreans beleived intently that everything could be explained in nice neat whole number ratios. Then there was this triangle... Two sides of 1 and a hypotnuse of the square root of two, which is an irrational number, which completely goes against their beliefs. So they kept it a secret. As Mr. K said any good secret gets out and the person who let it out was killed. Then Pythagoras the leader of the religion met an untimely end in Athens when he would not run through a bean field to escape a couple of guards chasing him for escaping. Anyway back to the math of the triangle. We are taught to rationalize the denominator, so we do so when using sine and we end up with arm lengths of sqr(2)/2. While explaining these triangles he made this circle diagram similar to the one below.





The corner points on the unit circle are in gold. These show that the highest and lowest numbers for sine and cosine are 1 and -1.
I think I got most of what was in class today. The next scribe is...








23 comments:

vincentr said...

Hey Graeme! Good job! Although I was confused with my notes and your post because for pi / 3 ( 1/2 , root of 3 / 2), i have both of them as positive values since they're both in quadrant 1. I'm guessing that your is right? So should it be (1/2, -rootof 3/2) like what it said on your post? I`m not sure.

Kasiaw said...

No, I think Vincent is right...but anyway, Just wondering where you go the cool font and what program you did the circle in? Would help me greatly when it comes to be my turn.

vincentr said...

HAHA, there are some math stuff you can download where you can actually create graphs, circles, parabolas, equations without any difficulties. I'm going to search for some of them then I'll post up the link where we all can download it. This way, everyone can do their scribe post to their fullest potential. Also, if you have Adobe 7.0, it willalso be easier. I have that program and that's what I'm going to use.

sandy ♥ said...

good job graeme. keep up the great work. (:

vincentr said...

I have a very simple question, how do you actually post something up here? I see no button to hit that says "POST or WHATEVER".. HAHA, and by the way, has there been any terms for our Math Dictionaries because I wanted to start that but I don't know the terms involved.

aichelle s. said...

Wow! A-ma-zing! haha. That was an excellent post. It was very detailed and the graphics were quite attractive. Just pro photo shop! haha! Good Job Graeme!

aichelle s. said...

oh and clever headline for your post.

Ricky said...

You really DONT know how to POST YET Vincent.... AHHH... Hey Graeme.. Are we going to do that thing again.... Hahaha. Well you just go to new post on the dashboard page and then type what you want and press publish.

Ricky said...

Yes excellent Job Graeme too. I want adobe photoshop.. too. Ha Ha Ha

Grey-M said...

Yes Vincent you are most definitly right, I'll fix it up soon enough and last year he told us exactly when to write in our dictionaries off the board. If you want to make a post Vincent go to Blogger click on New Blogger" enter your gmail and password and it will take you to your "Dashboard". There will be a the button that your looking for called "New Post". And Kasia I used a program called Euclid to make the circle. It's at the steinbach school site Euclid, takes about half a second to download and is fairly straight forward. As for the font I have photoshop and a tablet and too much time on my hands.
And Richard we should just make a link to that post.

Grey-M said...

Richard how did you get the avatar?

Ricky said...

Ok this was a long time ago.... But it's still as good on new blogger as old blogger. I'll post a new one later. Hahaha But this is like the ultimate blogger help from way back BLOGGERS R US
Graeme, you can put the avatar on when you edit your personal profile.

Ricky said...

Oh yeah theres also graeme and richard's quick guide to html located here

vincentr said...

Hey guys, the chatbox isn't here yet? AWW! Anyway, i used to use those HTMLs on my pages long time ago! HAHA.. cool. And thanx for that direction for blogging. LOL

m@rk said...

Good job for today's scribe!! I personally like the link that you put about Euclid.Euclid seems to be a nice program to make images.

e said...

Hi Graeme,

When I saw your name I though you were trying to tell us how smart you are: "grey matter" :)

Anyway, I know Mr. Kuropatwa said I was going to be asking questions so I better start! You said:

Sine is a function of y.
Cosine is a funciton of x.
Tan is Sine/Cosine


My question is: why is a sine function of x, and cosine a function of y? And is tan really sine/cosine or is tan(x)=sin(x)/cos(x)? And why do you think I asked?

One more thing: what happened to #10? I was little confused about this sign: <>.

This is the last thing: good job! Keep it up. I really liked how you taught your fellow students about the blogger as well as about math.

e

Grey-M said...

Hello e,
From the wording of your question I'm guessing that what I told them to remember was mixed up, as I said that:
sine was a function of y
cos was a function of x
In your question "Why is a sine function of x, and cosine a function of y?" you infer that it is the opposite. That is why I think you have asked this, to correct me. I do think that I am right though because with triangles drawn within the unit circle, the sine of the angle in a triangle is the vertical line or the line opposite the angle when drawing triangles in the standard position. Since sine does not affect the x position of where the line meets the circle and it does effect the y position I believe it is a function of y. Same goes for cosine, as it is the adjacent arm in a triangle in the unit circle in the standard position it only effects the x position and not the y position of the point where circle and line meet and is therefor a function of x. Tan(x) is sine(x)/cos(x) because tangent is opposite over adjacent. From what I said earlier sine is opposite and cosine is adjacent. Substitute that into tan(x) = O/A and you get tan(x)=sine(x)/cos(x). Please tell me if I'm right about sine being a function of y and cosine as being a function of x because I would really hate to mislead my classmates. Thx in advance.

As for question 10, I typed it in wrong and it has been fixed.

jann said...

Fantastic use of images on your scribe Robert! Your scribe was pretty straight forward and I'm sure you covered most of the stuff we did for that class. Keep up the good work.

e said...

Graeme,

I wasn't inferring anything. But here is another lesson: you shouldn't assume immediately that you're wrong and I'm right ;) I mistyped your claim, but it matters not.

I think I understand now what our misunderstanding is all about. See, when you say that something is a function of x, that means that x in an independent variable, it is a variable on which the output depends. For example, my weight is the function of my calorie intake, so my calorie intake would be my x, and most often the output, or dependent variable, in this case my weight, is denoted by y.

Back to our functions, both sine and cosine are functions of the angle. Most often we denote the independent variable with x (you called it theta, but it really doesn't matter), so both sine and cosine are functions of angle, x (or theta if you stick to it).

I think what you meant to say is that the value of sine is read on the y-axis, and the value of cosine is read on the x-axis. Both are functions of angle, so if you choose your angle in such a way that one of its sides is the positive part of x-axis, then its other side is going to intersect the unit circle you guys talked about. The point of intersection is the point whose x-coordinate is the cosine of your angle, and whose y-coordinate is the sine of the angle.

Let me know what you think of this.
e

Grey-M said...

Thank you, I think that I get it now. Better wording for that section of mine would then be:
sine as a function of theta yields the arm length of the triangle that lies parallel to or on the y-axis and therefor the y-coordinate of the intersection of the triangle and the unit circle
cosine as a function of theta yields the arm length of the triangle that lies on or parallel to to the x-axis and therefor the x-coordinate of the intersection of the triangle and the unit circle
Tan(theta) = Sin(theta)/Cos(theta)

e said...

Yay!

Tim_MATH_y said...

this concept discussion is quite interesting.

e said...

I'm glad you approve :) If you ever feel like we could all benefit from discussing concepts, you should point it out, and I'll be happy to participate!