Involute!

Involute of a Circle

I picked up my Calculus textbook again today, because I was having fun with polar curves and parametric equations.

Man, parametric equations still make me scratch my head sometimes! I get it, it just takes a while to really get what's going on.

Anywho. I came across this problem:

Problem:

A string is wound around a circle and then unwound while being held taut. The curve traced by the point P at the end of the string is called the involute of the circle.

If the circle has a radius r and centre O and the initial position of P is (r, 0), and if the parameter θ is chosen as in the figure, show that the parametric equations of the involute are
...

I'm purposely leaving this out, because I can figure it out on my own, thank ye very much!

The problem also came with the diagram to the left (I made this in Graph but had to define r, so here, r=1).




Solution

The first thing I did was find the path of the point T.

x = r * cosθ

y = r * sinθ

To get the parametric equations for the path of P, something must be added or subtracted to the path of T.

Re-draw the figure as triangles:

The distance from T to P is the same as the arc length for an angle θ. So, that distance is

S = rθ = distance from T to P

I made some reference points:

C := the point on the radius, with same y-value as in point P

D := the point with the x-value of point T and the y-value from point P.

x_p := distance along x-axis, from D to P. Add this to the path of T to get path of P.

y_p := distance along y-axis, from T to C. Subtract this to the path of T to get path of P.

From these new references, start defining the parametric equations for path of P:

x = r * cosθ + x_p

y = r * sinθ - y_p.

Next determine the values of x_p and y_p using the right-hand triangle from above:

From here, we can divide the triangle along line TD to get a similar triangle:

From here, we get:

sinθ = x_p / rθ

x_p = rθ * sinθ

cosθ = y_p / r

y_p = rθ * cosθ.

Plug these back to get:

x = r * cosθ + r * θ * sinθ

x = r (cosθ + θ sinθ)

y = r * sinθ - r * θ * cosθ

y = r (sinθ - θ cos θ).

Thus, the parametric equations for the involute of this circle are:

x = r (cosθ + θ sinθ)

y = r (sinθ - θ cos θ).

When you graph it from 0 ≤ θ ≤ 2πr, it looks something like the blue curve (but here, r=1):


Woots!
--Charissa

Maaaath

Relationships

Based on private entry, 13 Jan, 2008 at 13:08 on bus

Been thinking.

Had lunch with James. Was strange. He talks more, though. But, either Istill make him nervous, or he's still that way--at least, around me.

Yet, I find, ... physically ... I think we fit together.

We still have a weak relationship. He talks about things I don't understand; and I talk about things he can't comment on. When I ask him to explain, he usually can't, or, not very well.

Maybe our roles are too loose. If it were a planned or structured get-together / event, maybe it would flow better.

I think that's the biggest issue: we have nothing to talk about. I ask about him, and he doesn't say much. I ask a little more, and then he talks more; but of course he'll start talking about programming, and I won't understand much of it. I ask him to explain, and he becomes hesitant and stumbles over the explanation (or maybe I'm just difficult to explain to?). When I talk about myself, he won't be interested, or won't comment. I guess, the other thing: I jump from topic to topic, and he likes to stay on one topic at a time.

It's so strange to have that physical connection, yet no other connection. I can read his body very easily; but not his mind or heart. And it may be that his body has very contrary "thoughts" from his mind/heart. Afterall, he has much self discipline.



Math

At work, I got bored. So, I figured out that, for each area code, there are 4,251,528 possible phone numbers (8*9*9 *9*9*9*9). Assuming there are 60 people working on that area code at any given time, and each worker can make 90 calls per hour; it will take roughly 13 hours to dial an area code.

Of course, I was at work at the time, so my numbers were totally wrong. It's very hard to find 9⁷ when you can work on it only 20 seconds at a time, with 70 seconds in between, which are spent doing something completely different.

(Just for comparison, I got: 5,373,459 possible numbers, and 1.15 hours, assuming 60 workers and 80 calls/hour. I was 21% off on the phone numbers, and a whole order of magnitude off on the hours--but I think that was just stupid messy-writing error.)

My next task will be to find out how many possible phone numbers in the US and Canada. I've made a list of all area codes that don't exist. It'll be a matter of finding:

(729 - [number of area codes that don't exist]) * 5,373,459

Woo!


Teaching

I mentioned, last post, that we've started transformations. I think, most of the kids don't get it.

Well, I want to make a bonus question on their next assignment:

1. Determine algebraically whether y = f(x) = sinX is even, odd or neither.

2. Determine algebraically whether y = f(x) = cosX is even, odd or neither.

I'll even do this right now, off the top of my head!

1. f(x) = sinX
Step 1: Find f(-x).

f(-x) = sin(-x)
= sin(0 - x)	Trig ID: sine sum-angle identity.
= sin0 * cosX - cos0 * sinX
= 0 * cosX - 1 * sinX	Simplify.
= - sinX	Which is also....
= - f(x)	Therefore, this function is odd. Hoorays!

2. f(x) = cosX
Step 1: Find f(-x).

f(-x) = cos(-x)
= cos(0 - x)	Trig ID: cosine sum-angle identity.
= cos0 * cosX + sin0 * sinX
= 1 * cosX + 0 * sinX	Simplify.
= cosX	Which is also....
= f(x)	Therefore, this function is even. Hoorays!

If we do this in class, then we've just taught them another trig identity, which is always good.

I really want to show the kids the process of how to solve those "Even, odd or neither" problems. Augh. It's so frustrating, the way the teacher teaches... Sigh.


Okay, sleepy time.
--Charissa

Last day of school

Today was the last day of school before "Winter Break", and I got to lead the class. It was great!

Something I learned: Saying "Yo!" or "Word!" or "I'm-a get ma shizzle ON!" etc is a very easy way to get the attention of grade ten students. Indeed! One student, in particular, was embarrassed for me :). But it works. It's so outrageous, that they immediately cease all other activity and turn their heads.

We went over the kids' exam first (as planned here). Half the class got less than 50%. The average was 53%. We got through the first two pages before recess. I wanted to finish quickly so we'd have time for their "presents" afterward, so I just picked one of the hyperbolas to graph. They actually stayed the first minutes of recess to see it through.

Recess.

After, I handed back their tests. Not happy. I copied down a summary of How to Complete the Square; after, I gave them the flow-chart.

Handed out the "presents": condensed notes on sketching conics; and "Why Conic Sections are Cool!". Talked about "formula sheets", study guides, condensed notes.

I passed around my first "formula sheet" I ever made. It has taped edges to prevent tearing; sprayed with hairspray to prevent graphite smudging (write in graphite so that you can erase and position everything better); everything is labeled; colour and indents help titles to pop out... Only two things about this are dumb: One, I used pencil crayons, which the hairspray dissolved. Two, I spent too much time making it. On the next formula sheets, I smarted up. No colour, no tape; but the indents keep everything orderly.

I told them about speaking "Ukrainian Math Wizard", and (let's call him) Vasil, my prof for Honours Calculus.

Then, I gave them the One Million Beans problem (but with fixed values), saying that if anyone could solve and prove it, I'd bring doughnuts next class (solution below). The top students (grade-wise) couldn't get it, but two girls who were interested but determined they could not figure it out ended up solving it! Well, not necessarily proving it, but close enough. So I owe them doughnuts next class (January 12).

While they were working on it, I told them an Engineers vs. Mathematician joke; they're on the train, one ticket... They laughed--they got it! Just before class ended, I told them the joke about Mathematicians reducing everything to problems they've already solved whereas Engineers can solve "new" problems with originality. If they don't get it now, I'm sure they'll get it later!


Proof of "One Million Beans" Problem

We know that after the beans have been moved back and forth, each jar still contains P number of beans. Now let's look at the number of red and green beans in each jar:

Jar A has

(P - n) green + (m) red beans = P.

Jar B has

(P - m) red + (n) green beans = P.

Set up equality:

P = P	Number of beans in Jar B = number of beans in Jar A.
(P - m) + n = (P - n) + m	Cancel (P) on both sides.
- m + n = - n + m
2n = 2m	Cancel (2) on each side.
n = m

Therefore,

| n - m | = 0

for all Natural m, n, P, Q < P.


Okay, I have to get up early tomorrow, then work an 8-hour shift. Sigh. Shouldn't have committed to it...

But I'll mention quickly:

I've been blessed with amazing Math teachers over the years, which probably explains a lot about me. Hopefully, this will allow me to pass along that experience to others.

One student remarked how great it was to have a teacher who didn't mumble (which is especially funny because my father teaches the level below and some kids had him last year). Another said I was exciting and that she was having fun. A few others generally remarked that I explained well and was interesting.

And these are the three (recent) things that have made me feel so worthwhile, in chronological order:

1. Finding out my army-boss has harassment issues (it's not everyone--and we're not necessarily bad untrained privates!).
2. My army-boss telling me I look good (dress/deportment).
3. Hearing that the kids enjoyed my teaching.

--Charissa

Cars vs. Goats

There was an episode of NUMB3RS (113 - Man Hunt) that included a brief explanation of the Monty Hall Problem.

Monty Hall's "Let's Make A Deal" is gameshow where you choose a door to win either a goat or a car--the objective being to win the car.

Wikipedia has a pretty thorough article on the Monty Hall Problem, but I want to make my own little problem and proof using PMI, because I wuvvy wuv wuvs PMI so much, yes I do!


Cars vs. Goats (AKA: Monty Hall Problem)

Suppose there is a gameshow in which the player chooses a door. Behind any door is either a goat, or a car; and the objective is to pick the door with a car. After a door is chosen (but not opened), the host opens a door, behind which is a goat. Theplayer must then make a choice: stay with the chosen door, or switch to another door.

More specifically, let's deal with the case where there are c doors and only 1 car. That means, there are c - 1 goats, and the player can choose to switch to one of c - 2 remaining doors, with the objective of finding the car.

Scenario One: c = 3
In this case, there are three doors, and suppose the car is behind door three. There are three initial choices the player can make: door one, door two or door three. Then, the player can choose to either switch or stay.

Here are the results:

initial:	1	2	3
switch:	Car!	Car!	Goat
stay:	Goat	Goat	Car!

If we look at the switch row, we see that 2/3 result in cars.
If we look at the stay row, we see that 1/3 results in a car.

Thus, it is better to switch when there are three doors and one car.

Scenario One: c = 4
Let's suppose the car is behind door number four. The only real difference now is that there are two possible switch choices, but that's easy to take into consideration.

Results:

initial:	1	2	3	4
switch1:	Car!	Car!	Car!	Goat
switch2:	Goat	Goat	Goat	Goat
stay:	Goat	Goat	Goat	Car!

Looking at both switch rows, we see that 3/8 result in cars.
Looking at the stay row, we see that 1/4 results in a car.
Since 3/8 > 1/4, switching is still a better choice.

The remaining cases
Let's skip over the details of the next few cases. Let S be our probability of success in finding the car when choosing to switch, and c be the number of doors; still having only one car.

Summary:

c	3	4	5	6	7
S	2/3	3/8	4/15	5/24	6/35

Noticeable patterns:

• S_c = (c - 1) / [c (c - 2)]
  To simplify things let's separate the Numerator and the Denominator,
  
  let N_c = c - 1
  
  let D_c = c (c - 2)
• N_(c+1) = N_c + 1
  
• D_(c+1) = D_c + 2c + 1

Proof of pattern of S:

1. Prove that N_(c+1) = (c + 1) - 1

First, look at the best case of N_c = c - 1, ie: when c = 3

N₍₃₎ = (3) - 1

N₍₃₎ = 2

This is true!

Now prove all following cases are true; ie: prove N_(c+1) = (c+1) - 1

N(c+1) = Nc - 1
N(c+1) = (c - 1) + 1	Substitute.
N(c+1) = c	Simplify. Hoorays!

2. Prove that D_(c+1) = (c+1)(c+1 - 2)

First, look at the best case of D_c = c (c - 2), ie: when c = 3

D₍₃₎ = 3 (3 - 2)

D₍₃₎ = 3 (1)

D₍₃₎ = 3

This is true!

Now prove all following cases are true; ie: D_(c+1) = (c+1) (c+1 - 2)

D(c+1) = Dc + 2c - 1
D(c+1) = [c (c - 2)] + 2c -1	Substitute.
D(c+1) = c2 -2c + 2c - 1	Expand.
D(c+1) = c2 - 1	Simplify.
D(c+1) = (c + 1) (c - 1)	Difference of Squares. Hoorays!

Therefore,

S_c = (c - 1) / [c (c - 2)]

This doesn't really mean anything relevant yet, though. We've just proven that we always know our success rate, and not that switching is better than staying. Let's look at the staying data now.

Let T be the success rate of staying, and we'll compare with S.

c	3	4	5	6	7
T	1/3	1/4	1/5	1/6	1/7
S	2/3	3/8	4/15	5/24	6/35

Pretty dismal, eh? Let's prove that S > T. But first, we must prove that T always follows a pattern...

Patterns noticed in T:

T_c = 1 / c

T_(c+1) = (T_c) * c / (c + 1)

Prove that T_(c+1) = 1 / (c + 1)

Best case: c = 3

T₃ = 1 / 3

This is true!

All following:

T(c+1) = (Tc) * c / (c + 1)
T(c+1) = [1/c] * c / (c+1)	Substitute.
T(c+1) = c / [c (c + 1)]	Multiply through.
T(c+1) = 1 / (c + 1)	Simplify. Hoorays!

Prove that it is always better to switch:

Here's the final piece! We must prove that S_c > T_cProof by Contradiction.
Go!

Sc	<	Tc	Assume.
(c - 1) / [c (c - 2)]	<	1 / c	Substitute.
c (c - 1) / [c (c - 2)]	<	c / c	Multiply by c.
(c - 1) / (c - 2)	<	1	Since c > 2, this is most definitely false! In fact, (c-1)/(c-2) cannot even be equal to 1.  It is always greater than 1.   Since all our steps after the initial assumption of Sc <>c, that assumption must have been incorrect. Therefore, Sc > Tc  for all integers c > 2. Hoorays!

Thus, it is always better to switch.


I'll write something more journal-like later.
--Charissa

Math math math math militia

So, a while back, I applied to be a teaching assistant at a Maths school. This Maths school runs on Saturdays, and was originally meant to be an enrichment program: for kids to get ahead, or to just--yeah, "enrich" their understanding of Mathematics.

I want to write a paper on how awesome Math is.

It's just, that's been done so many times already; and, from what I've read, unsatisfactorily. I'm reading Mathematical Sorcery in my spare time right now. It's good, I guess; but there's this one section that tries to induce the reader to look at an equation as one might look at a piece of art, and the author describes how he gets tingles down his spine. I'm not a writer, but I think that was unnecessarily personal; if something is awesome, the reader can figure it out for himself (gender-inclusive "himself").



Tower of Hanoi

Back in Honours Calculus last year, we studied the Principle of Mathematical Induction and then did an example problem using the Tower of Hanoi.

Here is the basic setup: There are three poles onto which you can place some discs. The discs are of different sizes, and start out at one tower, arranged by width: the widest disc at the bottom and the smallest disc at the top.

The objective is to move all the discs from the first tower to form another tower. Usually, you start off with the discs at the far left and must move them to the far right.

A further objective would be to make as few moves as possible.

Rules: Only one disc (the top-most one) may be moved at a time. No larger disc may be placed atop a smaller disc.

Here is what we found (probably):

discs:	0	1	2	3
moves:	0	1	3	7

Let m_d be the minimum number of moves, and let d be the number of discs to be moved. We noticed that m_d = 2m_{(d - 1)} + 1
We hypothesized that m_d = 2^d - 1 for all non-negative integers m.


Proof:

1. We shall test the first/best case scenario, where p = 0.
Plug in values:

m = 2^0 - 1
m = 1 - 1
m = 0

If there are no discs to move, there are no moves to make.
This is true!

2. We shall now test if the following scenarios are true; ie, whether or not m_{(d + 1)} = 2^(d + 1) - 1 Go!

m(d + 1) = 2 md + 1
m(d + 1) = 2 [ 2^d - 1 ] + 1	Substitute md = [ 2^d - 1 ]
m(d + 1) = 2^(d + 1) - 2 + 1	Multiply through. There's a common base of 2, so add one to the exponent to simplify.
m(d + 1) = 2^(d + 1) - 1	Hoorays!

So, by the Principle of Math Induction, we have proved that m_d = 2^d - 1 for all non-negative integers of m. We could even add to the table:

discs:	0	1	2	3	4	5	. . .	n
moves:	0	1	3	7	15	31	. . .	2^n - 1

Sometimes, a story goes with the Tower of Hanoi. There are variations, but the idea is that, at the beginning of the world, there were these monks who had this game set up in their temple. When they had finished the game, the world would end.

Supposing it takes one second to move each plate (not very realistic), and there are 64 discs, when will the world end?

d = 64

m = 2^64 - 1
m = 18 446 744 073 709 551 615

So, the world would end in roughly 585.442 billion years. The universe is roughly 13.7 billion years old, +/- 1%. Hehehe.




Being a Teaching Assistant

Being a teaching assistant is conflicting right now. On the one side, it's like job-shadowing, except I get paid--quite well. On the other side, it's incredibly tedious and painful to see what I would consider bad teaching, and not be able to do much about it.

I got such fulfillment today when one student called me over to explain something.


The way I teach

When a student puts up his hand, I come over. If there is a seat available, I'll sit down; if not, I'll kneel on the ground (this usually puts me just below his eye level). Then I ask, "How can I help you?"

After the student has explained his difficulty, I will confirm that I understand it. This helps me to be clear what I'm trying to do; and also gives me a little thinking time.

I used to be afraid of stuttering, so I'd take silence over stammering or filler. But when I teach, I have to remind myself it's okay to show that I don't know everything or that I'm not infallible; so if I'm having difficulty, I might make a short, "Um," to let him know I'm thinking.

I try to use at least two different methods to explain or solve a problem. I have a habit of asking, "Does that help?" afterward. I think maybe I do this too quickly sometimes; but I found it to be very helpful.

If someone else wants my attention at the same time, usually, I will tell him to wait; because if a student calls for my help, I want to show I am available.

After, if I feel that the student has time or interest, I'll go a little further and tell how it relates to other fields, where else it can be seen, or other cool things that can be done with it.



I think that's all for tonight. I'll just mention that on Tuesday, I'm going to be enrolled in the Militia. They asked me to bring a small recent photo of myself (photo booth at a mall works well) for my temporary ID card; so I went to the mall, and, wow, not one of those four shots turned out well. At least, I'm identify-able, don't look terrible, and I'm wearing clothes I like.

I'm very excited, though.
--Charissa