## Monday, March 10, 2008

### SCRIBE MANNN SCRIBE. for today

Okaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaay. So I guess I am the blogger today! Or scribe. Whatever. So ! Today! .. Is 4 days away from Pi day! Alright, cool.

Alright we started off class ready to learn. Although we couldn't because our fingers were frozen. Except Rguy204, he was Rence's hero, because he had a shirt on, and he was perfectly fine. Practically sleeping. Gosh he should be living in the North Pole. Kidding, anyways.

Instead of starting off continuing the Saskatchewan Sunrise Problem, Mr. K decided to help the class get more familiar with transformation questions. Which I thought was really helpful and should do it more often in every unit. *HINT HINT* COUGH COUGH. I HOPE YOU READ THIS MAN. IT'S IN CAPS LOCK. Thanks, if you do. Cheers.

IN SLIDE 2 .. Yes that is my awesome super flashy graph that everyone loves. With some sprinkles on the side. This is the graph of y = |2-x| .. Absolute values cannot be negative. If you had a table of values, the Y coordinates go down to 0 then go up. 2-x has a negative slope, when it reaches 0, values of Y go positive, thus creating the graph you see on the slide.

SLIDE 3 .. I did this one also. Unfortunately, it's not so awesome. Sorry fans.
y = 1 / x^2 + 1
This graph has a history behind it. Although I'm too lazy to explain it, but it's also known as the which. Click Here, it's a link of a wikipedia about it.
Anyways, now for the explaination.
SORRY.. but the red line of the parabola is SUPPOSED to be dotted.
Then you must find where Y = 1 and -1 .. Because the parabola is biggering, the recriprical must be smallering. The curve towards the x axis is more smooth. Remember it gets infinitely big and never touches.

SLIDE 4 .. Good job BenOfSchool! (the red) but he made a minor mistake that a very mean mean mean mean mean green mean person had to find. (the blue)

y = |f(x)|

because the y value cannot be negative, because of absolute values, it is flipped onto the positive side, thus creating that bird looking graph.

It's not as beautiful as I would have graphed it though. But cheers, because ya'll dig.

SLIDE 5 .. WOO HOO! Go me! Look how nice that graph is! Check out those curves. OhhH dayum! Yes, anyways this graph is a tricky one.
y = 1 / f(x) ..
As you can see the right side of the graph has a stopping point (5, 2), and the bottom is continuously going to the left. Good to know! (Sorry for my poor math language here Mr. K)

To being graphing the recriprical of this function you find where y = 1 and -1

Those are the invariant points

Then you find the roots ( asymptotes )

In this case (-2,0)

From point (-3, -1) to (-2,0) f(x) is smallering negatively, thus the recriprical is biggering negatively.
The recriprical of 2 is 1/2 which 1/f(x) lines negatively to the left on -1/2 ..
As for the other curve, it curves infinitely big, biggering negatively.

from (-2,0) to (-1,1) f(x) is smallering, thus 1/f(x) is biggering, the line curves up infinitely.
As for the other, it lines positively to the right and stops at point (5,1/2)

OH GOD. I HAVE NO CLUE WHY MR. K HAS 21 SLIDES TODAY.
..
what did I ever do to you man ..

SLIDE 6 & 7 are familiar, so I'll explain them both. Everyone in class knows how a sin graph looks like ( sin(x) to be exact ) .. and so there are 4 asymptotes ( or roots ) in between of those asymptotes there is a wave to relate to.

In slide 7 you see the first wave on the left, it starts from 0, reaches it's maximum then goes back to zero. It may look like a parabola, but it starts from the wave's maximum then curves up and towards the asymptote, but never touches.

Mr k, you're a great artist by the way. (extra marks!)
-slide 8 is a must see

-------

So we continue our problem solving on Saskatchewan Sunrise Problem

we all agree that 3.25 is 3:15 Am, because 25 is a quarter of 100, which is 15 to 60. Yes? good.

ON SLIDE 10 you see all the calculations and setups done.
Now we move on to slide 11.
Why did we start the graph with -10?
Because Dec 21 is 10 days before January.

Dec21 to Jan21 is 172 days.

Why is the minumum 3.25 and maximum 9.25?
because 3:15am is the earliest sunrise and 9:15am is the latest sunrise

Now how can we find A, B, C, and D?

To find D ( vertical shift )
We simply did this ..
(3.25 + 9.25 / 2) which gave us ..
6.25

To find A ( amplitude)
we simply did this ..
(9.25 - 6.25) which gave us ..
3

To find B ( period )
We simply did this ..
We did some thinking. 172 is half the period.
but because we started off at -10, half the period is 182.
KEEP THAT IN MIND.
182 x 2 = 384

So parameter b is 2Pi/364

To find the others, we found a quarter of 364.
Which is 91 days.

To imply that on the graph, we subracted 10, because we started on -10.

-10, 81, 172, 263, 354

IN SLIDE 12
we wrote the equations of the graph in sin and cosine.

SLIDE 13
predict what time it'll be on April 6 using one of the equations.

We simply added up all the days from jan to april 6,
which is 96 days.
Then we replaced all (d) with 96.
We came to a time of 5:28:55 am

SLIDE 14
Simply look at parameter D.

SLIDE 15
as you can see, we inserted 7 on the y axis.
(RUSHING, GETTING SLEEPY)

we
then let theta = 2Pi/364(d + 10)

-0.25 = sinTheta

Negative! So we know the answer will be in Quadrants 3 and 4

ARC sin(-0.25) = Theta

-0.2529 = Theta

Now we input it to find out d

-0.2529 = 2Pi/364(d+10)

we multiply both sides with 364/2Pi

which became

-14.6384 = d-81

66.3616 = d

J 31
F 28
= 59

66 - 69 = 7
March 7

Which gave us ..
3.3945

Then we did the same process to find which month and day.

(SLIDE 17)

2
77.6384 = d
278 = d

278-273=5
October 5

THAT'S ALL FOR NOW FOLKS! APOLOGIES FOR THE LATE POST. PRE-TEST TOMORROW. TEST ON WEDNESDAY. PI DAY FRIDAY.

THE NEXT BLOGGER IS. DIMPLE.