BOB likes to eat icecream [cones]

I feel pretty confident for the test tomorrow, I might not get 100% but I believe I'll do better than what I have been doing so far. I liked this unit because it was pretty straight forward and I had no real trouble understanding any concepts.

I think that for this unit I just need to make sure I know the formulas really REALLY well. Let's review shall we ^__^

A PARABOLA:
The equations:

If the x is squared, the parabola is vertical (either opens up or down) and if the y is squared the parabola opens horizontally (opens left or right). p is the distance from the vertex to the directrix and the focal point. If 4p is a big number, the stretch of the parabola is also big. If the number is smaller than one the parabola will be skinny.

A CIRCLE:
The equation:


r is the radius. In general form, the coefficients of x and y are squared and are the same. That's how you know it's a circle!

AN ELLIPSE:
The equations:

When a (the bigger number) is under x, the ellipse is horizontal. If it's under y then its vertical.
a is the length of the semi major axis. b is the length of the minor axis. Use c^2 = a^2 - b^2 if you have to find the distance between the origin and the focal points. In general form, the coefficients for x and y are different.

A(N) HYPERBOLA:
The equations:


When x is positive the hyperbola is horizontal (opens right and left). When y is positive it is vertical (up and down). a is the length of the semi transverse axis, while b is the length of the semi conjugate axis.

use c^2 = a^2 + b^2 to find c, the distance from the origin to the focal points.

hyperbola graphs have asymptotes! use the slope formula to figure out their equations.

To be alright for the test, all we need to know are the formulas and how to graph them. I'm sure everyone will do fine! ;D

Make sure to eat healthy, drink plenty of water and get some rest so you can be in top shape tomorrow morning!

~Bye bye! :D

Today's Slides: May 28

Here they are ...

Pre-Cal 40S May 28, 2009

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BOB for Conics

hi everybody, one more unit is done and i was almost forget about the BOB for this unit. T_T .
Anyhow, i just do this BOB briefly.

I think this unit is not really, you just need to know the formula and understand the graph so you could know which formula is parabola, ellipse, or hyperbola.

About the formula, i think some of us already posted it, so i am not going to repeat it. :D

Anyway, Good luck all guys, test on Friday and don't forget your delicious link [which i also almost forgot :-"] ...

BOB BOB BOB BOB :)

Hey everyone!
This unit went by really fast, and since it's the end of the school year, I'm becoming more lazy, so that means that I really haven't been paying much attention to a lot of things happening in class. So I really need to study a lot and make sure I do well on this test! The only thing I really need to review is the day when we learned about graphing parabolic equations since I wasn't in class for that lesson. I have a feeling it's not a hard thing to pick up, so hopefully I learn everything I need to know before our test, which is on friday. Other than that, I think I'll be okay. I'll just have to read through the slides to refresh my memory, and hope that I do well. I guess that's all, good luck! :)

BOBS on Conics?

Well this unit was relatively easy for me. The only things I think I will have problems with is reading the question properally and putting it in the right way.


So the first thing we learnt about was parabolas. We learnt another formula for a parabola which is :


P is the distance from your foci to your vertex or your distance from your directrix to your vertex.

If you want the horzontal graph, you just switch the (x-h) and (y-k).



The next thing we learned was circles and the formula is the same as the one we learned before so I'm not going to go into detail with that.



Then comes the elipses with the formula:

Where "a" is the length of the semi major axis and "b" is the semi minor axis.

If you want the vertical graph you just exchange the "a" and "b"

Then there's hyperbolas (parabolas on steriods, according to Mr. K)

Where "a" is your semi transverse axis and "b" is your semi congurant axis. If you want your vertical graph you just make "x" negative instead of "y".

So that's basically the simple summary of what we learn, time for my nap. Night.

Reflection on Conics

Well that was a quickfast unit. I found it so much easier than any of the previous units. :) But lately I've been lazy sooo... I need to get back with the program and practice, practice, practice! >_< Oh and I still need to look over the formulas a few times to make sure I don't mix them up. Remember that the transverse axis connects the two vertices of the hyperbola and can be either horizontal or vertical.

Good luck to everyone!

Is it a parabola? is it an ellipse? no its.. BOB

Well this unit was pretty short and sweet and fairly simple. I'm going to get right into a review because I've got to work on DEV.

##############################################

A parabola is a locus of points a fixed distance from a fixed point (focus) and a line (directrix).

Vertical Parabola: (x-h)^2=4p(y-k)

Horizontal Parabola: (y-h)^2=4p(x-k)

The coordinates (h,k) are the vertex of the parabola. "p" is the distance from the vertex to the focus or the vertex to the directrix.  These distances are both equal. When 4p is greater than one, the parabola is wider and when it is less than one, the parabola is narrower.

##############################################

A circle is the locus of points equidistant from a fixed point, the center and the distance to any one point is called the radius.

Circle: (x-h)^2 + (y-k)^2= r^2

The coordinates (h,k) are the vertex of the circle. "r" is the radius of the circle.

##############################################

An ellipse is the locus of points which the sum of the distances to the foci is known.

Horizontal Ellipse: [(x-h)^2/a^2] + [(y-k)^2/b^2] = 1

Vertical Ellipse: [(y-h)^2/a^2] + [(x-k)^2/b^2] = 1

In an ellipse addition is used because the locus of points is the sum of the distances. The center of the ellipse is the coordinates (h,k) and is halfway between the two foci. The semi major axis is equal to "a" and the whole major axis is equal to "2a". The semi minor axis is equal to "b" and the whole minor axis is equal to "2b". The distance of either foci to the center is distance "c" and can be determined by c^2=a^2-b^2.

##############################################

A hyperbola is the locus of points which the difference of the distances to the foci is known

Horizontal Hyperbola: [(x-h)^2/a^2] - [(y-k)^2/b^2] = 1

Vertical Hyperbola: [(y-h)^2/a^2] - [(x-k)^2/b^2] = 1

The transverse axis is the distance between the vertices of the hyperbola. This is distance "2a". The distance from either vertex to the center is "a" and is the semi transverse axis. The conjugate axis is perpendicular to the transverse axis and it's distance is "2b". Half the distance, "b", is called the semi conjugate axis. The distance of the foci can be calculated c^2= a^2+b^2.

##############################################

Well that is all so I'm to work on DEV! I hope everyone did well on the test!

 

What are the chances this is a scribe post?

Hey everyone!! I'm back for round 3 of this exciting scribe postage!!!

We started off class today with Mr. K. teaching us how to do his snappy thing he always does. It starts with a snap, then another on your other hand and finally a clap/pop.

Here's how....

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Anyways, on to our actual math class!

Our first slide of the day depicts a scene of 3 coins being flipped. We need to determine the sample space which is not the amount of outcomes we can have but a list of those outcomes. The number of possibilities is the SIZE of the sample space. 

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We learned about the words "and" and "or". If, for example, you say you want this to happen AND this to happen AND this to happen, you multiply the probabilities of each of the events. If, for example, you say you want this or this or this to happen, then you add together the possibilities of those events.

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We then learned how to plug our outcomes into a chart. We drew a chart for the outcomes of 2 dice ( or die?) being rolled. There were 2 ways to show this on the chart. We either wrote the numerical values of both dice (die?)...

or we put the added value of the dice (die?)...

It just depends what we are looking for and what would be easier. For example, if we were trying to find the ways that we rolled doubles we would perhaps use the first chart but if we were looking for how many ways could you roll 10 then you would use the second chart. 

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The bus and train question on the next slide, can be a bit confusing at first but once you read it it becomes quite easy to understand. In a nutshell, you have all the times the bus can arrive at the station and you have all the times the train arrives at. A wise math teach once said, something along the lines of, math is not the crunching of numbers but the study of patterns, when you find a pattern in one question, you can apply to a ton of others. This is but one of those many times where we can find a pattern. We make a chart, quite similar to that of the two dice being rolled (OMG is that the pattern??) where along one side we have the times the bus arrives at and the top we have the times the train arrives. From here we can clearly see all of the ordered pairs of times that the two vehicles arrive allowing to solve the following questions very easily. I will not show the answers on my scribe post here tonight but just remember, think patterns!

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We then went over independent and dependent events. Independent events are those that have no effect on each other like flipping two coins, or after drawing a card from a deck, replacing that card before you draw the next card. Dependent events are those which do have an effect on following events such as not replacing a card into a deck after drawing it because in this situation the probability for the second event has changed due to a different amount of cards in the deck.

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That's about all, I just have a few small comments before I finish up:

·Good luck on the test tomorrow!! Study hard, do your BOB's and get that link!

·PATTERNS!!!!!!!!!!!!

·Mr. K, there will be several, I believe 5 or 6 of us, not here tomorrow for the test just as a reminder!

·Congratulations to Ann and Rebecca for finishing their DEV already!

·And lastly the next scribe will be......... John!

BOB

Hi everybody its Alex, hope you're all doing WonderfullyExcelent. I have to admit that I don't feel very confident going in to this test. I found that I missed a lot of days and that we didn't do as much practice problems in class. Usually these practice problems help me a lot, but this unit I didn't feel there were enough of them. I understand that we are pressed for time so I'm not complaining.

I know we have to know about ellipses and hyperbolas and that we need to know how to graph them, but I feel that there are going to be a lot of problems on the test that will require us to use our indepth knoweledge of these ellipses and hyperbolas, that I just won't know. I'm expecting a test that will be very tricky for me, with lots of curve balls coming from no where.

I hope that I and my fellow class mates can get enough studying in to do WonderfullyExcelent on this test, and I hope everyone has more knowledge than me on this unit going into this test.

Best of Luck! Alex.

BOB for Conics

Hooray we are done our DEV! :D
Now its time to do the OTHER math homework that I have been procrastinating on...

Well this is my Bob for the conics unit. I found this unit pretty easy, except for the moments where I was just confused... or couldn't remember the equation for a circle... I think the test will be pretty easy as long as I don't do anything stupid like get an ellipse and a hyperbola mixed up.

Hmmm well what is important for this unit?

1. Equations for...
parabola: (x-h)^2 = 4p(y-k) (vertical) and (y-k)^2= 4p (x-h) (horizontal)

Circle: (x-h)^2 + (y-k)^2= r^2 (standard form)

Ellipse: (x-h)^2 / a^2 + (y-k)^2 / b^2 = 1 (horizontal) and (x-h)^2 / b^2 + (y-k)^2 / a^2 = 1 (vertical)

Hyperbola: (x-h)^2 / a^2 - (y-k)^2 / b^2 = 1 (horizontal) and (y-k)^2 / a^2 - (x-h)^2 / b^2 = 1 (vertical)

So for the test i think we just need to make sure to know all the parts of shapes and how everything fits together. Remember how to find the vertices, foci, center, radius and length of major and minor axis! I think that's all I have to say for this unit. Just remember to get lots of sleep, (something I need), DON'T eat a ton of candy and study the formulas! :D

Today's Slides: May 27

Here they are ...

Pre-Cal 40S May 27, 2009

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Scribe and Bob Post

Delicious For Conics – Click here! Quite a number of people have asked to see my Counting test, so I just scanned it and decided to post it here. Page 1 Page 2 Page 3 Page 4 I found this unit extremely easy. The stuff looked like nonsense and garbage to me from the start, but when I actually looked at the stuff, I realized that I was wrong. The folding of the papers did not help me however. Nevertheless, it was Mr. K’s method to help us understand every one of the conic sections. You must basically understand the anatomy of each conic section. That is the key to solving the questions. Other than that, the unit test should be easy. Well, I hope it is. WATCH OUT WHEN DOING/DRAWING THE VERTICES AND FOCII. IT’S OBVIOUS THEY’RE NOT THE SAME, BUT EVEN PEOPLE THAT KNOW THAT CAN MAKE MISTAKES WHEN DRAWING THEM. SOME PEOPLE USE THE FOCII AS THE VERTEX BY ACCIDENT OR ADDRESS THEM INCORRECTLY WHEN ANSWERING QUESTIONS !!!!! Hyperbola The transverse axis extends from the vertices of the hyperbola. This does not have a fixed orientation. It can be either horizontal or vertical. It can be found by using “2a.” Two semi-transverse axes complete a full transverse axis. The conjugate axis is the perpendicular bisector of the transverse axis, meaning it cuts the transverse axis in half at a 90 degree angle. This does not have a fixed orientation. It can be either horizontal or vertical. It can be found by using “2b.”Two semi-conjugate axes complete a full conjugate axis. For the conjugate and transverse axes, watch out for what the question asks for. Some may ask for the “semi axis” while some may ask for just the “axis.” Video on Hyperbola Worksheet on Hyperbola My question One focus is at (-5,0) and the nearest vertex is at (-3,0). What is the orientation of the hyperbola? Horizontal Given a hyperbola, let’s say you have the points of the conjugate axis. You also have a point (99,1000) which is located on the hyperbola. How would you find the equation? Use the equation of the hyperbola to plug in your x and y values with (99,1000) while you obtain A or B with the conjugate axis. This leaves you with one variable, which is A if you chose B or B if you chose A. Isolate that variable and bam. EllipseVideo on Ellipse Worksheet on Ellipse The major axis is the longest of the axes. Two semi-major axes complete the major axis. The minor axis is the shortest of the axes. Two semi-minor axes complete the minor axis. Circle Video on Circle Worksheet on Circles The circle is like the ellipse’s brother. They’re pretty much the same except the circle has radii equidistant from the circumference to the center of the circle. Parabola Video on Parabola Worksheet on Parabola The parabola is very easy to understand. There is one focus which is the same distance as the directrix from the vertex of the parabola. Question If the focus is to the left of the vertex, what direction does the parabola open? It opens to the left. Probability The next scribe is DION!!

Today's Slides: May 26

Here they are ...

Pre-Cal 40S May 26, 2009

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CYCLE 3!

Well its Anthony again. I will be starting off the 3rd cycle. Today in class we learned about the anatomy of a hyperbola. We began by looking at our paper folding homework. We were suppose to go home and fold 30 points on the circle to the outside dot (Points were allowed to be anywhere as long as they were spread around the circle's circumference.)

The result of our folding is a hyperbola. After we found this out we went on a quest to find any patterns within, just like what we did with the ellipse.



Mr.K told us to pick a point anywhere on the hyperbola and find the distance from F1 to P and F2 to P. We then take the difference of these two numbers. We did this several times and found out that the difference was constant no matter where you put the point. We also found out that this constant value is exactly the same as the distance between the two vertices's of the hyperbola.



The distance between the two vertices's is known as the Transverse axis. The conjugate axis is the perpendicular bisector of the transverse axis. This intersection point between the transverse axis and the conjugate axis is known as the center.

These are notes you will need to understand so you can draw these graphs with ease. The technical terms are not all that important.

Below is the standard form for the Hyperbola equations along with the similarities and differences between them.

We are now back to the anatomy of the hyperbola.

Just like the ellipse, this also has a Pythagorean triangle within it. The distance of the hypotenuse is the same distance as the distance between the center and one of the focus points. With this new knowledge we can create hyperbolas with limited amounts of information.

HOMEWORK!

(i) We did the standard form equation in class as it shows on this slide. If you need help understanding, what we did was take that 225 on the right side of the equation and divide by that number throughout the whole equation. This will help us get the preferred number 1. After we just simplify the equation and get...

(x^2 / 9) - (y^2 / 25) = 1

(ii) The transverse axis equals 2a. We can find a from the number under x^2. If you noticed this term is positive therefore making this equation a horizontal one. a^2 is under the x^2 in this type of graph. So if we take the square root of 9 we can find a. So a=3. Therefore the transverse axis is 6.

Now to find the conjugate axis's length. The conjugate axis is 2b. Under the y^2 term we can find b^2. If we take the square root of 25 we will get what b equals, multiply by 2 and the conjugate axis's length is 10.

In order to find the coordinates of the foci and the vertices's we must know where the center is. Since the x^2 term and the y^2 term have nothing within it to shift the graph; the center is (0,0). So the vertices's is the distance of away from the center. Therefore the vertices's must be (3,0) and (-3,0). To find the foci we must find c. Which requires the Pythagorean theorem.
a^2+b^2 = c^2 will be used in this case. (Remember it is not always a^2 + b^2 = c^2, it is actually (leg1)^2 + (leg2)^2 = (hypotenuse)^2) With this step we get c to be a value of root(34). Therefore the foci are located at (root(34),0) and (-root(34),0).

Finding the asymptotes. I found the asymptotes by drawing the rectangle as done when constructing a hyperbola. When I found the corner points I was able to figure it out. The points I got were (3,5) , (3,-5) , (-3,-5) , (-3,5). In addition all of the asymptotes always intersect with the origin. So I used (3,5) and (0,0) to find the positive asymptote and (-3,5) and (0,0) to find the negative asymptote. So with these points I just plugged it into the two point formula which is...
(y2-y1)/(x2-x1) = (y1-y0)/(x1-x0)

This is a formula we learned way back in grade 10 but it's still useful. I ended up getting two asymptotes that were y=5/3x and y = -5/3x. There are probably different ways but I chose this way.

The graph should look something like this.

Sorry for my poor drawing xD. Well that's about it. If any information on here is incorrect I will try to change it as soon as possible.

The next scribe will be Aldrin S!!! WOOT! show us your mad skills xD.

Today's Slides: May 25

Here they are ...

Pre-Cal 40S May 25, 2009

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Today's Slides: May 22

Here they are ...

Pre-Cal 40S May 21, 2009

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May21st/09

Today we started on the ellipses part of the unit.
if you take the cone again and cut it across at an angle it creates an ellipse.
Reference Jessi.
The word ellipses comes from Greek: (ἔλλειψις ellipse) Means a "falling short". ...
By definition: The locus of points for which the sum of the distances from each point to two fixed points is equal.

Then Mr.K showed us the tip of drawing ellipses: use 2 pin and a elastic. (it's in the slides #4)

In this diagram there is few point to remember about ellipses:

• O is the centre
• the line A1 to A2 is the length of the major axis, the length is 2a.(Major axis is the longer axis of the ellipses)
  
• PF1& Pf2 are the focal radii of the ellipses
• OA1=OA2 are the length of the semimajor axis with length a. (Semimajor means half of the Major axis)
• B1B2 is the length of the Minor axis. it's length is 2b.(Minor axis is the shorter axis of the ellipse)
• OB1=OB2 is the length of the Semiminor axis with length b. (Semiminor axis means half of the minor axis).
• F1 and F2 are called the foci, they are c units from then centre. ( foci because it's more than one)
• A1 and A2 are the vertices of the ellipses.
• B1 and B2 are the endpoints of the minor axis.

Difference between vertical and horizontal ellipses:
Vertical-major axis is going up and down
-minor axis is going left to right
horizontal-major axis is going left to right
-minor axis is going up and down

Remember it's not always a^2+b^2=c^2 in this case it's a^2+c^2=b^2
In an ellipses a endpoint of the minor axis connect with a focus point Will form a right triangle.


the 2 highlighted formula are ellipses for both horizontal and vertical.
In the diagram shown how the Pythagorean works for the ellipses.


In this problem we used the formula fist we find the vertex: which is (3,-1) same as circle.
then 25 is a^2 and 9 is b^2 so the distance of semimajor is 5 and semiminor is 3.
from there we fin c using a^2+c^2=b^2. therefore c is 4 the foci are 4 units left and right of the centre point.

And basically that all we have done today, and be careful don't forget to do ur home work!!!!

Next scribe is misuiz