His story is spellbinding. There's a really important message for you in the last two minutes ... but you won't really understand it unless you watch the whole thing.
Click on the picture. (18 min. 48 sec.)
December 29, 2006
Ben Saunders
Arctic explorer Ben Saunders recounts his harrowing solo journey to the North Pole, complete with gorgeous images, amusing anecdotes and previously unseen video footage from the Pole. At 26, Saunders became the youngest person ever to ski solo to the North Pole, updating his blog daily during the trek. He's now planning the next journey, SOUTH, an unprecedented, roundtrip expedition across Antarctica and back.
His story is spellbinding. There's a really important message for you in the last two minutes ... but you won't really understand it unless you watch the whole thing.
Click on the picture. (18 min. 48 sec.)
His story is spellbinding. There's a really important message for you in the last two minutes ... but you won't really understand it unless you watch the whole thing.
Click on the picture. (18 min. 48 sec.)
December 28, 2006
Bono - TED Prize Winner
For the past 20 years, members of the TED community have gathered together to share ideas and passions that are big enough to change the world. Each year they will honor a maximum of three individuals who have shown that they can, in some way, positively impact life on this planet.
Rather than simply receiving financial support, winners of the TED Prize are granted something extraordinary: something which children dream about, but which adults assume is merely the stuff of fairy-tales.
They are granted three WISHES to change the world.
They may wish for anything. And TED will seek to make their wishes come true.
Bono won the TED prize in 2005.
Rock star and activist Bono accepts the 2005 TEDPrize with a riveting talk about our moral obligation (and economic incentive) to help lift Africa out of poverty. He unveils his TEDPrize wishes by challenging the TED community to help build a social movement of more than one million American activists for Africa; to tell people one billion times about the ONE campaign; and to connect every hospital, health clinic, and school in one African country, Ethiopia, to the Internet.
Click on the picture. (28 min. 37 sec.)
For more: Read the update on Bono's wishes.
Rather than simply receiving financial support, winners of the TED Prize are granted something extraordinary: something which children dream about, but which adults assume is merely the stuff of fairy-tales.
They are granted three WISHES to change the world.
They may wish for anything. And TED will seek to make their wishes come true.
Bono won the TED prize in 2005.
Rock star and activist Bono accepts the 2005 TEDPrize with a riveting talk about our moral obligation (and economic incentive) to help lift Africa out of poverty. He unveils his TEDPrize wishes by challenging the TED community to help build a social movement of more than one million American activists for Africa; to tell people one billion times about the ONE campaign; and to connect every hospital, health clinic, and school in one African country, Ethiopia, to the Internet.
Click on the picture. (28 min. 37 sec.)
For more: Read the update on Bono's wishes.
December 27, 2006
Richard St. John
Why do people succeed? Richard St. John compacts seven years of research into an unmissable 3-minute slideshow on the real secrets of success (Hint: Passion, persistence, and pushy mothers help).
Click on the picture. (3 min. 40 sec.)
Click on the picture. (3 min. 40 sec.)
December 26, 2006
Gregory Colbert
Photographer Gregory Colbert shares the remarkable images and film footage from his exhibit, "Ashes and Snow," and announces his founding of the Animal Copyright Foundation, which will require that royalties be paid when images of nature are used for commercial purposes.
The pictures are just stunning.
Click on the picture. (18 min. 42 sec.)
The pictures are just stunning.
Click on the picture. (18 min. 42 sec.)
December 25, 2006
Hans Rowling
Hans Rowling is a professor of of international health at Sweden's world-renowned Karolinska Institute.
Watch how he displays and talks about statistics. He'll make you laugh and he'll make you think.
Click on the picture. (20 min. 35 sec.)
December 22, 2006
Pi and e ... their first date
You know enough math to get all the jokes in this now ... ;-)
Have an excellent holiday!
Cheers,
Mr. K.
Have an excellent holiday!
Cheers,
Mr. K.
Flickr Assignment Rubric v1.0 - We're Out of Beta!
Here is the rubric we've settled on together.
Thanks to everyone who helped put this together. I found this to be a great experience for me as a teacher I hope it was also valuable to you as a student.
Anyway, here it is, version 1.0 fresh out of beta. ;-)
Flickr Assignment Rubric
It is paramount that the picture be in tune with the purpose of the assignment. It should show, first of all, the student's understanding of how the photo is related to mathematics. The hot spots are important too, because that's essentially your way of teaching other people. Creativity is a factor, because keeping one's interest in the photo contributes to the learning process. Finally, the picture quality should be kept in mind too. If we can't see the picture, it's going to be hard achieving all the other requirements.
Tags
The picture must be tagged properly with the course tag and assignment tag. If tags are misspelled or no tags are present the photo cannot be graded and will receive a grade of ZERO. Not tagging your photo properly and accurately is analogous to not handing in your work or not putting your name on it.
| Classification | Mathematical Content (50%) | Hot Spots (35%) | Photograph (15%) |
| Level 4 | Packed with mathematical concepts/facts. (Minimum 7 concepts/facts.) | All hot spots accessible; i.e. "smaller" hot spots are "on top" of larger ones, they do not obscure each other. All hot spots are actually labels and relate to parts of the photo (not on blank space with filled in notes). One or more hot spots include a link to a relevant supporting resource on the internet. Minimum 7 hot spots. | In focus or appropriately focused for effect. The subject of the picture occurs "naturally," it is not a contrived shot. Really makes the viewer "see" math in a place they hadn't realized it existed. (Example: trigonometry) |
| Level 3 | Significant number of concepts/facts included. (Minimum 5 concepts/facts.) | All hot spots accessible. Most hot spots are actually labels and relate to parts of the photo. Not more than one hot spot on blank space. One or more hot spots may include a link to a relevant supporting resource on the internet. Minimum 5 hot spots. | In focus or appropriately focused for effect. The subject of the photo has been "set up" or contrived yet still illustrates math found in "the real world." (Example: derivative) |
| Level 2 | Some effort to include content evident. (Minimum 3 concepts/facts.) | Most hot spots accessible. Most hot spots are actually labels and relate to parts of the photo. More than one hot spot is on "blank" space. May or may not include links to relevant supporting resource on the internet. Minimum 3 hot spots. | In focus or appropriately focused for effect. Although it is a "real world" picture, objects have been used to "draw" the math. An obviously contrived shot. (Example: trigonometry) |
| Level 1 | Very scarce content related to assignment. | Less than three hot spots are visible or have information related to the theme of the assignment. | It is evident that little effort went into finding and shooting a picture that reflects the theme of the assignment. |
| Level 0 | Content unrelated to theme of assignment. | No hot spots or mostly unrelated to the theme of the assignment. | Out of focus and/or otherwise difficult to look at. |
Creativity (up to 5% bonus)
The maximum possible mark for this assignment is 105%. You can earn up to 5% bonus marks for being creative in the way you approach this assignment. This is not a rigidly defined category and is open to interpretation. You can earn this bonus if your work can be described in one or more of these ways:
- unique and creative way of looking at the world, not something you'd usually think of;
- original and expressive;
- imaginative;
- fresh and unusual;
- a truly original approach.
December 21, 2006
Scribe post
the 3rd scribe post of ruschev
note:
- "1" is the highest possible value you can get ; a 100% chance ; more than 1 will result in an incorrect answer ; when you add up all of your probability the numbers you should only get is equal to or less than 1 or zero.
PROBABILITY
note:
- Mutually Exclusive : unable to be both true at the same time ; one event taking place prevents the other event from taking place ex) flipping a coin is a mutually exclusive since it can not be heads at the same time.
- drawing a tree diagram will help allot when solving probability questions
- drawing a Venn diagram will also help
- adding probability. P(a)+(b) when the question is an "or" question. ex) 2 boys are running what is the probability of one of these boys win if one boy has 1/4 of winning and one has 2/3 of winning.
-multiplying probability. P(a)*(b) when the event is independent. ex) a boy flips a coin and gets a heads, he flip it again and get a tails. independent does not change the outcome of the next occurring event.
- (~) is the symbol for "not"
- (~) is the symbol for "not"
- "1" is the highest possible value you can get ; a 100% chance ; more than 1 will result in an incorrect answer ; when you add up all of your probability the numbers you should only get is equal to or less than 1 or zero.
- changing percent into a decimal will make calculation easier
- fraction is our Friend!
Questions:
1. man of war and secretarial are in a horse race, man of war has a 2/5 chance of winning and secretarial has a 1/3 chance of winning..
a) are these events mutually exclusive?
b) what is the probability that one of these horses wins the race?
c) what is the probability that both win?
answers:
a) no
b) probability of man of war winning is 2/5 ; probability of secretarial winning is 1/3 so...
(2/5) + (1/3) = (11/15)
c) (2/5) x (1/3) = (2/15)
Questions:
10 grade 12....40% of the students take math, 35% take history and 15% take both. If a student is randomly chosen. what is the probability that the student takes..
a) math but not history?
b) neither math or history?
Answers:
a) change percent into decimal. P(math) = .40 P(history) = .35 P(~m)= .60 P(~h) = .65
P(M x ~H)= P(.40 x .65) = .26
b) "green" = students taking math ( 40%-15% = 25% or .25) "brown" = students taking history (35%-15% = 20% or .20) "blue" = students taking both subjects ( 15% or .15) *add up all of our values ( .20+.15+.25 = .60) meaning that there are .40 students that are not taking neither subjects (1-.60 = .40)
Question:
you are dealt a 5 card poker hand, what is the probability that you have "at least one" heart?
Answer:
52 cards in a standard deck of cards and there are 13 hearts * 1 - none leaves all the possibility of "at least one"* *1- none leaves all the possibility of "at least one"* * 1 subtract none leaves all the possibility of "at least one"* * 1 take away all the cards that are not hearts will leave all the cards that are hearts*
39C5 = 39 ( the number of cards that are not hearts) 5 ( 5 card poker hand)
52C5 = 52 (total number of cards) 5 ( 5 card poker hand)
1- 39C5/52C5 = .77
numerator: 39C5 <- 39 cards that are not hearts choosing 5 random cards
denominator: 52C5 <- total cards choosing 5 random cards
"1" <- the highest possible value that you can get ; 100%
First things first? One day left to go--
Wondering if you're having trouble putting first things first in these last days before winter break?
And
Wanting to wish each of you happiness, peace, and joy during this season! Be safe and enjoy!
Best,
Lani
December 20, 2006
BOBing the conic way :D
Looks like its that time again folks... that's right BOB time, which
means a test is really close... and this was a short unit, but that
didn't mean we didn't learn a lot :D ... conics was not my favourite
unit, but it was far from the worst, so overall its not too bad. I
thought i was doing ok... until i hit the online quiz... but i figure
my barely passing mark was because of some poor
concentration.. i like doing the quizzes at home, but this
time there were just too many distractions, but if i had paid
attention (i just went and reviewed my mistakes... jeeze i made
a lot of dumb ones) i could have done really well... so this means
that during the test i really need to just buckle
down, stay calm and believe that i know what I'm doing... then i will
do just fine... other than that the hardest part for me is just trying
to make everything work out correctly without missing steps because
I'm rushing.. again if i stay calm and take my time I'll end up with
a much better result.. well that's just about it from me, and soon
for conics too (until the going for gold assignment and exam anyways :D )...
So good luck everyone, and I'll see you on test day :D *learn hard!!* :P
December 19, 2006
Blogging on Blogging
Alright well, I'd like to say, conics was okay, it didn't last very long, but I like this unit, But I hope I do good tomorrow. I usually don't do as good at online quizes, But I believe that's a ood way of reviewing, or testing what you may know already, and see where you need improvements. Anways I wish everyone good luck tomorrow, and myself as well! Bye.
Conic's Blogging on Blogging
It's the end of the conics unit, and I think this one was one of the hardest units yet! I think I'm okay with it though, I just hope I don't blank out or anything. Hopefully I'll be able to remember how to apply the rules for the test tomorrow.
Hopefully everyone does well on the test tomorrow!! ^_^
Hopefully everyone does well on the test tomorrow!! ^_^
bob
well here's my last minute blog...almost forgot. For this unit I thought it was going fine and then we had the group work for parabolas which kind of scared me because I couldn't get the answer since I forgot what to do...but then afterwards i remembered which was good...we should have handed in the second parabola question..my group actually got that one correct..yayy..well anyway back to the test, the quiz for some odd reason i didn't do too well in which really sucks because I thought I knew everything so now I think I don't know anything.I think it'll be okay I just have to try my best..I think I got all of my thoughts for this test out so see everyone tomorrow and good luck on the test.
Probability
Sorry guys this came late, and I gotta sleep soon too, My net's very .. bad .. and I haven't gotten on till now, but yes .. In the morning we were given 3 questions to do on the board.
1. Find the sample space for the genders of 3 children?
2. What is the probability of 2 children in a family both being boys?
3. What is the probability of drawing a red card and then a black card out of a deck of 52 playing cards if:
a) The first card is replaced before the second is drawn?
b) The first card is not replaced before the second is drawn?
The answer to this question will have to be drawn, but it would have to be listed and these are it:
B= Boys
G= Girls
BBB
BBG
BGB
BGG
GBB
GBG
GGB
GGG
2. The answer to this question is well 1/4 because when you want something to happen you have to multiply. It would help making a diagram =).
3.a) The answer for the first one would be 1/4.
b) The answer for this would be
P(RB)= (26/52)(26/52)
which would equal 0.2549
For the last part of class we wrote a bit of definitions in out math dictionary:
Dependant events: 2 ( or more ) events are dependant if the outcome of the second event depends on what occurred in the first event.
Example: A bag contains 3 red and blue marbles. A marble is drawn without replacement. The probability that a second marble will be blue depends on what the first marble was.
Independent events: 2 ( or more ) events are independent if the outcome of the first event has no effect on the outcome of the second event.
Example: Given the same bag of marbles. A marble is drawn with replacement. The probability that the second marble is blue is unaffected by which marble was drawn first.
Alright well, that's all everybody, sorry there are no diagrams for the moment, I'm trying to fix my computer, it's hard enough trying to fully fix our connections. Anyways, it's that time again, the NEXT SCRIBE WILL BE PATRICK. GOOD LUCK ! =).
1. Find the sample space for the genders of 3 children?
2. What is the probability of 2 children in a family both being boys?
3. What is the probability of drawing a red card and then a black card out of a deck of 52 playing cards if:
a) The first card is replaced before the second is drawn?
b) The first card is not replaced before the second is drawn?
The answer to this question will have to be drawn, but it would have to be listed and these are it:
B= Boys
G= Girls
BBB
BBG
BGB
BGG
GBB
GBG
GGB
GGG
2. The answer to this question is well 1/4 because when you want something to happen you have to multiply. It would help making a diagram =).
3.a) The answer for the first one would be 1/4.
b) The answer for this would be
P(RB)= (26/52)(26/52)
which would equal 0.2549
For the last part of class we wrote a bit of definitions in out math dictionary:
Dependant events: 2 ( or more ) events are dependant if the outcome of the second event depends on what occurred in the first event.
Example: A bag contains 3 red and blue marbles. A marble is drawn without replacement. The probability that a second marble will be blue depends on what the first marble was.
Independent events: 2 ( or more ) events are independent if the outcome of the first event has no effect on the outcome of the second event.
Example: Given the same bag of marbles. A marble is drawn with replacement. The probability that the second marble is blue is unaffected by which marble was drawn first.
Alright well, that's all everybody, sorry there are no diagrams for the moment, I'm trying to fix my computer, it's hard enough trying to fully fix our connections. Anyways, it's that time again, the NEXT SCRIBE WILL BE PATRICK. GOOD LUCK ! =).
jessica.'s bob.
hmm, conics. i think this unit was pretty straight forward. i pretty much know what i'm doing. i've actually learned a different way of doing the equations. it's all good (: goodluck on the test tomorrow.
gooday&goodnight.
gooday&goodnight.
Tennyson's BOB
man , i am trying my best on this one , everytime we do a test , i always mess up. I know how to do the questions , but somehow, the question always different , very little is the same that we do on our assignments. Like i do well in homework , but when the test comes , its like everything is different then what i have done before. Well like Mr k said , " your mark determines , how much time you put into it". This unit is ok , its not bad, very short.well i hope i pass this test good luck everyone
JessicaJill's BOB
Whoa, I almost forgot to do this. Well, the unit has come to an end, and the question is, have I learned anything? Indeed I have. A quick and easy unit I thought, but there were new things for me to pick up and learn. All in all, a good unit. Now I'll just head to class. =)
December 18, 2006
Ashley's BOB on Conics
So this unit Conics is my absolute favourite! I actually understand what I'm doing and I finally know how to find the foci because for the longest time now I had no idea what the focus was:P I think I'm doing in this unit but I don't want to get too happy because the test could be hard. Some questions still confuse me, I mean I tend to make little mistakes and forget that y^2 = x that means that the parabola is opening sideways. Stuff like that I always forget so I need to learn how to read the question properly but other than that I'm doing fine. Good luck everyone!
Probability and math dictionary notes on it
Hey it's Ashley and I'm the scribe for today!
In the morning we started our new unit Probability. Mr.K said that Pascal invented Probability and its the same Pascal that did the Pascal's Triangle.
Mr.K asked us "What can happen if you roll six-sided die?"
Well you could get 6 possible answers. (i forgot what he said oops)
Soon we were asked "What can happen if you flip a coin twice?"
What you could do is draw a tree diagram
Next question was a family of two children - "What is the probability they have one of each, a boy and a girl?"
Draw the tree diagram.
That question and the flipping of the coin question are the same.
Next was "What if you roll a die twice?"
You could for the question draw a tree diagram but that would be too complicated so instead you could draw a table.
Last question was "What if you flip a coin and roll a dice?"
Mr. K made us draw a table and a tree diagram.
Towards the end of the period Mr.K handed us 2 booklets. One was yellow and one was green. Their both from the June 2005 exam. The title for this assignment is drum rolls please..... Go for Gold!!!
For this assignment you must:
- get 100% on it.
- complete the two booklets with no mistakes or errors.
- show all work including for the multiple choice questions either by writing a sentence explaining yourself but don't write a book. If there is no work shown or a sentence you get 0.
- must write in pencil! no pen or else you get 0.
In Go for Gold there are only two possible marks available. 0 and 100. It's worth 5% on our final class mark. Due on the Friday we get from winter break and you may get help from others and Mr.K.
In the afternoon Mr. K went over the questions on the quiz. If you want to see which ones you got right or wrong you could check now because it will show the answers. Mr.K got a question wrong on "Determine the foci of the hyperbola (y2/16) - (x2/25) = 1" by placing the answer on the wrong letter so that question won't count.
Next we took some notes down into our Math Dictionary
Probability
Probability: the branch of mathematics that deals with chance.
Sample Space (Ω): the set of all possible outcomes for a given "Experiment" represented by capital omega Ω.
Event (E): A subject of the sample space. A particular occurence in a given experiment.
Simple Event: The result of an experiment that is carried out in a single step.
Example: Flip a coin. The result is heads
(a simple event)
Compound Event: The result of an experiment carried out in two (or more) steps.
Example: 1. Flip a coin and roll a die - 6
The result is {H,6}
2. Flip a coin twice.
The result is {H,T}
Probability of an Event: (suppose to be sides to the box below but i don't know how to draw a rectangle so just put a box around it, i only managed the top and bottom borders)
In the morning we started our new unit Probability. Mr.K said that Pascal invented Probability and its the same Pascal that did the Pascal's Triangle.
Mr.K asked us "What can happen if you roll six-sided die?"
Well you could get 6 possible answers. (i forgot what he said oops)
Soon we were asked "What can happen if you flip a coin twice?"
What you could do is draw a tree diagram
Next question was a family of two children - "What is the probability they have one of each, a boy and a girl?"
Draw the tree diagram.
That question and the flipping of the coin question are the same.
Next was "What if you roll a die twice?"
You could for the question draw a tree diagram but that would be too complicated so instead you could draw a table.
Last question was "What if you flip a coin and roll a dice?"
Mr. K made us draw a table and a tree diagram.
Towards the end of the period Mr.K handed us 2 booklets. One was yellow and one was green. Their both from the June 2005 exam. The title for this assignment is drum rolls please..... Go for Gold!!!
For this assignment you must:
- get 100% on it.
- complete the two booklets with no mistakes or errors.
- show all work including for the multiple choice questions either by writing a sentence explaining yourself but don't write a book. If there is no work shown or a sentence you get 0.
- must write in pencil! no pen or else you get 0.
In Go for Gold there are only two possible marks available. 0 and 100. It's worth 5% on our final class mark. Due on the Friday we get from winter break and you may get help from others and Mr.K.
In the afternoon Mr. K went over the questions on the quiz. If you want to see which ones you got right or wrong you could check now because it will show the answers. Mr.K got a question wrong on "Determine the foci of the hyperbola (y2/16) - (x2/25) = 1" by placing the answer on the wrong letter so that question won't count.
Next we took some notes down into our Math Dictionary
Probability
Probability: the branch of mathematics that deals with chance.
Sample Space (Ω): the set of all possible outcomes for a given "Experiment" represented by capital omega Ω.
Event (E): A subject of the sample space. A particular occurence in a given experiment.
Simple Event: The result of an experiment that is carried out in a single step.
Example: Flip a coin. The result is heads
(a simple event)
Compound Event: The result of an experiment carried out in two (or more) steps.
Example: 1. Flip a coin and roll a die - 6
The result is {H,6}
2. Flip a coin twice.
The result is {H,T}
Probability of an Event: (suppose to be sides to the box below but i don't know how to draw a rectangle so just put a box around it, i only managed the top and bottom borders)
_________________________________________________________
P(E) = number of favourable outcomes
Total number of possible outcomes (sample space) ________________________________________________________
Probability can be expressed as:
- a Ratio
- a Fraction
- a Decimal
- a Percent
(rectangle box below but i only got the top and bottom border, not the sides)
IMPORTANT: Probability of any event is always a number between 0 and _______________________________________________________________
Certain Event: an event whose probability is 1
example: roll a die - 6 and get a result less than 10
Impossible Event: an event whose probability is 0.
Impossible Event: an event whose probability is 0.
example: roll a die - 6 and get a 7
Complimentary Event: the compliment of E is E',
If P(E) = a then P(E') = 1-a
example: Given a standard deck of cards, a card is drawn at random.
P(spade) = (13/52) = (1/4) P (not a spade) = 1 - (1/4) = (3/4)
And that was the end of the math dictionary notes
Mr. K recommended we visit these sites:
math40s.com because it has the entire curriculum on it and some other stuff on it which I forgot but apparently it's a very good site.
webct.merlin.mb.ca because it too has the entire curriculum on it. you need to log on to this site.
username: demo
password: demo
it asks if you trust this site I think and you say yes.
both sites are good for reviewing.
** Don't forget to check the rubric for the Flickr assignment and leave your comments and what you like and what you want to change because the Rubric will be finalized on Friday. To leave a comment just press the Insert Comment button. The next Flickr assignment is on Exponential function which is due the Tuesday after winter break. The tag will be expfunc.
So that's it for my scribe.
Homework assignment is Ex.39 by the way.
Next scribe is........ well you know who you are *cough*Oliver*cough*
December 15, 2006
scribe post
Hey it's Anh for your last scribe of the cycle. Class started with a discussion of the flickr assignments. We talked about how the assignment would be marked. The smaller hotspots should be on top of bigger hotspots so that Mr.K is able to access it. Also the photos should be clear. Yesterday Mr.K sent everyone an email invitation to a document where we can have a input to the marking rubric for the flickr assignments. So please participate.
*** TEST IS MOVED TO WEDNESDAY!
*** QUIZ IS ACTIVE TILL MIDNIGHT SUNDAY!
We pretty much did our pre test all class.
CONIC PRE-TEST
The pictures are still being stupid.
I guess this concludes my scribe post...umm good luck on the quiz.
Next scribe: Ashley :)
Online Conics Quiz
Your online conics quiz
Remember the deadline to write is Sunday night at midnight; which means you need to start it no later than 11:15 Sunday night. There are 10 multiple choice questions and 5 true or false questions. The quiz is timed. You will have 45 minutes to complete it once you begin.
BTW, the light in the picture forms a hyperbola; two of them actually. The light itself is a circle. ;-)
daphne's SAIS blogging on blogging
The CONICS was a very interesting, mind-squizzing, and mental blocking unit (...for me!!!). It was a lot of "It seems like a complicated problem" remark but in reality, most of Mr. K's problem are the basics. Unfornately in my part, I found it hard to solve word problems especially when we are looking for the equation such as the ellipses and hyperbola. Parabola equations are way easier to solve than ellipses and hyperbola because I sometimes mixed up the information needed to find the equation. Futhermore, complicated word problems sort of gave me a headache specifically when I'm in the part of solving the equation then suddenly I'm stock because I do not know what will be my next step but in all fairness, Mr. K did a great job in teaching us this unit. He explained every unsolved problems in details and he never gave up on us until we are confident enough to solve CONICS problems with a guaranteed high mark. In addition, I would to thank those who helped me in this unit because without I may not be able to understand some of the topics.
Go pre-cal 40S!!!!!!!!!!!!....We can all do it on top!!!!!!!!!!!!!!
Go pre-cal 40S!!!!!!!!!!!!....We can all do it on top!!!!!!!!!!!!!!
December 14, 2006
azn_chilly's style of bloggin on itself
this unit about conic is somewhat easy but i still need to understand how to get those equations from those problem questions but with the full equation out and i have to make the graph its easy to understand it but it all the problem solving, its giving me a hard time and its giving me a negative feeling that i did all i can on the question and trying to figure it out soo.. i think i have to say for this unit, and one thing good luck to everyone on the big test
Word Problems with Conics
Hey kids, I was yesterday's scribe post. This is going up a day late because I had to work late last night and I went straight to bed as soon as I got home. Anyways yesterday was a one period class, and we worked on some conic word problems from previous exams. Here are the five questions given to us on the board:
1. Find the equation of an ellipse with major axis AB and minor axis CD if the following coordinates are the vertices: A(2,7) B(2,-1) C(0,3) D(4,3).
2. Identify each conic:
3. Find the equation of the circle with the centre (4,0) which passes through the orgin.
4. y²-2y-3=0
a) Find all intercepts.
b) Sketch a clearly labeled graph.
5. A balloon arch in the shape of one branch of a hyperbola is made for a wedding party. The arch is to span 4m and have a max clearance of 2.2m.
a) Find an equation for the hyperbola.
b) What would be the height of the arch 0.8m from one end?
ANSWERS:
1. This is what I know. AB have the same x-coordinates, so that means that the line of AB runs vertically. So I just added the absolute values of the y-coordinates to get the length of AB: -1 + 7 = 8. It's an absolute value because distances don't have negative values. So AB is 8 units long, and we divide it by 2 to get the value of a, which is 4. So, we can say that line of CD runs horizontally and we add together their x-coordinates: 0 + 4 = 4. We divide 4 by 2 to get the value of b and we get 2.
1. Find the equation of an ellipse with major axis AB and minor axis CD if the following coordinates are the vertices: A(2,7) B(2,-1) C(0,3) D(4,3).
2. Identify each conic:
3. Find the equation of the circle with the centre (4,0) which passes through the orgin.
4. y²-2y-3=0
a) Find all intercepts.
b) Sketch a clearly labeled graph.
5. A balloon arch in the shape of one branch of a hyperbola is made for a wedding party. The arch is to span 4m and have a max clearance of 2.2m.
a) Find an equation for the hyperbola.
b) What would be the height of the arch 0.8m from one end?
ANSWERS:
1. This is what I know. AB have the same x-coordinates, so that means that the line of AB runs vertically. So I just added the absolute values of the y-coordinates to get the length of AB: -1 + 7 = 8. It's an absolute value because distances don't have negative values. So AB is 8 units long, and we divide it by 2 to get the value of a, which is 4. So, we can say that line of CD runs horizontally and we add together their x-coordinates: 0 + 4 = 4. We divide 4 by 2 to get the value of b and we get 2.To get the centre, we take the y-coordinate of A and subtract the value of a from the y-coordinates, just as we take the x-coordinate of D and subtract the value of b from the x-coordinates.
If you wanted to add the value of a to find the y-coordinate of the centre, you'd have to add from B's y-coordinate. To add the value of b to find the x-coordinate of the centre you'd have to add it from C's x-coordinate.
Subtracting
AB: 7 - 4 = 3
CD: 4 - 2 = 2
Adding:
AB: -1 + 4 = 3
CD: 0 + 2 = 2
So the center is (2,3).
Now we can start plugging in our values to an equation.
2. a) We can say that it is a hyperbola, because we have the subtraction sign between a fraction of x² and a fraction of y².
b) In the expansion we have positive squared values for x and y, which tells us it can't be a parabola or a hyperbola, but they have different coefficients so that tells us it is has to be an ellipse.
3. The equation of a circle is: (x-h)² + (y-k)² = r²
(h,k): (4,0)
point: (0,0)
From this we can tell that both points rest on the x-axis, and have are 4 units away from each other. Hence, we can say that the radius of this circle is 4 units.
(x-4)² + y² = 4²
(x-4)² + y² = 16
4. We take our eqation and change it to standard form.
y² - x - 2y - 3 = 0
(y² - 2y + 1 ) = x + 3 + 1
(y-1)² = x + 4
a) To find intercepts we just plug in zero for our variable.
x--intercept; y = 0
(0 - 1)² = x + 4
(-1)² = x + 4
1 = x + 4
-3 = x
y-intercepts; x = 0
(y - 1)² = 0 + 4
(y - 1)² = 4
y - 1 = ±2 [square root both sides]
y -1 = 2
y = 3
y - 1 = -2
y = -1
Therfore, our x-intercept is at -3, and our y-intercepts are at 3 and -1.
b)
If you wanted to add the value of a to find the y-coordinate of the centre, you'd have to add from B's y-coordinate. To add the value of b to find the x-coordinate of the centre you'd have to add it from C's x-coordinate.
Subtracting
AB: 7 - 4 = 3
CD: 4 - 2 = 2
Adding:
AB: -1 + 4 = 3
CD: 0 + 2 = 2
So the center is (2,3).
Now we can start plugging in our values to an equation.
2. a) We can say that it is a hyperbola, because we have the subtraction sign between a fraction of x² and a fraction of y².
b) In the expansion we have positive squared values for x and y, which tells us it can't be a parabola or a hyperbola, but they have different coefficients so that tells us it is has to be an ellipse.
3. The equation of a circle is: (x-h)² + (y-k)² = r²
(h,k): (4,0)
point: (0,0)
From this we can tell that both points rest on the x-axis, and have are 4 units away from each other. Hence, we can say that the radius of this circle is 4 units.
(x-4)² + y² = 4²
(x-4)² + y² = 16
4. We take our eqation and change it to standard form.
y² - x - 2y - 3 = 0
(y² - 2y + 1 ) = x + 3 + 1
(y-1)² = x + 4
a) To find intercepts we just plug in zero for our variable.
x--intercept; y = 0
(0 - 1)² = x + 4
(-1)² = x + 4
1 = x + 4
-3 = x
y-intercepts; x = 0
(y - 1)² = 0 + 4
(y - 1)² = 4
y - 1 = ±2 [square root both sides]
y -1 = 2
y = 3
y - 1 = -2
y = -1
Therfore, our x-intercept is at -3, and our y-intercepts are at 3 and -1.
b)
5.
After realizing how lengthy question 5 was, we got to doing some group work. I kind of lost the sheet with the question on it, so I don't have much else to post up. I'll fix this up when I do find it, I just kind of forgot I never finished this. Sorry for the huge delay!
After realizing how lengthy question 5 was, we got to doing some group work. I kind of lost the sheet with the question on it, so I don't have much else to post up. I'll fix this up when I do find it, I just kind of forgot I never finished this. Sorry for the huge delay!
Analytic Geometry ( Conics )
Hello everybody!, I hope everyone is doing fine tonight. I am today's scribe and well here's a short summary of today's class. In the morning we came in class, like any other day and then Mr. K put up 3 kinds of problem solving questions on the board and we had to answer them. He gave us sometime then we took up the questions. Here are some of my answers. The first question was:
An earthquake observation centre is located at coordinates (-2,5). An earthquake epicentre is located at a distance 120k, from the observation centre. What is the equation of the circle that defines where the epicentre ca be located?
** I would just like to say, I couldn't get most of the explanations on how we got the answers for these questions, because in the morning there was 2 fire drills. Very sorry about that.
Well the answer for this first quesion was (x+2)^2 + (y-5)^2 = 120^2.
This is the second quetion:
A reflector dish has a diamtere of 1.8 and is 24cm deep. Model the dish as aparabola opening up with its vertex at the origin. Find its equation and the coordinates of the focus.
Well around this time, we got the fire drill, and then we stood outside for a bit, so I didn't have too much time to copy the answers. We gt back in and then we somehow got into the convo about the sun, and I learned that in the galaxy our solar system is caled the "orion's arm" or something like that.
In the afternoon, we were supposed to have a pre-test but we finished up the last question from the morning, and this was when tennyson came up to the bored, and it was quite funny actually. Then a little after that we got into groups and we did some problem solving questions together. I personally found the questions, not hard, but confusing, because for some reason we kept mixing up numbers and what not. Anyways I hope everyone is ready fo the test on monday, and tomrrow is our pre-test! Good luck Everyone! That is all. Till next time, BYE !
THE NEXT SRIBE POSTER IS, **ANH !!=)**
An earthquake observation centre is located at coordinates (-2,5). An earthquake epicentre is located at a distance 120k, from the observation centre. What is the equation of the circle that defines where the epicentre ca be located?
** I would just like to say, I couldn't get most of the explanations on how we got the answers for these questions, because in the morning there was 2 fire drills. Very sorry about that.
Well the answer for this first quesion was (x+2)^2 + (y-5)^2 = 120^2.
This is the second quetion:
A reflector dish has a diamtere of 1.8 and is 24cm deep. Model the dish as aparabola opening up with its vertex at the origin. Find its equation and the coordinates of the focus.
Well around this time, we got the fire drill, and then we stood outside for a bit, so I didn't have too much time to copy the answers. We gt back in and then we somehow got into the convo about the sun, and I learned that in the galaxy our solar system is caled the "orion's arm" or something like that.
In the afternoon, we were supposed to have a pre-test but we finished up the last question from the morning, and this was when tennyson came up to the bored, and it was quite funny actually. Then a little after that we got into groups and we did some problem solving questions together. I personally found the questions, not hard, but confusing, because for some reason we kept mixing up numbers and what not. Anyways I hope everyone is ready fo the test on monday, and tomrrow is our pre-test! Good luck Everyone! That is all. Till next time, BYE !
THE NEXT SRIBE POSTER IS, **ANH !!=)**
Blaging on blagging-Jefferson style
CONICS IS SO FUN AND EASY TO UNDERSTAND!. Unfortunately I'm having a lot of problems on the word problem questions. i just don't seem to know where to start. after one particular word problem is solved, i begin to think why i wasn't able to solve it with ease. There is nothing different about the idea. But if it's rewritten in a way that seems like it's from another language then it's really really hard.
Richard's 6th BOB
In this unit, CONICS, ahhh! I find it difficult.
(How many times did I say the word difficult in my BOB?)
I somehow understand the lesson but still I don't have enough background in this unit.
Maybe because I didn't take grade 11 here. It's really hard for me to adjust.
There are lots of formula to memorize and lots of critical cases.
This unit is so painful..but I still believe in "No pain, No gain"
I should really work harder for this unit. It's so sad that I don't have the time to ask Mr. K (or rather shy).
Anyway...good luck everyone to the test! We still had a long time to prepare.
December 13, 2006
Blogging on Blogging
this unit about conics is very easy.................but.......I seem to be having a trouble on vertical parabola and vertical Ellipse type of questions. I keep on forgetting that the formula for vertical is different from horizontal parabola and Ellipse, like the one we did today in class and I don't really get that word problem question about the balloon with the arc circle and angel...arg..ugh..ah..(brain shuts down....reboting...back online). other then that, I think I understand most of the horizontal type of questions. good luck on the test on friday
ruschev~~out
ruschev~~out
Blogging on Blogging
Might as well get this done before I forget, conics I'm having an easy time with if you can believe it. The only problems that I seem to be having are just a few of the problems similar to the ones involving a semi-ellipse or half of a hyperbola like the questions in class today. Well that and the random bits of mistake that I make on occasions when I forget that an equation like "y^2 = x-4" is suppose to open vertically. Other then that's there's nothing too important to mention about how I'm doing for this unit.
December 12, 2006
Next Scriber?????????????
i dont know if this is a new cycle of scribers or not but the person who will be the next scribe for tomorrow and i choice, choice..... plus i will be not finished by 12am i came home at 10 pm and am like 3/4 done typing it all out and starting my graphs soo.... the next person will be..... jessicajill who was to be today's scriber (mad mood) and now she own me 20 dollars in cash tomorrow hahahha j/k
azn_chilly's scribe post (problems + learning)
This morning we started off with three questions off the bored and i was late for that class and was who willing to the scribe today hahaha!!!!!!. so we stared off by this .....
graph this conic section and label all information points
4x(square)+9x(square)-16x+18y-11=0
4x(square)-16x + 9x(square)+18y =11
(4x(square)-16x+4-4) + (9x(square)+18y+1-1)=11+16+9
4(x-2)(square)/36 + 9(y+1)(square)/36 = 36/36
(x-2)(square)/9 + (y+1)(square)/4 =1
(x-2)(square)/3 + (y+1)(square)/2 =1
A(square) - B(square)=C(square)
3(square) - 2(square)=C(square)
9-4=C(square)
5=C(square)
(square root)5=C
find the equation of an ellipse with center at (2,-5), major axis at length 6 units parallel to the x-axis and major axis 2 units long
(x-2)(square)/9 - (y+5)(square)/1=1
a tunnel has a semi-elliptical cross-section it is 20m wide and 5m at its highest point, find the equation of the ellipse and use it to find the height of the tunnel 2m away from one end
(x-h)(square)/a(square) + (y-k)(square)/b(square)=1
x(square)/100 + y(square)/25=1
8(square)/100 + y(square)/25=1
y(square)/25=100/100-64/100
y(square)/25=36/100
(25)y(square)/25=36/100(25)
y(square)=9
y(square)-9=0
(y+3)(y-3)=0
y=-3 y=3
y=-3 is rejected cause it will be in the negative side
in the afternoon we talked about of the construction of the hyperbola and its properties...
(x-h)(square)/a(square) - (y-k)(square)/b(square)=1 is the hyperbola
(x-h)(square)/a(square) + (y-k)(square)/b(square)=1 is the ellipse
hyperbola: a locus of points that moves in such a way so that the absolute value of the distance(focal radii) from two fixed points(foci) is constant
|PF1-pf2|= 2a[a constant]
the anatomy of a hyperbola
0 is the centre
A1 + A2 are the verticies
A1A2 is the transverse axis its length is 2a
B1 +B2 are the end point of the conjugate axis its length is 2b
F1 + F2 are the foci they are for the centre
PF1 and PF2 are the focal radii
|PF1 - PF2|= 2a
the picture of the hyperbola is horizontal and centred at the the origin, the equation of the of the asymptotes are: y= b/ax and y=-b/ax
the transvers and conjugate axis are lines of symmetry for the hyperbola
graph this conic section and label all information points
4x(square)+9x(square)-16x+18y-11=0
4x(square)-16x + 9x(square)+18y =11
(4x(square)-16x+4-4) + (9x(square)+18y+1-1)=11+16+9
4(x-2)(square)/36 + 9(y+1)(square)/36 = 36/36
(x-2)(square)/9 + (y+1)(square)/4 =1
(x-2)(square)/3 + (y+1)(square)/2 =1
A(square) - B(square)=C(square)
3(square) - 2(square)=C(square)
9-4=C(square)
5=C(square)
(square root)5=C
find the equation of an ellipse with center at (2,-5), major axis at length 6 units parallel to the x-axis and major axis 2 units long
(x-2)(square)/9 - (y+5)(square)/1=1
a tunnel has a semi-elliptical cross-section it is 20m wide and 5m at its highest point, find the equation of the ellipse and use it to find the height of the tunnel 2m away from one end
(x-h)(square)/a(square) + (y-k)(square)/b(square)=1
x(square)/100 + y(square)/25=1
8(square)/100 + y(square)/25=1
y(square)/25=100/100-64/100
y(square)/25=36/100
(25)y(square)/25=36/100(25)
y(square)=9
y(square)-9=0
(y+3)(y-3)=0
y=-3 y=3
y=-3 is rejected cause it will be in the negative side
in the afternoon we talked about of the construction of the hyperbola and its properties...
(x-h)(square)/a(square) - (y-k)(square)/b(square)=1 is the hyperbola
(x-h)(square)/a(square) + (y-k)(square)/b(square)=1 is the ellipse
hyperbola: a locus of points that moves in such a way so that the absolute value of the distance(focal radii) from two fixed points(foci) is constant
|PF1-pf2|= 2a[a constant]
the anatomy of a hyperbola
0 is the centre
A1 + A2 are the verticies
A1A2 is the transverse axis its length is 2a
B1 +B2 are the end point of the conjugate axis its length is 2b
F1 + F2 are the foci they are for the centre
PF1 and PF2 are the focal radii
|PF1 - PF2|= 2a
the picture of the hyperbola is horizontal and centred at the the origin, the equation of the of the asymptotes are: y= b/ax and y=-b/ax
the transvers and conjugate axis are lines of symmetry for the hyperbola
December 11, 2006
Scribe Post
The morning started out with the usual problems nothing new about that.
1) Find the center and radius of the circle: 4x2-12x+4y2-30 = 0
For what value of k will the circle x2+y2-2x+4y+k = 0 have a radius of 4?
1) Find the center and radius of the circle: 4x2-12x+4y2-30 = 0
4[(x2-3x+(3/2)2-(3/2)2)]+4y2 = 30
4[(x-3/2) 2-9/4]+4y2 = 30
4(x-3/2) 2-9+4y2 = 30
4(x-3/2) 2+4y2 = 39
4[(x-3/2) 2+y2 = 39
(x-3/2) 2+y2 = 39/4
The center is at (3/2,0) while the radius is the sqr. root of 39 over 2
For what value of k will the circle x2+y2-2x+4y+k = 0 have a radius of 4?
(x2-2x)+(y2+4y) = -k
(x2-2x+1)+(y2+4y+4) = -k+1+4
(x-1) 2+(y+2) 2 = -k+5
Since we want a radius of 4 we add it in
42 = -k+5
16 = -k+5
k = -11
Find the equations of the ellipse with center (3,-1), major axis of length 10 and parallel to the y-axis, minor axis is 4 units long.
(x-h) 2 + (y-k) 2 = 1
b2 a2
(x-3) 2 + (y+1) 2 = 1
4 25
Foci:
C2 = A2-B2
=52-22
=25-4
=21
C =sqr(21)
The foci is measured from the center of the ellipse moving up and down hanging in between sqr(16) and sqr(25) since sqr(21) is the next perfect square above sqr(16) and below sqr(25).
The second half of the class was us working on the hyperbolas that we drew last week, the changes I'm afraid you'll have to ask people in class about since the new additions wouldbe a little confusing for me to show on computer.
Next scribe is JessicaJillDecember 10, 2006
Conic Scribe Post
Hey everyone it's tennyson in the house!!!!
During the morning we started to work out some problems that we had from the last homework that everybody had troubles with.
From the homework Exercise sheet
#2.
Determine the equation of the circle which passes throguht the points
J(-3,2) k(4,1) L(6,5).
Points:
J(-3,2)
k(4,1)
L(6,5)
this is an example of how we know where the centre is
( spelling error know = knowing )
we have to find the midpoint.
So we used Rise/Run
Rise/run
m(KL)= 4/2 = 2
m(AL)= -1/2
midpoint
midKe=(5,3)
y-3=-1/2(x-5)
After that the whole class has to do the paper folding again, this time for the "Hyperbola"
Here was a sample of what it turn out to be:
After that we had to calculate what line measures up to.
Pf1 - Pf2 = 56mm-135mm = 79
Qf1-Qf2 =146mm-69mm = 77
Rf1-Rf2 =143mm - 69mm = 76
A1A2= 81
( note: my calculations are different from other students)
During the afternoon mr.K was away , so we just did a quiz & homework on exercise 38
well by3 everyone , next person would be : JOHN!!!!!
During the morning we started to work out some problems that we had from the last homework that everybody had troubles with.
From the homework Exercise sheet
#2.
Determine the equation of the circle which passes throguht the points
J(-3,2) k(4,1) L(6,5).
Points:
J(-3,2)
k(4,1)
L(6,5)
this is an example of how we know where the centre is
( spelling error know = knowing )
we have to find the midpoint.
So we used Rise/Run
Rise/run
m(KL)= 4/2 = 2
m(AL)= -1/2
midpoint
midKe=(5,3)
y-3=-1/2(x-5)
After that the whole class has to do the paper folding again, this time for the "Hyperbola"
Here was a sample of what it turn out to be:
After that we had to calculate what line measures up to.
Pf1 - Pf2 = 56mm-135mm = 79
Qf1-Qf2 =146mm-69mm = 77
Rf1-Rf2 =143mm - 69mm = 76
A1A2= 81
( note: my calculations are different from other students)
During the afternoon mr.K was away , so we just did a quiz & homework on exercise 38
well by3 everyone , next person would be : JOHN!!!!!
Did you know?
Did you know I can see your classroom from two windows?!
My first window is your blog. I am excited by what I see and hear! I never cease to be amazed by the quality and sophistication of your scribes; you constantly achieve new heights in illustrating and annotating your scribes. More than that I am so impressed when you celebrate each others’ learning, when you are creative, and when you critically reflect upon your learning in your BOBs.
Did you know Mr. K’s blog is my second window? I admire and respect what I see and hear here too! Did you know that Mr. K celebrates your learning on his blog? that he reflects upon what best helped you to learn and why? that he unselfishly shares all he knows with those who read his blog? that he learns from the conversations on his blog? that he writes with passion and is creative? and that he commits many random acts of kindness by honoring other teachers’ accomplishments in his posts? Did you know all he expects of you, he shares those same expectations for himself?
Did you know that because of all that and more, Mr. K.’s blog has been nominated for “Best Teacher Blog 2006” on the Edublog Awards website?
I just happen to think that no one deserves this honor more than Mr. K.
What about you?
December 07, 2006
Jefferson's scribe post
Today Mr.K was away so we had a substitute. All we did today was work on a stencil for the rest of the class and if you didn't finish it in class, it's for homework.
The next scribe will be Tennyson
The next scribe will be Tennyson
December 06, 2006
Conics: The Parabola, Circle and Ellipse (Cheek's 3rd Scribe post)
Hello its cheeks again working on today's scribe post on Conics: The Parabola, Circle and Ellipse. We started off by talking about our website reaching its max on 300 mb due the cause of all the images we posted up. So for now on we will be using photobucket to upload our images into our blog. We then watched some videos that were quite hilarious based and mathematics in relation with the real world. Our final video was about a tire then went loose as the driver was driving and how it rolled, flyed, bumped into other cars and reaching its final destination back into the drivers car again and the probability of that was 0.0000001%. Think its possible? Comment it. Our teacher then wanted all of us to comment about how mathematics is used in our other subjects. Think it is? Comment that too.
We continued our lesson where we left off from the anatomy of the parabola...
Our lesson began and ended with our dictionaries:
Deriving the standard form for the equation of a parabola
__ __
PF=PD (by def'n)
√[(x-h)²+[y-(k+p)]]² = √[(x-x)²+[y-(k-p)]]² (Distance formula)
(x-h)²+[y-k-p][y-k-p] = [y-k+p][y-k+p] (Square both and simplify)
(x-h)²+y²-ky-py-ky+k²+kp-py+kp+p² = y²-ky+py-ky+k²-kp+py-kp+p² (expan)
if we were to bring the terms that cancel from the left to the right side we simplify to:
(x-h)² = 4py - 4pk (balance)
(x-h)² = 4p(y-k) (factored)
***Standard form for equation of a vertical formula
__ __
PF=PD (by def'n)
√[[x-(h+p)]²+(y-k)²] = √[[x-(h-p)]²+(y-k)²] (Distance formula)
[x-h-p][x-h-p] +(y-k)² = [x-h+p][x-h+p] (Square both and simplify)
(y-k)²+x²-hx-px-hx+h²+hp-px+hp+p² = x²-hx+xp-xh+h²-hp+px-ph+p² (expan)
if we were to bring the terms that cancel from the left to the right side we simplify to:
(y-k)² = 4px - 4ph (balance)
(y-k)² = 4p(x-h) (factored)
***Standard form for equation of a horizontal formula
__________________________________________________________
Circle: A locus of points that move such that they are a fixed distance (radius) from a fixed point (centre).
Deriving the standard form for the equation of a circle
__
OP = r (by def'n)
√[(x-h)²+(y-k)²] = r (Distance formula)
(x-h)²+(y-k)² = r² (standard form for a circle)
__________________________________________
Ellipse: A locus of points that move such that the sum of the distances (focal radii) from fixed points (foci) is constant.
The anatomy of the ellipse
A1 and A2 are the vertices of the ellipse
A1A2 is the major axis of the ellipse. It is a line of symmetry and its length is 2a
B1B2 are the endpoints of the minor axis.
B1B2 is the minor axis of ellipse. It is a line of symmetry and its length is 2b.
F1 and F2 are teh foci of the ellipse each focus is C units from the centre.
PF1 and PF2 are called the focus radii. Their sum is always 2a, the length of the major axis.
The standard form for the equation of an ellipse
Standard form for a horizontal equation
Standard form for a vertical equation
The focal radii property
Suppose P is at A
PF1 + PF2 = A1F1 + A1F2 (by def'n)
PF1 + PF2 = A2F2 + A1F2 (by symmetry
PF1 + PF2 = A1 A2
PF1 + PF2 = 2a
The Pythagorean property
Suppose P is at B
PF1 + PF2 = B1F1 + B1F2
By symmetry we know that
B1F2 = B1F1
there fore...
PF1 + PF2 = B1F1 + B1F1
PF1 + PF2 = 2B1F1
By the focal radii property
PF1 + PF2 = 2a
therefore..
2a = 2B1F1
a = B1F1
and by Pythagorean theorm
b² + c² = a² or c² = a² - b²
Thats all for tonight the next scribe is Jefferson
We continued our lesson where we left off from the anatomy of the parabola...
Our lesson began and ended with our dictionaries:
Deriving the standard form for the equation of a parabola
__ __
PF=PD (by def'n)
√[(x-h)²+[y-(k+p)]]² = √[(x-x)²+[y-(k-p)]]² (Distance formula)
(x-h)²+[y-k-p][y-k-p] = [y-k+p][y-k+p] (Square both and simplify)
(x-h)²+y²-ky-py-ky+k²+kp-py+kp+p² = y²-ky+py-ky+k²-kp+py-kp+p² (expan)
if we were to bring the terms that cancel from the left to the right side we simplify to:
(x-h)² = 4py - 4pk (balance)
(x-h)² = 4p(y-k) (factored)
***Standard form for equation of a vertical formula
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PF=PD (by def'n)
√[[x-(h+p)]²+(y-k)²] = √[[x-(h-p)]²+(y-k)²] (Distance formula)
[x-h-p][x-h-p] +(y-k)² = [x-h+p][x-h+p] (Square both and simplify)
(y-k)²+x²-hx-px-hx+h²+hp-px+hp+p² = x²-hx+xp-xh+h²-hp+px-ph+p² (expan)
if we were to bring the terms that cancel from the left to the right side we simplify to:
(y-k)² = 4px - 4ph (balance)
(y-k)² = 4p(x-h) (factored)
***Standard form for equation of a horizontal formula
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Circle: A locus of points that move such that they are a fixed distance (radius) from a fixed point (centre).
Deriving the standard form for the equation of a circle
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OP = r (by def'n)
√[(x-h)²+(y-k)²] = r (Distance formula)
(x-h)²+(y-k)² = r² (standard form for a circle)
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Ellipse: A locus of points that move such that the sum of the distances (focal radii) from fixed points (foci) is constant.
The anatomy of the ellipse
A1 and A2 are the vertices of the ellipse
A1A2 is the major axis of the ellipse. It is a line of symmetry and its length is 2a
B1B2 are the endpoints of the minor axis.
B1B2 is the minor axis of ellipse. It is a line of symmetry and its length is 2b.
F1 and F2 are teh foci of the ellipse each focus is C units from the centre.
PF1 and PF2 are called the focus radii. Their sum is always 2a, the length of the major axis.
The standard form for the equation of an ellipse
Standard form for a horizontal equation
Standard form for a vertical equation
The focal radii property
Suppose P is at A
PF1 + PF2 = A1F1 + A1F2 (by def'n)
PF1 + PF2 = A2F2 + A1F2 (by symmetry
PF1 + PF2 = A1 A2
PF1 + PF2 = 2a
The Pythagorean property
Suppose P is at B
PF1 + PF2 = B1F1 + B1F2
By symmetry we know that
B1F2 = B1F1
there fore...
PF1 + PF2 = B1F1 + B1F1
PF1 + PF2 = 2B1F1
By the focal radii property
PF1 + PF2 = 2a
therefore..
2a = 2B1F1
a = B1F1
and by Pythagorean theorm
b² + c² = a² or c² = a² - b²
Thats all for tonight the next scribe is Jefferson