Wednesday, May 31, 2006 

Scribe

All right, so I guess it's up to me to break the tie of hall of famers in a row. It's never a good idea to leave this sort of stuff up to me. :P

The first thing we did in this morning's class was hover around Mr. K's computer and look at our sexy wiki. Remember guys, our constructive modification + significant contribution are due this FridayMONDAY. Mr. K gave us a bit of an extension (again!) because he's a really cool teacher. [/suckup] Seriously, thanks Mr. K.

Then we got onto sequences. Mr. K wrote two examples on the board for us to solve.

We got to use our calculators today, I am not sure if they were used the other day. With that said, make sure the calculator is in sequence mode!


-All you have to do is press mode and arrow over to seq. Then yeah, hit enter.

Question the first:

My cat is sick. His name is Little John and he has a cold. The vet has given him some medicine. Each day he gets a pill with 35 mg of medicine. His body eliminates 25% of the medicine each day and then he gets another pill.

a. How much medicine will be in his body in 5 days?
b. Will the amount of medicine in his body stabilize? How many days will it take and how much medicine will be in his body?

Answers:



a. After 5 days, there will be 106.77 mg of medicine.
-In the second column, the reason you multiply by .75 is because Little John's body eliminates only 25% of the medicine, leaving the other 75% in his body. Then, when he has another pill (35 mg), it's added with the medicine still in his body.


-.75 represents what you multiply U by, U being the term. N is the column. I believe so, anyway. Also, to get to that screen all you have to do is press Y=.

-I really like this, because you can go up as many terms as you'd like and then store them into one of the lists. 1 represents the first term, and 50 is the term I chose to go up to. I would assume you guys know all about this, as we did it quite a bit with periodic functions. The only difference is that we have it in sequence mode and we put the u. Ohh, to get the U it's 2nd mode 7.

b. The medicine does stabilize after 36
days, at 140 mg.


-This represents the first week of medication.


-This is where it stabilizes. Hurray! :D



- This is the graphed data, as you can see, it stabilizes

Question the second:

A small forest pf 4000 trees is under a new forestry plan. Each year 20% of the trees will be harvested and 1000 new trees are planted.

a. Will the forest ever disappear?
b. Will the forest size ever stabilize? If so, how many years and with how many trees?

Answers:

-.80 is what you multiply by the first term, as 20% is being cut down, leaving the other 80% in tact. 1000 is what we'd add on to what we're left with after the first year. 4000 is what we started with, as it's stated in the question.


-See the difference? The "ipart" means that we won't have answers with decimals. We can't have part of a tree.

a. As long as they're replanting, the forest won't disappear. In fact, the amount of trees inceases before it stabilizes.

b. The forest size does stabilize at 27 years with 4996






This afternoon:

We worked in our textbooks this afternoon, starting on page 264. I'll go over the first three questions, as I think that's where the majority of the class was by the end of class. I could be wrong, but too bad. :P

Une:
1. Use a calculator to generate the following sequences and find the indicated term.
a) 5, 7, 9, 11 ...; the 14th term

nMin=0
u(n)=u(first term)(n-1)+2 (The sequence goes up by 2)
u(nMin)={5} (Our sequence starts at 5)


-To get the 14th term you go to the home screen, u 2nd function, 7. Then (14) ... You could do (1, 14) ... but the first way is faster. Oh yes, and the answer is 31.



b) 5, 15, 45, 135, ...; the 12th term


nMin=0
u(n)=u(first term)(n-1)*3(multiplying by 3)
u(nMin)={5}(Again, our sequence begins at 5)


- The answer is 885735. To get the answer you do the same thing you'd do for the first problem, except instead of 14 it'd be 12. I'll spare you the play by play from here on. ;)



c)3.7, 2.3, 0.9, -.5, ...; the 15th term


nMin=0
u(n)=u(n-1)-1.4
u(nMin)={3.7}


- The answer is -15.9



d) 2.4, 3.6, 5.4, 8.1, ...; the 9th term


nMin=0
u(n)=u(n-1)*1.5
u(nMin)={2.4}


- The answer is 61.51



e) 12, 3, .75, .1875, ...; the 10th term


nMin=0
u(n)=u(n-1)/4
u(nMin)={12}


- The answer is .000046


Deux:
A salesperson at a car dealership receives a base salary of $275 per week, plus $50 for every car sold. Each week, the salesperson sells one more car then the previous week. In his first week, he sold one car.



a. Create a graph to show the salesperson's earnings each week.
b. Use your graph to determine the week in which earnings will exeed $1000, if the pattern continues.


-This table is simple enough, all you do is add the base salary to the money made for each car sold. Week 15 is when more than $1000 is made.

c. solve the problem using another method.

nMin=0
u(n)=u(n-1)+50
u(nMin)={275}


-Quit to the home screen.
u(1,15)
{325 375 425 475 525 575 625 675 725 775 825 875 925 975 1025}



Trois:
A colony contains 100 ants at the start of May. Suppose the population doubles every month and no insects die.
a. How many insects are there at the beginning of July?
I did this in my head. May-July is a span of 2 months. There will be 200 at the beginning of June, and therefore 400 at the beginning of July.
b. How many insects are there at the beginning of September?
Again, did this in my head. July-September is again, a span of two months. We had 400 insects at the beginning of July, therefore by August we would have 800. This means that by the beginning of September we'd have 1600 insects.
If you wanted to, you could do 100x24. That would give us 1600 as well.

Two terms to keep in mind are:
Arithmetic Sequence:
There is a common difference between terms. For example; 3.7, 2.3, .9, -.5. The common difference is -1.4 as you're taking -1.4 from the one term to get the next, or adding 1.4 to get one term from the previous.
Geometric Sequence:
There is a common ratio between terms.
For example; 2, 4, 8, 16. The common ratio is 2, as you're multiplying by 2 to get the next term, or dividing by 2 to get the previous term.

And finally:
Our homework is finishing up Checking Your Skills on pages 264 through 265.

The scribe for tomorrow is: Mr. M. Sherifsucker!










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Tuesday, May 30, 2006 

Scribe Post - 05/30/06

Sequences
The Scribe Post - May 30, 2006

- We only have one class today, since Caitlin wasn't in our class, Stephanie decided that I should go instead but anyways. Here is what we did.......


Mr. K directly putted up 4 Sequences while the announcers are speaking. He asked.

Find the next three Terms.....


4, 7, 10, 13, ?, ?, ?
3, 6, 12, 24, ?, ?, ?
32, 16, 8, 4, ?, ?, ?
2, 3, 5, 8,12, ?, ?, ?
The Answers.....

4, 7, 10, 13, 16, 19, 22
3, 6, 12, 24, 48, 96, 192
32, 16, 8, 4, 2, 1, 1/2
2, 3, 5, 8, 12, 17, 23, 30
How did we find out?

It is actually easy to find out, what is the next number of these values, IF YOU HAVE COMMON SENSE; we're not actually talking common sense in here, but I would like to say we're talking about math and here how it goes.

We see in 4, 7, 10, 13, 16, 19, 22 this values have all the same differences, and the difference between this numbers is THREE ( 3 ). It is easy to see that 3 is the difference, there is also another way to do it.


One of the way we can solve the number is by 3 (The Differences) multiply by the numbers right up the value and add 1.




The right equation for this Sequence is tn = n3 + 1 This is the RULE.


If you would to find out what is the next term without finding the TERM BEFORE THAT, this is what we called the RECURSIVE DEFINITION. The other definition for that is ..... Do the same thing to get the next answer is the ReCursive Definition.


For EXAMPLE, if you would like to find the 7th term without finding out what is the 5th, and the 6th terms. The equation for that is........


We need to put in the First term then we add the Number of Columns then minus 1 from it since there is only 6 in between and in between that's the differences and the Difference for this value is 3.


TO SOLVE THE EQUATION BY FINDING THE TERM 7...this is how is goes.

T7 = 4 + (7-1) 3
--- = 4 + (6) 3
--- = 4 + 18
--- = 22


( The SEQUENCE THAT HAVE A COMMON DIFFRENCE IS WHAT WE CALLED ARITHMETIC, and one example is right above*)


FOR THE NEXT SEQUENCE.


3, 6, 12, 24, 48, 96, 192

(This sequence is what we called GEOMETRIC, by having COMMON RATIOS in all the values*)


This is how we solve the equation, by finding the term without looking for the number before that value.







Tn = the term you are looking for
T1 = the first term
r = the ratio
n-1= the number of columns minus 1


T7 = 3 (2)^7-1
T7 = 3 (2)^6
T7 = 3 (64)
T7 = 192


The last two Sequence are also GEOMETRIC, because they don't have a common differences.


Then later on that period Mr. K told us about the Implicit and Explicit.


Implicit is a cuirrent inside suggestion

Explicit is a high current outside.


There is one more way we can find it, the LAST ONE IS IN OUR CALCULATOR. First we need to check if our mode is in SEQUENCE, so CLICK MODE, Go to the 4th row then choose SEQ.Then we enter our information in our Y= screen, and then we need to check our window for a better view.


To generate:
Go Second Quit
Enter in your calculator:
u by 2nd, 7 the bracets (First Term, and the up to what term would you like to see)
Then we store it in our L1.


MR.K also SAID THAT ADD UP SEQUENCES IS CALLED SERIES.


I don't know if I'm doing good at this scribe post this time, damn I shouldn't have wait for a day to fix this up now my memory is kind of loosing it.




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Monday, May 29, 2006 

SCRIBE: Goodbye Periodic Functions, Hello Sequences!

Hi everyone. I have good news and more good news. Today we finished Periodic Functions. Good?! I don't know about you guys, but I'm a bit relieved -_-' And finally, we're on our last unit: Sequences. So after this unit, we have the exam, aaand then we're DONE!! Hope that was a bit motivating...

Before I talk about what we did in class today, I just wanted to remind everyone that the Wiki assignment has been extended to the end of the week: Friday, June 2nd (midnight at the latest). Mr. K just wanted to say that you shouldn't be afraid to edit someone else's work because any previous work is saved in the history..which shouldn't be hard for Mr. K to find. But to make sure he knows whose work is whose, then you should remember to sign in with your name. Another reminder: For the constructive modification, you cannot edit your own work!!

So today we had two classes. For third period, We did a Group Problem.

Group Problem - Periodic Functions 5 // Tsunami
A tsunami (commonly called a "tidal wave" because its effect is like a rapid change in tide) is a fast-moving ocean wave caused by an underwater earthquake. The water first goes down from its normal level, then rises an equal distance above its normal level, and finally returns to its normal level. The period is about 15 minutes. Suppose that a tsunami with an amplitude of 10 metres approaches the pier at Honolulu, where the normal depth of the water is 9 metres.

Before continuing with the question, it's best to sketch a diagram/graph:


From the sketch and information from the question you can figure out the equation:

A = 10, B = 2π/15, C = 0, and D = +9

And the equation would then be: y = -10 sin ( [ 2π/15 ] x ) + 9

* Parameter A is NEGATIVE because "the water first goes down from its normal level.." which can also be seen from the sketch.

(a) Assuming that the depth of the water varies sinusoidally with time as the tsunami passes, predict the depth of the water at the following times after the tsunami first reaches the pier.

This can be found by plugging in the equation into y= , graphing the equation, and to find each time you would go to: 2nd CALC > [value] and plug in the time asked. (I also rounded to two decimal places).

i. 2 minutes ii. 4 minutes iii. 12 minutes
1.57m -0.95m 18.51m



(b) According to your model, what will be the minimum depth of the water? How do you interpret this answer in terms of what will happen in the real world?

To find the minimum, you'd go to 2nd CALC > [minimum], left-bound, right-bound, guess anywhere and from there you should get the value: -1. So this means that the "lowest" depth of the water would have no actual wave. Technically /mathematically the height is -1 but you can't go beyond zero because there's no where to go. All that's left in reality is the ocean bed.

(c) The "wavelength" of a wave is the distance a crest of the wave travels in one period. It is also equal to the distance between two adjacent crests. If a tsunami travels at 800 kilometres per hour, what is its wavelength?

800km/hr = 800km/60min ; you can figure it out like that.

OR

800km/60min = x km/15min

The answer is : x = 200km



After the group problem we had a tiny introduction into Sequences:

The numbers in grey are the 'nth' terms. The blue two's under the arrows is the common difference between the black and red numbers in the sequence. Surprisingly there is an equation that you can extract from the sequence. You can imagine that x=n , m(slope)=2, b(intercept)=4, and y=value. And so:

y = mx+b equals y = 2n+4

*Not every sequence has a common difference.

Here are the three other sequences that we didn't get a chance to look at. I'm not sure if my answers are right, but I'll post them anyway:

b) 1, 2, 4, 8, 16, 32

c) 2, 3, 5, 8, 12, 17, 23, 30, 38

d) 4, 2, 1, 1/2, 1/4, 1/8, 1/16

For fifth period, we did our test on Periodic Functions. So that means no homework tonight, just Wiki if you haven't already finished it.

And of course, before I forget: tomorrow's scribe will be C-A-I-T-L-I-N =)




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Ma blogging and ma unblogging(periodic functions)

This unit so far has given me a big headache, the problem with this unit is that it makes sense when the teacher is doing a question but later when i am trying it its hard to find a way to solve these kind of questions specially the problem ones. Almost every one so far is saying that the unit went easily and i agree with them but still i have difficulties in understanding the word problems. I hope someone would listen to me and agree to postpone the test.

CHEERRSSSS




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B.O.B. - Periodic Functions

This unit was fairly easy. From converting degrees to radians and vice versa, to knowing the sinusoidal function. Like always, problem solving is the toughest part in this unit since it can be very confusing if you don't read the question carefully. What's also sort of confusing is figuring out the horizontal shift on the sinusoidal axis. Hopefully I'll do okay on the test. That is, unless I somehow blank out. But I'm sure everyone else will do okay on the test.




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BOB

almost forgot! ha, well test is already tomorrow, and so far, this units been okay...it was easy in the beginning. like converting degrees into radians and radians into degrees, i understood that. i also understood how to look at a graph and find the y= equation from it. also i understood how to graph with the given equation. i thought that was okay, then when it came to the applying it to problems. i understood them when we did them in class, but trying the homework was kinda tricky. kind of different, but i did enjoyin figuiring out the problems in our group activities. welps, its late...good lucky everyone!

-CORRIE




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BOB

Aight, all i have really got to say about this unit is that its basically all calculator functions. I found the unit pretty easy. I aslo found that the unit went by super quick. Thats probably because it did. The only problem that I think i may have a slight problem with drawing the graphs. I understand it pretty well but some times i make stupid mistakes. If I avoid those mistakes I think I will do pretty well on the test.

Cya,




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Sunday, May 28, 2006 

3D Tic-Tac-Sunday!



Better late than never. ;-)

I missed the Sunday Funday post last week and I really don't want to make that a habit so this week I'm posting two versions of this week's game ... 3D Tic-Tac-Toe. Here are the instructions for the game pictured on the left (may take a little while to load) ...

The object of 3D Tic Tac Toe is to get four in a row horizontally, vertically or diagonally on one plane or across all four planes.

You are Red, the Computer is Blue. In the first game, you go first. In subsequent games, the loser goes first. If the game ends in a tie, whoever went first will do so again.

The moves of the game are notated on the right. Moves that threaten a win are noted with an *. The Computer plays a strong, but not unbeatable game. Good luck!!!


These are the instructions for the game on the right ...

This game is basically a 3D tic-tac-toe, except that the rules have been changed to keep the first player from winning all the time. Here you have to form two rows that meet at a right angle (in the shape of an "L").


Have Fun!!




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BOB

the periodic functions unit for me was half review and half learning. there were some things in the unit that i already knew how to do because i've done it before in other classes. and the new things that i learned were mostly calculator functions. hopefully i do well on this test. make up for my last test. goodluck to everyone else tomorrow.




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In Depth Analysis of the Friday class. Applied Math 40s

I hope u guys didn't have any problem with thinkfree online beta. also if you can't get the files on there, after u go to NEW DOCS. try the search box. on top of the screen (right) side. search as: contruction1, contruction2, Contruction 3, Contruction4. and Friday.doc. or scroll down the page and look for my name, as Mr. K say: theres many ways to skin a cat, but don't skin a cat, cau'z thats cruel. back to business.

ok. first you have to calculate the Area of driveway, you can do that as follows:

Option 1: devide into triangle & trapezoid

Area 1 of driveway ( Rectangle)

Area: 98' by 18'----- 98*18=1692ft^2

Area 2 of driveway ( trapezoid)

((16+21)4)/2=74ft^2





OR you can devide into rectangles and triangle:

Rectangles:
A: 73'*18'= 1314'

B: 5'*18'= 90'

C: 22'*16'= 352'

Triangles:

D: (4*5)/2=10'

Total Area of driveway= 1766ft^2

Concrete & Gravel ( first method)

Then use the area to calculate volumes & cost. change inches into ft. 12''=1 ft

Example : 12''---- 12/12= 1 ft.

For first method as follows: For second method as follows:

Gravel : 1 ft Limestone: 1.00ft Lime stone: 1.00ft

concrete:0 .33ft sand: 0.29ft sand: 0.29ft

concrete: 1.33ft

paving stone: 1.50ft

To calculate volume changed ft. into cubic ft. then cubic ft. into cubic yard to calculate cost.

First method volumes & Cost: Gravel= 1766*(12/12)= 1766ft^3/27 = 65.41 yd^3

Concrete=1766(4/12)= 588.67ft^3/27= 21.80 yd^3

Volume excavation= gravel + concrete= -------------------------------------= 87.21 yd^3

Gravel = 65.41 yd^3 ( $12.00/yd^3) = $784.92

Concrete= 21.80 yd^3 ($130.00/yd^3) = $2834.00

Excavation= 87.21yd^3($60.00/yd^3) = $5232.60

Subtotal:----------------------------------------------------------------------= $8851.52

Taxes: concrete and gravel= $2834+$784.92= $3618.92*14% =$ 506.65

Excavation $5332.60 * 7% ----------------------------------------- =$ 366.28

* Cu'z, only GST is calculated for excavation

Total Taxes= $ 872.93

Total cost= subtotal plus taxes------------------ Total Cost= $ 9724.45

Paving Stones (Second Method)

Second method Volume & Cost: Paving Stone = 1766*(2.5/12)= 367.92ft^3/27= 13.62 yd^3

Lime Stone = 1766*(12/12)= 1766ft^3/27 = 65.41 yd^3

Sand= 1766*(3.5/12)= 515.08ft^3/27 = 19.08 yd^3

Volume Excavation= volume lime stone + volume paving stone + volume sand = 98.11 yd^3

Paving Stone= 13.62 yd^3 change into ft^2 = $4730.85

Lime Stone= 65.41 yd^3($15.00/yd^3) = $981.15

sand= 19.08 yd^3 ($14.00/yd^3) = $76.32

Excavation= 98.11 yd^3($60.00/yd^3) = $5886.60

Subtotal= $11314.92

Taxes: (cost sand + lime stone + paving stone) * 14%--------------------------- =$ 759.97

(Excavation * 7%)------------------------------------------------------ = $412.06

Total Taxes= $1172.03

Total cost= subtotal plus taxes Total Cost= $12486.94

*** Therefore paving stones cost more by---------------------- $1286.94-9724.45= $2762.50

That's All Folks. have good weekend next scribe i will pick in class on monday.




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NOTICE

i have tried to upload the files, it keep saying error uploading. which sucks. but no problem. i saved the file on thinkfree. which is the best ever online office on earth. it works. really it works. so go to www.thinkfree.com and click on NEW DOCS. then the files class is all there. i did spreedsheet part. click on Contruction1, Contructions2, Contruction3, Contruction4. and if you need to Get the questions part Click on Friday.doc, which will explain what went on friday class. i will try to do on paper explaining how all the method we used to get those answers. brb.boys and girls, or should i say ladies and gentlemen.




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Saturday, May 27, 2006 

Friday's Post DMCI Applied Math 40s.

I am trying here to post the scribe. but i am having trouble uploading files. from paint and microsoft word, it is a long scribe, so it will take a while.




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Friday, May 26, 2006 

Bob on Periodic Functions

On this chapter. we worked with formulas to solve problems. the main formulas that we have to know is DABC. which Y=AsinB(X-C)+D. the problems where easy to at the beginning as we only had to change degrees in radians, and vise versa. but when we get to solving real problems using DABC. things where little hard. but hope the test won't be that hard. thats all folks.




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Thursday, May 25, 2006 

B.O.B(Ivy 05/25/06)

The Periodic Function Unit (B.O.B)

hmmm....This unit for me at first was easy..but when we got to this problems now I'm in trouble...I didn't find A,B,and D hard to get but C is. About C first I was cool with it, by doing the sketch by looking at the formula y=AsinB(x-C)+D and by converting formula to sketch it was cool. I also like doing the converting Radians to Degrees and the other way. I really do need to review, like whoa...for the last units we have, I never did study and I ended up failing the tests, and I really do feel bad about it, one thing is I also keep hustling, and just let my studies go for a passing mark, oh whatever.. but yeah, need to focus this is the last time I'm going to study math.. ahehehehe. Oh. Yeah..I also feel bad about missing the last class this afternoon




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Scribe: Periodic Functions Problems

Today we had 2 periods and we were supposed to have a pre-test and we had but not the actual pre-test instead we answered 2 problems as our pre-test. So whoever missed the class here are your pre-test.

We did two problems and here is the first one:

LIFT OFF!!!

When a spaceship is fired into orbit from a site such as Cape Kennedy (which is not on the EQUATOR), it goes into an orbit that takes it alternately NORTH and SOUTH of the equator. Its DISTANCE FROM THE EQUATOR is approximately a SINUSOIDAL FUNCTION of time. Suppose that a spaceship is fired into orbit from Cape Kennedy. 10 min. after it leaves the Cape, it reaches its farthest distance NORTH of the equtor, 4000 km. HALF A CYCLE later it reaches its farthest distance SOUTH of the equator (on the other side of the Earth) also 4000 km. The spaceship COMPLETES an orbit once EVERY 90 min. (completes a cirlce around the world).

Let 'y' Be the number of kilometres the spaceship is north of the equator (consider distance south of the equator to be negative). Let 't' be the number of minutes that have elapsed since liftoff.

Tip: When trying to answer this kind of question try to write what you are given.

- 4000 km. (max and min) NORTH and SOUTH (this is the "A" )
- 10 min., from CAPE, it reached the farthest NORTH
- 90 min. to COMPLETE A CYCLE (this is the "B")
- HALF A CYCLE (90/2 = 45) it reached the farthest SOUTH (it took 45 min. to reach SOUTH from NORTH)
- EQUATOR as the SINUSOIDAL AXIS (this is the "D")


A.) Sketch a complete cycle of the graph of y versus t, and write the sinusoidal equation expressing y in terms of t.

This is how I got the C: (refer to the graph)
In 10 min. the spaceship reach the farthest NORTH and 45 min. to reach the farthest SOUTH from NORTH.
10 + 45 = 55 (half circle)
the middle number between 55 and 10 is 32.5
10 + 55 = 65 then 65/2 = 32.5 (quarter Circle)
now since I know the quarter circle
32.5 - 10 = 22.5
10 - 22.5 = -12.5 (C)

A = 4000 (max and min range)
B = 2(3.14)/90 (Period) (3.14 = pie)
C = x - -12.5 (Starting point)
D = 0 (Sinusoidal axis)

y = 4000sin((2*3.14/90)(x - -12.5))+0

window: Xmin=-20 Xmax=100 Xscl=10 Ymin=-5000 Ymax=5000 Yscl=1000















B.)Use your equation to predict the distance of the spaceship from the equator when:

Since we didnt talked about this in class here is what I came up.

Solution:(use calculator Ti-83)
Put in the equation "y = 4000sin((2*3.14/90)(x - -12.5))+0)" on "Y="; then "GRAPH"; then "2nd FUNCTION" "TRACE"(CALC); then choose "VALUE"; then plug in whatever the number is asked.
window: Xmin=-20 Xmax=100 Xscl=10 Ymin=-5000 Ymax=5000 Yscl=1000

i.) 25 min
ans.= 2000 km.

ii.) 41 min.
ans.= -2236.77 km. ( - = going south)

iii.) 24 hours. (adjust Xmax window setting = 1500)
24 x 60(min.) = 1440 min.
ans. 3064.17 km.

C.) Calculate the distance Cape Kennedy is from the Equator by solving for Y when t = 0

Solution: same as what I did in B.
Plug in 8 as your x value
ans. 3064.17km


Here is the second problem now:

EAST COAST HARBOUR

On May 3rd, the DEPTH of the water in an east coast harbour will vary over time as described by the equation:
"y = 2.3sin0.506(x+3.1)+2.8"
where x represents the time (hours), and y represents the depth (metres) of the water.
The time at x = 0 is midnight of May 2nd, and x = 12 is noon of May 3rd.

Since we are given the equation already:
A.) What are the minimum and maximum depths of water in the harbor?
D + A = MAX
2.8 + 2.3 = 5.1

D - A = MIN
2.8 - 2.3 = .5

B.)What is the average depth of the water in the harbour?

- you can refer to the SINUSODAL AXIS = 2.8
- or use this way (max + min)/2 (5.1 + .5)/2 = 2.8

C.)How much time is there between two high tides

Write the equation "y = 2.3sin0.506(x+3.1)+2.8" in "Y="; then "GRAPH"; then "2nd FUNCTION" "TRACE"(CALC); then choose "MAXIMUM";
Then "left bound and right bound and guess"; The X coordinate is the time
x = 12.4217 since there are no decimal hour: .42 X 60 = 25

window settings: Xmin=-2 Xmax=(30 or 50) Xscl=6 Ymin=-1 Ymax=6 Yscl=6

ans. 12:25 p.m. 12 hours and 25 minutes (from 0 hours)
















D.) What is the depth of the water at 8:00 AM?

With the same equation; press "GRAPH"; then "2nd FUNCTION" "TRACE"(CALC); then choose "VALUE"; then plug in 8 for x
ans. 1.377















E.) At what time in the afternoon is the water at it's lowest? How deep is the water at this time? (This is the time that little Ziggi likes to go kayaking in the harbour because there are no large boats moving at this time.)

With the same equation; put the .5 (lowest) in Y2(or in any Y); then "GRAPH"; then "2nd FUNCTION" "TRACE"(CALC); then choose "INTERSECTION"; it'll will give you the x coordinate which is
ans. 6.21 ; .21 X 60 = 6:12 am, 6:12 pm. etc...

D.)A commercial boat requires at least three metres of water to more around the harbour. Describe how you would determine when it is safe for the boat to operate in the harbor.

With the same equation; put a the 3 (required level) in Y2(or in any Y); then "GRAPH"; then "2nd FUNCTION" "TRACE"(CALC); then choose "INTERSECTION"; it'll give you Y coordinates. In able for the boat to operate he should ONLY operate in hours in the blue line.

















So that all folks. Sorry my scribe is too long. Anyways gotta sleep and Mohamed you are the scribe for tommorow.




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Wednesday, May 24, 2006 

B-O-B ; Periodic Functions

So here's my blogging on blogging for Periodic Functions. I remember this unit from last semester but I never did like it. Although I'm somewhat comfortable with the concept..I don't think I feel too strongly about the word problems. I can figure out parameter D, A, and B. But C always seems to stump me at times. Figuring out parameter C is probably my only problem right now. Hopefully someone can show me before Friday?!




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SCRIBE

Today we only have one period, the unit is done and we are going to have a pre-test on thursday and our unit test would be on friday ... but before anything else here is what happened on our class today.. We did couple of problems regarding on applying periodic functions on a data collected..

*Problem Solving
---> Based on the data collected over the last 30 years, the average high temperature for regina are shown.

Part I
A. sketch the graph
B. find the equation for the data by approximating values for each parameter
A,B,C and D
C. Use your function to calculate the temp. on
Apr. 15th (using calculator)
Part II.
A. Find the equation of the data(using table of
values)
B. Use your equation to calculate the temp. on
Apr. 15th
C. Why is the value different from the one
above?












Part I
A.





Equation: Y = 18.65 sin[2ii/ 12(x-3.75)] +7.65
B.) A = 18.65
.....B = 2ii / 12
.....C = 3.75
.....D = 7.65

C.) Using the 2nd calc.-->VALUE button in the Ti83 calculator we found the temp on April 15(4.5) is 14.78 degrees celsius.

Part II

A. using these values from the list: press STAT>CALC ^ SINREG enter 2nd L1, 2nd L2,Y2, press VARS> Y-VARS 1 Function enter Y2
Equation: 19.6 sin (.48X - 1.82) + 7.33
B. Using 2nd CAlC Value X= 4.5 the temp. for Aprill 15 == 14.78 degrees celsius








Group Work Problem

*Cottage owner Brad measures the depth of the water in his dock 10 times during the course of one day. The water keeps changing because of the tides. The times are based on the 24-hour clock, and written in decimal form (i.e 5:45AM is 5.75 hours) The charts shows the times and the water depths A. Sketch a rough graph and write the appropriate equation for the values indicated.B. Use the calculator to find the equation of the values.
......Store the TIME and DEPTH values in L1 and L2. Then press STAT > CALC ^ SinReg Enter...SinReg L1,L2,Y2...then the equation is .................Y = 0.814 sin(X - 0.862) + 3.25

C. What is the depth of the water at 23:00 hours(11pm)?
......Graph the equation in the calculator then press 2nd CALC, Value 0f X = 23 Enter
......depth at 23 hours is 3.125 meters
D. What is the median depth of the water?
16 m + 5.75 m / 2 = 10 .88m



next scribe is jesse




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Tuesday, May 23, 2006 

* unit's almost done.


Well, I have just looked over everything that I have to include into this scribe post for today and well....theres alot! So bare with me, sorry if its late, if i missed anything, or did anything wrong, cause i'm still learning myself. First off, we started the week off with 2 classes.

In the first class we went over a equation that we worked on last week.


So the equation was
y=-3sin4(sinx-π/8)-2
On the graph here I have shown the
y=sinx.

So when drawing out this equation you'd want to first look at
D and A of our equation. D and A of our y=AsinB(x-C)+D formula. Thats shown in red, above.

From our equation we see that
D is -2, and D is our sinusoidal axis, that is were it shows if the graph shifts either up or down x units. -2 means, that our graph shifts down 2 units. ( as shown above).

Now we look at our
A, which is our amplitude , it shows us how far away the max. and the min. are from the sinusoidal axis. From our equation our A is -3. With the negative (-) sign in front of the 3, this means that our graph starts from the minimum. So you go down 3 units. That shows the distance from the minimum to the axis.

The next thing we look at is
B, and in our equation this is 4 . So, now we need to look for the period, to do this we taken the formula 2π/B.
period=2π/B
=2π/4
=π/2 <- makes one wave.

So now, that we've got D, A, and B down, C is left. C is just where we have to shift our graph left or right on the sinusoidal axis. It shows that our C is -π/8. So we take our graph and shift it right π/8 units.

So thats, that. After that he had assigned us two problems to do at the spot. One, we were given the equation and we had to graph that. Then the second problem, we had a graph, which we had to find the functions, A, B, C, and D. And also, write the equation out for that graph. I'm not going to bother putting them up on here, because theres so much more I have to do, but if anyone does want to see the two equations, just leave a comment or something to let me know, and i'll make sure to post them up.

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2ND CLASS

So if anyone missed class this afternoon, we wrote some more in our dictionaries. Here it is.

Applications of Periodic - an example

A ferris wheel has a radius of 25 meters. It;s center is 26 meters above the ground. it rotates once every 50 seconds. SUppose you get on at the bottom at t=0.

a) graph your height above the ground during the first 2 minutes of the ride.
b) write an equation for the graph, showing heigh as a function of time.
c) find how high above the ground you are at:
i) 10 seconds ii) 20 seconds iiii) 60 seconds
d) find two times when you are 50 meters above the ground.

THE SOLUTION:
A) Draw a picture of the situation and then sketch the graph. It is also helpful to figure out the value of each parameter A, B, C, D.



1 m. above ground (min. height)

diameter is 50 m.

max. height is 51 m.






y = A sin B(x-C)+D
A=51-26=25 (radius)
B= 2π/50 [2π/period]
C=12.5 [starting point moved right 12.5, see graph]
D= 26
y=25sin[2π/50(x-12.5)]+2


The he just told us, if we wanted to find out for 2 minutes. We would just continue on with our graph, with the pattern. Yeah gets?!

------------------------------------------------------------------------------------------------

Then after our dictionaries, we got into groups and started to do an activity.

Pebble Round and Round
As tiy stio tiyr car at a traffic light, a pebble becomes wedged between the tire treads. As you start to drive, the pebble remains stuck in the tire tread, and the distance of the pebble from the pavement varies sinusoidally with the distance you drive. The period is, of course, the circumference of the wheel, and tghe wheel has a diameter of 24 inches.

(a) Sketch a graph of this function.





















(b) Write the sinusoidal equation of this function.

y=12sin(2π/75.398)(x-12.50)+26

How we got this.
Circumference = 2πr
=75.398 or 24π
D= 12
A=12
B=2π/period
=2π/75.398
C=18.8495 or 6π

(c) Caculate the distance the pebble is from the pavement after you have driven 15 inches; 100 inches; 200 inches.

So you enter the equation y=12sin((2π/75.398)(x-18.8495))+12 into [y=] in your calculator.
Then you would now have to go to your homescreen so [2nd][MODE]. From there hit the [VARS] -> Y-VARS [Functions][1: Y1]

Once Y1 is on your homescreen, put in brackets (15)
Y1(15) = -9.30
Then, with (100)
Y1(100)=-1.04
Then, (200)
but it has to be (200)(12), dont know why?! Got lost there!

EEK! Kay, I don't know if im doing this last part right, Mr. K was in such a rush I couldnt really get it, something like that? Not sure. Sorry!

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So, that was pretty much our day back from the long, long weekend! I hope you guys get this, its not the best, not 'hall of fame' material. It's okay, I just want this over and done with. So, PATRICK, your next scribe okay, make it a good one! =)

- CORRIE






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