AP Calculus AB

The Adventure Continues ...

2006-06-29T15:49:00.000-05:00

Our adventures in blogging continue....

Watch for 3 new blogs going live September 6, 2006 ...

Pre-Cal 30S (Fall '06) (Grade 11)
Pre-Cal 40S (Fall '06) (Grade 12)
AP Calculus AB 2006-07 (Grade 12)

So Long ...

2006-06-28T18:15:00.000-05:00

We had our graduation exercises today. A gentle push into the world for all of you. I hope you're leaving with the keys to your future in your hand.

I'm so glad we've had this time together,

Just to have a laugh or learn some math,

Seems we've just got started and before you know it,

Comes the time we have to say, "So Long!"

So long everybody! Watch this space in the fall for pointers to new blogs for each of my classes.

Farewell, Auf Wiedersehen, Adieu, and all those good bye things. ;-)

Student Survey Results

2006-06-28T18:02:00.000-05:00

The exam is long over and we did a little survey in class. The results are below; 4 students participated. Here are the result. Please share your thoughts by commenting (anonymously if you wish) below .....

How prepared were you to write this exam? (Average score out of 100)

75%

How much effort did you put into preparing for this exam? (Average score out of 100)

67.5%

How good a job did your teacher do preparing you for this exam? (Average score out of 100)

92.5%

Did you have enough preparation using your calculator?

Yes 100% No 0%

Did you have enough preparation without using your calculator?

Yes 50% No 25% Middle 25%

Was I too hard or too easy on you??

Semester 1 was easy. Semester 2 was really hard for me. It killed me actually.
Just right.
Too easy.
Too hard.

What was the best learning experience you had in this class?

Blogging (2)
Mini exams (2)
Group Work
Wiki
Pre-Tests
Teaching and explanations were very clear and easy to understand

What was the worst learning experience you had in this class?

Homework (2)
None
Blogging On Blogging before tests

What suggestions can you share for next year?

More wiki assignments from past exams.
More exam-like long answer questions in class.
Have students doing boardwork in class.
The blog didn;t help me that much.

It's interesting to compare the items that were considered both the worst and best learning experiences. Also, take a look at the list of worst learning experiences compared to suggestions for next year. Help me do a better job next year by commenting on what you see here ....

The Places You'll Go!

2006-06-09T19:09:00.000-05:00

You have brains in your head.
You have feet in your shoes.
You can steer yourself
Any direction you choose. --Dr. Suess, Places You'll Go

It's almost graduation--what direction are you choosing?

Passing it forward--

2006-05-29T20:09:00.000-05:00

Some 43 years ago, when I was about your age, my father who used to travel a lot, wrote me a letter enclosing a small article clipped from a magazine. In that letter, he wished for me a life of purpose and joy. He shared with me the clipping which described such a life and quoted George Bernard Shaw whose words you can read above.

I carried that article with me for years; unfortunately, somewhere in the many moves, it was lost. But not the thought and the power of those words. They have become a part of who I am. I know that my career as a teacher, and the mentoring I do now for teachers seeking National Board Certification are but "drops in the ocean" in this often violent, globalized world of ours but my life has been filled with joy, happiness and love.

I'm sure that my father writing and sharing those thoughts adds to their meaning for me, especially now that Alzheimer's prevents him from recalling what occurred. But his belief in me, in mankind in general, shaped my world. I'm passing that forward --my belief in you, my belief in mankind, and my wish that you find the real joy in life!

Wishing you success ahead!

2006-05-26T18:58:00.000-05:00

I am thinking that these days each of you, your family, and your friends are looking to your future and wishing you success. I'd like to do that too!! To wish for you all that Emerson describes-- Would this meaning of success be one you'd be willing to adopt?

Two roads diverged in a wood---

2006-05-18T19:25:00.000-05:00

THE ROAD NOT TAKEN by Robert Frost

Two roads diverged in a yellow wood,
And sorry I could not travel both
And be one traveler, long I stood
And looked down one as far as I could
To where it bent in the undergrowth;
Then took the other, as just as fair,
And having perhaps the better claim,
Because it was grassy and wanted wear;
Though as for that the passing there
Had worn them really about the same,
And both that morning equally lay
In leaves no step had trodden black.
Oh, I kept the first for another day!
Yet knowing how way leads on to way,
I doubted if I should ever come back.
I shall be telling this with a sigh
Somewhere ages and ages hence:
Two roads diverged in a wood, and I-
I took the one less traveled by,
And that has made all the difference.

This has always been one of my favorites as I've found myself faced with those "two roads diverged" so many times in my life. I'm thinking you may now be viewing "two roads diverged" now as you are about to graduate and I know you will be many other times in your lives.

My choosing ( It wasn't necessarily an easy choice.) Earlham College was one of those times I "took the road less traveled" and it has made "all the difference" in my life. My years at Earlham have had a profound impact on who I am today, how I see the world, and what I believe.

Have you taken/will you be taking a road "less traveled by" and has it/will it made/make "all the difference"? Or does it matter?

Congratulations-- What's next?

2006-05-09T07:24:00.000-05:00

Congratulations on completing your AP Calculus exam!! That is quite an accomplishment!!! When you look back on the process, did you surprise yourself? Did you take some time to celebrate?

After celebrating, was your first thought "What's next?"

What will be the next immediate challenge? Isn't that the incredible part of living? That once you complete one challenge, another awaits. More hard work, more frustration, more hard work, always something.

Below is one of my favorite quotes (when I was teaching, we began each class with a quote; often ones that my students suggested; we felt each one captured something essential about living and life) ; Ara shared some of hers and since I've lived my life by a quote as you'll find in a future posting, I'll continue with this one here:

"Every day you may make progress. Every step may be fruitful. Yet there will stretch out before you an ever-lengthening, ever-ascending, ever-improving path. You know you will never get to the end of the journey. But this, so far from discouraging, only adds to the joy and glory of the climb." --Sir Winston Churchill

Your climb has just begun! Your AP exam was one great step!! What's next?

Kakuro Sunday

2006-05-07T11:12:00.000-05:00

'A Kakuro consists of a playing area of filled and empty cells similar to a crossword puzzle. Some black cells contain a diagonal slash from top left to bottom right with numbers in them, called “the clues”. A number in the top right corner relates to an “across” clue and one in the bottom left a “down” clue.

The object of a Kakuro is to insert digits from 1-9 into the white cells to total the clue associated with it. However no digit can be duplicated in an entry. For example the total 6 you could have 1 & 5, 2 & 4 but not 3 & 3. Sound simple? Be warned it gets hard and is as addictive as Sudoku.'

Click here for more Kakuros.

(Thanks again to Think Again!)

Running Up That Hill

2006-05-02T22:29:00.000-05:00

Take a look at this:

You're not the only ones using our blog to get ready for tomorrow. ;-)

If you're an AP Calculus student reading this now ... go to bed! You've got to be ready for tomorrow. Your brain is an organ; take good care of it. It needs sleep, a good breakfast tomorrow (like cheese, fruit, eggs and juice) and lots of water. Don't forget to do some exercise when they give you breaks so you can get the blood from your bottom into your brain. ;-)

Lastly, remember that luck has nothing to do with it. It's all about doing your best ... Learn Hard!

Answers to Mini-Exam #6

2006-05-02T09:45:00.000-05:00

(1) D
(2) D
(3) A
(4) B
(5) A

free response question

(a) Because g is the derivative of the function ƒ, ƒ will attain a relative minimum at ta point where g=0 and where g is negative to the left of that point and positive to the right of it. This occurs at x=6.

(b) Bcause g is the derivative of the function ƒ, ƒ will attain a relative amximum at a point where g=0 and where g is positive to the left of that point and negative to the right of it. This occurs at x=3.

(c) We are trying to find the area between the graph and the x-axis from x=-3 to x=6. From x=-3 to x=3, the region is a semicircle of radius 3, so the area is 9π/2.
From x=3 to x=6, the region is a semicircle of radius 3/2, so the area is 9π/8.
We substract the latter region from the former to obtain: (9π/2) - (9π/8) = (27π/8)

(d) Because ƒ''(x) = g'(x), we are looking for points where the derivative of g is zero. This occurs at the horizontal tangent lines at x=0, x=4.5, and x=7.5.

Just do it!!!!

2006-05-01T18:35:00.000-05:00

On the day that I was to take my "big" test, I was just about to leave for the testing center when I asked my husband to wish me luck. "No," he said, "I won't do that." I was crestfallen. I felt like I needed one last boost before the "big" one.

Then he continued, "You don't need luck. You're smart. You're prepared. You're good. I believe in you. Go out there and just do it!!!! I'll be here when you get back."

It's time for me to pass that forward to you! You don't need luck. You're smart. You're good. You're well prepared (thanks to your hard work and Mr. K). I believe in you. Go out there and just do it!!!!

Your second wind--

2006-04-30T18:20:00.000-05:00

Most people never run far enough on their first wind to find out they've got a second. Give your dreams all you've got and you'll be amazed at the energy that comes out of you. --William James

I wonder, would this apply to calculus too?

From ZZZZZ's to A's

2006-04-28T18:38:00.000-05:00

The bottom line: Teens need 9.25 hours of sleep per night

from Do Teens Get Enough Sleep?

In experiments done at Harvard Medical School and Trent University in Canada, students go through a battery of tests and then sleep various lengths of time to determine how sleep affects learning. What these tests show is that the brain consolidates and practices what is learned during the day after the students (or adults, for that matter) go to sleep. Parents always intuitively knew that sleep helped learning, but few knew that learning actually continues to take place while a person is asleep. That means sleep after a lesson is learned is as important as getting a good night's rest before a test or exam.
from Adolescents and Sleep

At the risk of sounding "mom-ish", have you taken this into consideration in your preparation for your upcoming test? I saw in a scribe that Mr. K had mentioned it!

Asking only because, when I was sleep deprived, I know I wasn't fully aware of how much more difficult problem solving and remembering was. I never fully realized how sleep deprivation changed my abilities and me until after I started getting adequate sleep.

Another factor in your preparation to be your very best for your test??

Your self portrait!

2006-04-26T18:46:00.000-05:00

How is your self portrait coming?

Visualizing excellence in calculus!

2006-04-24T19:48:00.000-05:00

"Almost all of the world-class athletes and other peak performers are visualizers. They see it; they feel it; they experience it before they actually do it. They began with the end in mind. You can do it in every area of your life. Before performance, a sales presentation, a difficult confrontation, or the daily challenge of meeting a goal, see it clearly, vividly, relentlessly, over and over again. Create an internal "comfort zone." Then, when you get into the situation, it isn't foreign." --Steven Covey

Are you an athlete? Do you visualize already? Maybe you've already thought about this before--

I wanted to share because I think that with your success on your mini exam (saw that in one of Sarah's comments! Congratulations to you all!) and all that you are sharing, and reading, and problem solving on this blog, you really can visualize excellence for May 3! You are doing everything the quote from Stephen Covey suggests! Now with visualization for May 3, you'll be in your comfort zone and on your way to a peak performance.

I had mentioned earlier in an earlier post that I had been faced with a major 3 hour assessment. I sense that I prepped in some of the same ways that you are. I "relentlessly" researched and reviewed all I could find that could help me with the six test questions. The organization provided generic test questions and scoring guides for each question. I practiced answering the question within the half hour framework. I practiced with the software that I would be using in the testing center. I visualized how each question might be phrased and how I would respond. I can't honestly say I was in a comfort zone when I entered the testing center, but I know that when I took a deep breath and began, all that I had visualized and practiced seemed to flow from my brain, through my fingers and into the testing software.

Do you think visualizing could be helpful to you too?

Ara's Blog Assignment

2006-04-24T10:15:00.000-05:00

An object in motion along the x-axis has velocity v(t) = (t + e^t) sin (t^2) for the interval 1, 3.

a. How many times is the object at rest?

Defining an object at rest means the slope is zero, that is, the same way as saying the derivative of the parent function is equal to zero. Therefore, since the velocity function or the first derivative is given, we only have to look for the points where v(t) is zero.

0 = (t + e^2) sin (t^2)

Plotting it into the graphing calculator, we obtain the roots at x = 1.7725 and x = 2.5066
Having these points, we say that at x = 1.7725 and x = 2.5066, the object is at rest.

b. When is the object moving to the left? Justify your answer.

A movement to the left, in my understanding, is similar to moving backward as opposed to forward. Therefore, the object is moving towards left on the interval where the velocity function is negative.

'velocity less than zero'

Looking at the graph, the points where the velocity is negative are pretty obvious. Having said that, on the interval [1.7725 , 2.5066] the object is moving to the left.

c. During the interval found in part b, when is the speed of the object increasing?

In this question, the acceleration or the second derivative is needed since the speed of the object is being asked.

a(t) = (1 + e^t) (sin (t^2)) + (t + e^t) (2t) (- cos (t^2))

*it is actually much easier if you guys just plug the function into the calculator at Y2 using nDeriv.

Finding the interval where the speed of the object is increasing also means looking for the positive values of a(t).

The roots of a(t) are located at x = 1.377 , x = 2.2161 and x = 2.8307.

Now, to the left of x = 1.377 the graph is positive and to its right negative. So on the interval 1 , 1.377 the speed is increasing.

at x = 2.2161, the graph is negative to its left and positive to its right. on the other hand, at x = 2.8307, the graph has its positive values on its left.

But, since the interval is specified, we only have to consider the roots between x = 1.7725 and x = 2.5066. Therefore, the object's speed is increasing on the interval 2.2161 , 2.5066

Steve's Blog Assignment

2006-04-23T17:03:00.000-05:00

The town of Calcuville has a water tower whose tank is a circular ellipsoid, formed by rotating an ellipse around its minor axis. The tank is 20 feet tall and 50 feet wide.
a) If there are 7.46 gallons of water per cubic foot, what is the capacity of the tank to the nearest 1000 gallons?
b)Calcuville imposes water rationing regulations whenever the tank is only one quarter full. How deep is the water in the tank when water rationing becomes necessary?
c) During peak usage, 5000 gallons of water are used per hour. How fast is the water level dropping if it is 12 feet deep when peak usage begins?

Equation for an ellipse ((x-h)²)/a² + ((y-k)²)/b² = 1

width is related to the x axis, and height is related to the y axis

50 = 2a 20 = 2b
a = 25 b = 10

Substitute the values of a and b into the equation and the coordinate (0,0) because no specific coordinate was mentioned and it is also easier to work with.

((x-0)²)/25² + ((y-0)²)/10² = 1
x²/625 + y²/100 = 1

Find volume by using the technique of slicing. Solve for x because ellipse is rotated around the minor axis which happens to be the y axis.

x² + 625y²/100 = 625
x² = 625 - 625y²/100
x = (+/-)√(625-(25y²/4))
You need both the positive and the negative of the square root because it is an ellipse and it exists in both the first and fourth quadrants.

V(y) = πΣ(0,10) √(625-(25y²/4))²dy
Σ takes the place of the integral sign and the numbers after it are the (a,b) of the integral.
That part covers only the top half or the first quadrant of the ellipse, but the bottom or fourth quadrant MUST be included in order to find the volume of the whole ellipse.
V(y) = πΣ(0,10)(√(625-(25y²/4)))² + πΣ(-10,0)(-√(625-(25y²/4)))²
Because the inner function is squared, it becomes positive which means that it becomes the same as the positive one and it can become one integral.
V(y) = πΣ(-10,10)(√(625-(25y²/4)))Â²
The square root and the squared cancel each other out. Leaving a parabola.
V(y) = πΣ(-10,10)(625-(25y²/4))
Antidifferentiate then evaluate
= π(625y-(25y^3/12) from -10 to 10
= π{[625(10) -(25(10^3)/12)]-[625(-10)-(25(-10^3)/12)]}
= π[(4166.6667)-(-4166.6667)]
= π(8333.3333)
= 26179.9388 ft^3
Use the constant of 7.46 gallons of water per cubic foot to calculate how many gallons of water there are.
V= (26179.9388 ft^3)(7.46 gal/ft^3)
V= 195302.3433 gallons
Answer to Part a) 196000 gallons of water
Part b) Requires changing volume from gallons to cubic feet and take one quarter of that amount.
0.25V= (1/4)(195302.3433 gallons/7.46 gallons/ft^3)
0.25V= 6544.9847 ft^3
Now take the integral from the bottom of the tank to y and set it equal to 0.25V
πΣ(-10,y)(625-(25y²/4)) = 6544.9847
Divide by π
Σ(-10,y)(625-(25y²/4)) = 2083.3333
Antidifferentiate the left side and evaluate it while still equaling 2083.3333.
625y-(25y^3/12) from -10 to y
[625y-(25y^3)/12]-[625(-10)-(25(-10^3)/12)] = 2083.3333
625y-((25y^3)/12)+4166.6667 = 2083.3333
625y-(25y^3)/12) = -2083.3333
Solve for y to find the depth.
Multiply by 12.
7500y -25y^3 = -25000
0 = 25y^3 -7500y -25000
0 = 25(y^3 -300y -1000)
Using a calculator, I found y to equal -3.472964, in terms of depth, -3.472964 is not the depth, but it is the y coordinate on the graph of the ellipse. For example, when y=-10, depth is 0 ft. If y =0, depth is 10 ft, and finally if y=10, depth is 20 ft, the max.
depth = 10-3.472964
Answer Part b) depth = 6.5270 ft
Part c) Given dV/dt = -5000 gallons/hour and the depth of the water is 12 feet, I can find the y value needed to find the rate of how fast the water level is dropping. y=2.
V = 7.46π[(625y-(25y^3)/12)-4166.6667]
Differentiate implicitly
dV/dt = 7.46π[625y'-(25y²y'/4) -0]
dV/dt = 7.46π[625y'-(25/4)y²y']
Substitute the values for dV/dt, and y to solve for y'.
-5000 = 7.46π[625y'-(25/4)(2²)y']
-213.3444 = 625y' -25y'
-213.3444 = 600y'
-.3556 = y'
Answer Part c) y' = -0.3556 feet/hour, the depth is decreasing at a rate of 0.3556 feet per hour when the depth is 12 feet.

The Da Vinci Code Quest Sunday

2006-04-23T15:34:00.000-05:00

It started last week. Google releases one puzzle each day for 24 days until the movie "The Da Vinci Code" is released in May. So far 7 puzzles have been released. You have to solve the puzzle to reveal a clue. Then you have to answer the clue question(s) to advance to the next puzzle. You can win a prize for solving all 24 puzzles. Now I realize this is all about marketing and they're really just trying to get as many of us as possible to go see the movie but the puzzles are really cool! Google searching often helps to find the answers. One of the puzzle questions can be answered using The Fundamental Principle of Counting and the very first (sudoku-like) puzzle uses a couple of mathematical symbols.

Challenge 1: What is the question that can be solved using The Fundamental Principle of Counting and how do you use the counting principle to find the answer?

Challenge 2: What mathematical symbol is used in the very first puzzle and what number does it represent? (Not the "delta," in a later puzzle it has a different meaning.)

You have to sign up for a Google Homepage in order to play, but that's a free and very useful service. After that you can begin the game. Click on the US button to start 24 days of fun! (Actually, 17 because you could work through the first eight today.) Don't forget to also find the answers to the Challenge Questions above!. ;-)

Blog Assignment 1

2006-04-23T12:46:00.000-05:00

A differentiable function f defined on -7 < 0< 7 has f(0)=0 and f'(x) = 2x sin x- e^(-x^2) +1
a) describe the symmetry of f.
b) On what intervals is f decreasing?
c) For what values of x does f have a relative maximum? Justify your answer.
d) How many points of inflection does f have? Justify your answer.

a) f is an even function.

b) f'(x) = 2x sinx - e^(-x^2) + 1

2x sinx- e^(-x^2) +1 =0

x=-6.2024, -3.294, 0, 3.294, 6.2024

+ - + + - +

--------1----------1----------1-----------1-----------1-----------

-6.2024 , -3.294 , 0 , 3.294 , 6.2024

If f'(x) less than zero,f(x) is decreasing, therefore, f(x) is decreasing at (-6.2024, -3.294) and (3.294, 6.0224).'

c) If f'(x) larger than zero, f(x) is increasing; If f'(x) less than zero, f is decreasing.

from increasing to decreasing, f reaches a maximum. f has relative maximum at x= -6.2024, and x= 6.2024

d) f"(x)=2x cos x+ 2 sinx + 2x e^(-x^2)
2x cosx+2 sin x +2x e^(-x^2) = 0
x= -4.9136, -2.0405, 0, 2.0405, -4,9136
f has inflection points when f"(x)= o, so, f has 5 inflection points

Sarah's Blog Assignment

2006-04-20T20:32:00.000-05:00

Question:
Let C represent the piece of the curve y = (64-16x^2) ^ (1/3) that lies in the first quadrant.
Let S be the region bounded by C and the coordinate axes.
a) Find the slop of the tangent line to C at y=1.
b) Find the area of S.
c) Find the volume when S is rotated arount the x-axis.
d) Find the volume when S is rotated arouind the line x=-2.

I have tried to find an equation editor that will show the following math signs properfly but the files seem to be given out incomplete. As for point, it really is hard to use and I have no time to fiddle with it, so bare with how my answers will be, and sorry in advance.

Answer:
a) slope @ y=1

1 = cube root ( 64 - 16x^2 )

cube both sides

1 = 64 - 16x^2

-63 = - 16x^2

63 / 16 = x^2

square root both sides

square root ( 63 / 16 ) = x

x = 1.9843

( 1.9843 , 1 )

y = ( 64 - 16x^2 )^(1/3)

y' = (1/3)( 64 - 16x^2 )^(-2/3) * ( -32x)

y' = (-32x) / [( 3 ) ( cube root [ ( 64 - 16 x^2)^2])]

therefore, y' (1.9843) = - 21.1538

(**NOTE: It is always good to store exact values into your calculator. If you are using programs like the Riesum programs, be careful where you store it, the alpha letter you may be using might be used by the program as well. **)

b) Area of S.

x - intercept ( 2 , 0 )

c) Volume of S when rotated around the x-axis.

Taking a slice, it would look like a disc. A circle. Area of a circle is the pi times the radius squared. Pi becomes a constant. The radius is measured by the function itself so therefore,

d) volume when S is rotated around the line x= -2.

When revolving it around the vertical line, we get a cylinder formed. Method that we use is cylindrical shells. Thinking of a piece of paper and just wrapping it around some center, How do we find the volume of that piece of paper? L * W * H. Height in this case is still the function value. Width is the itty bitty x difference, dx. Lenght though is a the circumference of the circle, 2 pi radius or 2 pi x. Therefore we have the integrand,

P.S. To that post, what else makes me focus? Other than candies, if I'm not at home its the music, not just any particular ones right now are korean r&b and pop. As long as it sounds good to me and something that I could never sing or know what its about. =D

See you all Monday.

The Main Thing

2006-04-19T19:40:00.000-05:00

Spring has sprung in Chardon, Ohio! Yes! We get lots of snow and winter seems so long that when the daffies bloom and the forsythia bursts forth with yellow, I'm estatic. (Last year this time, a late storm dumped 15 inches of snow on us) Is Winnipeg the same?

With spring, I'm finding that it is more difficult to stay with the course I'm designing for superintendents and principals now. I'd much rather be walking in the park, or out in the garden-- And so I've been using the thought above to help keep me focused.

What about you? What helps keep you focused?

endings-- beginnings--

2006-04-13T18:44:00.000-05:00

Ara's post really hit a spot with me. This entire notion of an end to something or not---
Do you think this quote applies as you work toward to your graduation?

today's last.. doesn't seem like it

2006-04-13T05:30:00.000-05:00

"Give a man a fish and you feed him for a day, teach a man to fish and you feed him for a lifetime"

Wow. It surprises me that I lasted this long in this course. I really had no idea that I am close to finishing it. I'm telling you guys, it was pretty tough along the way... the thought of quitting had crossed my mind hundreds of times already. I felt like through the entire span of this class I was always hanging on to the ropes. But Mr. K, he stretched his arm, reached for me and pulled me through. I wouldn't be surviving this class without you.

Although, this is actually my last post on blogging on blogging, it seems to me... it's not. As that Chinese proverb says, learning will never end. In every step I take and in every move I make (sounds familiar huH? ;D ), there will always be an instance where I'll remember and apply (well, somehow) some of the lessons we took in this course: the optimization, the related rates, the volumes by slicing (ooh... I hated that!), the slope fields and all. Could I be telling my friends the best time they should drink their hot cocoa? Or would I tell my mom the right dimensions in wrapping christmas gifts? Or maybe, just maybe, I can estimate the rate of how horrifyingly fast or sadly slow the airplane flies when I spend my holiday back to the Philippines... Hmm... I might consider that one! ;)

This could be the last, but it's never going to be the last. There are a lot more things to learn in this course and I am actually looking forward to struggling through them... *smiles*

A New Feed Window

2006-04-09T23:08:00.000-05:00

There is another Winnipeg AP Calculus class sharing their learning on a blog. You can peek in on what they're learning by checking out the new feed window way down there at the bottom of the side bar underneath the del.icio.us box. Take a look at their blog. Are they publishing anything that you find helpful? If so share it with us in the comments to this post.

I'll bet you like getting comments on our blog. Be a good netizen; drop in on them and leave them a positive comment too. ;-)

Differential Equations and Us

2006-04-09T15:03:00.000-05:00

Who knew that the last unit of AP calculus would be the best and the most fun. Granted it is difficult and the problems take up a page each, but it is definitely more interesting than the other units, probably because it relates to real life with the coroner problem and cooling problems. The hard part is when the problem has something like " the rate at which the population changes is directly proportional to the population at that time." Man that confuses me to no end. It is even harder when they a constant rate, because I'm not sure whether or not I should use that as k or make that number the coefficient of P and still include a k or not include a k. It's really hard sometimes, but at last were finished.

Four Colour Sunday

2006-04-09T14:01:00.000-05:00

You may have heard that any map can be coloured with four colours in such a way that neighbouring countries receive different colours. That it can be always done is one thing. How to do it is another. Are you ready to start colouring?

(Thanks again to Think Again!)

Hot Coffee, give it a few minutes

2006-04-05T15:57:00.000-05:00

Today in class we wrapped the last part of the course, except for the coroner problem which we will be doing next class. We did a problem that is similar to the coroner problem, the problem we did a problem about coffee cooling. The coffee's temperature is 190 degrees, and room temperature is 70 degrees, and coffee is "drinkable" at 120 degrees.

First thing you have to do is identify the differential equation that describes the cooling of the coffee.

T' = k(T-70)

dT/dt = k (T-70)

(1/(T-70))dT = k dt

Now antidifferentiate both sides of the equation.

ln(T-70) = kt+C

T-70 = e^(kt+C)
Now that the variables are separated, use the point (0,190) (T,t) to solve for one of the unknown variables.
T-70 = e^(kt)*e^C
190-70 = e^(k(0))*e^C
120 = 1*e^C
120 = e^C
T-70 = 120e^(kt)
Now that there is only one one unknown "k", use a second point (5,180) to solve the last unknown.
180-70 = 120e^5k
110 = 120e^(5k)
110/120 = e^(5k)
11/12 = e^(5k)
ln(11/12) = 5k
(1/5)ln(11/12) = k
k = -0.0174
T-70 = 120e^(-0.0174t)
Now that the equation has no unknowns besides T, and t, the problem can be answered. Use T=120

120-70 = 120e^(-0.0174t)
50/120 = e^(-0.0174t)
5/12 = e^(-0.0174t)
ln(5/12) = -0.0174t
(1/-0.0174)ln(5/12) = t
t = 50.3143
The coffee is drinkable after 50 minutes.

The next scribe is Sarah

Roboclaw Sunday!

2006-04-02T14:40:00.000-05:00

Move the robot arm to pick up the ball. Clean, simple design. I got to level 19. I died. It's a doozy!

Synergize? in calculus??

2006-04-02T12:31:00.000-05:00

I really liked Ara’s quotes and intro in her scribe! And wonder if there is any relationship between our willingness to embrace the “pain of a new idea” (step outside our comfort zone), and the habits we develop? Do you think these last four habits (that round out the 7 habits from Sean Covey’s book) have anything to do with willingness to embrace something new?

Habit 4: Think Win Win
How might this attitude help Goldilocks and/or affect your life?
Habit 5: Seek First to Understand, Then to be Understood.
Does Covey sum it all up with? “You Have Two Ears and One Mouth… Hello!
Habit 6: Synergize
I mentioned the lesson from the geese in a comment on Sarah’s blog. If I bring it here, does this suggest a strategy for success?
Habit 7: Sharpen the Saw
Subtitle—“It’s me time!” I’m the first to admit I’m not good at this. How can taking time for you really be important??

Of these 4, I have two favorites. What do you think? Are these habits of value? And if they are not new to you, then which of these helps you the most and how? Is it time for some synergy here?
Best,
Lani

Scribe for last Thursday

2006-03-31T10:03:00.000-06:00

Differebtial Equations-intro Euler's Method

Consider the differential equation dy/dx=y-x

Complete the table below, starting at the point (0,0) to generate a numeric solution to be the differential equation on interval (0,1) in 4 steps (so n=4 and △ x=0.25).

Nest scribe is Chris

You guys need to see this ....

2006-03-30T15:06:00.000-06:00

This is a video recording of a presentation called: "From Smart Toilets to Smart Schools." It was given by a world class educational technologist named Alan November. He travels all over the world teaching teachers about powerful uses of technology for their classrooms. He's been talking about us all over the world. In the video he talks about our whole class, and Ara, he mentions you by name. (How cool is that?) This particular presentation was given in Ohio on February 14th. If you didn't really believe me when I said you had a world wide audience reading and learning from your work, well, ..... watch the video.

Start watching the video at the 38 minute mark. I think it will go offline on April 14th so don't put off watching it. You'll need to have realPlayer media player installed on your computer (it's free). Just click on the link to [RealPlayer - Free]. Here's the video! You should be very proud of yourselves ... I'm proud of you. ;-)

Urgent, Important? First things first!

2006-03-29T18:41:00.000-06:00

I can certainly relate to Sarah’s feeling that there are too many big rocks for the bucket of our life, as she mentioned in a comment. In early 2003, my bucket seemed to overflow. I felt then that I had to find a way to make it all work. Achieving my goals was too important to me and my test (which if I passed would lead to a national teacher certification; only 40% of test takers passed) was only one of many goals. I was really trying to be superwoman—just as I imagine many of you work at being superteens. I thought I was prioritizing but ooohh---a to-do list with 20 items all the time!!!

What I struck me one day when I found myself in tears because I had not met a school deadline, was that I was not always prioritizing by what would help me meet my goals. Urgent “stuff” kept happening and I was always responding to that. Does that sound familiar?

That night I sat down and broke my test review into manageable chunks (I had 2 months until the test; I divided the topics into sessions for those two months) and put them into my planner. I did that first because the goal to pass the test was so important. And I planned to turn off instant message, not answer the phone, or have the TV on during the review sessions. I had tried to plan before but was always interrupted by the phone or my students on instant message with questions about our studies and I found that I didn’t accomplish half as much then.

Then I looked at my other rocks, and categorized them: vital, important, or nice. Then I took the vital rocks and categorized them again: vital, important, or nice. So my house wasn’t very clean during the process, and we didn’t have gourmet dinners. But the laundry was done and we had quickie suppers. My student’s work was graded but I didn’t plan any big field trips or projects during that time. I set aside a time every second night to evaluate their work. As I look back now, I prioritized, and then prioritized again. I did a mental daily check of my goals and made every effort not to be dragged down by urgent if it didn’t help me achieve my goals.

Of course, I had to be flexible at times. I couldn’t always follow the plan exactly. But since I knew where I was going and I had planned for time to get there, my review was accomplished by “the day”.

I truly believe, that with good preparation and putting first things first, you'll too feel that great rush of a job well done, and a goal achieved when you learn your scores. I share these experiences, knowing that you are planning and reviewing, but wondering is there one little piece here you could use to help you on your way? Or can you point us to some tips that are really helping you manage the rocks and put first things first?

New yet Old

2006-03-27T00:53:00.000-06:00

�One of the greatest pains to human nature is the pain of a new idea.�

Or is it? Sometimes new beginnings are just too hard to take. But some do give you something to think about. It's the same feeling as the day you packed up all your stuff to move in to a new house. Boy you really wished you never went to that neighborhood. But eventually you learned to like it. Or that time when babies had to stop drinking from their milk bottles and start acting like "grown ups." You see them whine and do every little thing they can to persuade you into giving back their tiny, precious milk containers. And you always wonder "What's the big difference?" Well, tell you my friend, no matter how alike two things are, you'll always see yourself looking for that same old thing you've grown to love. It's never the same.

But then, the case here is quite clashing to what you've just read. (Some introduction huh? ;P) I just put that up to make it a little bit more interesting. We do have a new topic (if we all would consider that) but it's not that heart-breaking or agonizing to take. It's pretty cool actually.

This day, we learned some new terms of those we've known in the past chapters. NEW WORDS, SAME CONCEPT.

Differential Equation - equations that relate derivatives to other functions
Order - refers to the highest-order derivative of the differential equation
General Solution - corresponds to a family of solution curves
Particular Solution - exact solution for the differential equation
Initial Condition - The values assumed by the variables in a component, system or model at the beginning of some specified duration of time
Initial Value Problem - the combination of an initial condition and a differential equation

These are just some and I know there's a bunch to come.

So as this point, saying DERIVATIVE or SECOND DERIVATIVE is prohibited. We all have to get used to the new ones. Because if not, it's never going to be a habit. And if it's not a habit, it's never going to run successfully in our system. And if it doesn't, then there was no learning.

�The greatest pleasure in life is doing what people say you cannot do.�

Are you ready for this? :)

Invading your space?

2006-03-24T13:11:00.000-06:00

Greetings AP Calculus!
I do hope that you don’t feel that I’ve invaded your space! That certainly is not my intent! But after reading your blog, I felt that we might have something in common—facing a major testing challenge. I made it through my testing successfully, thanks to hard work, mentors, and lots of encouragement-- and I hoped I could help to “pass it forward” so to speak. Thinking we all might learn from each other if we had an opportunity for conversations about dealing with challenging and stressful testing (and mine was more than stressful!), I just jumped right in without asking your permission (I had Mr. K’s of course). Let me know if you’d rather I’d not post up front. However, I’m not sure you can get me to go away completely, because it’s my sense that such a talented group as you should have a good cheerleading section on your way to May 3!

I asked you a question in my last posting and since I’ve had no takers with an answer in the comments, I’m going to answer it myself, given the time to May 3 is steadily dwindling! If you haven’t read that posting yet, do it now before you read the answer!

The answer is “Habit”.

I know from lots of experience that habits can ensure that I reach my goals or hold me back. Given that, there have always been 7 habits that if I worked at them have helped me be effective! Have you seen these 3? Do you think these habits can be of value as you look to “the test”?

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

I think this short autobiography is a good introduction to habit 1.

There's A Hole In My Sidewalk: AUTOBIOGRAPHY IN FIVE SHORT CHAPTERS

by Portia Nelson
Chapter 1
I walk down the street.
There is a deep hole in the sidewalk.
I fall in.
I am lost.... I am hopeless.
It isn't my fault.
It takes forever to find a way out.
Chapter 2
I walk down the same street.
There is a deep hole in the sidewalk.
I pretend I don't see it.
I fall in again.
I can't believe I am in the same place.
But, it isn't my fault.
It still takes a long time to get out.
Chapter 3
I walk down the same street.
There is a deep hole in the sidewalk.
I see it is there.
I still fall in.... It's a habit.
My eyes are open.
I know where I am.
It is my fault.
I get out immediately.
Chapter 4
I walk down the same street.
There is a deep hole in the sidewalk.
I walk around it.
Chapter 5
I walk down another street.

Habit 1: Be Proactive®
Take responsibility for your life.

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

Does this cartoon say it all for Habit 2?

Habit 2: Begin with the end in mind.®
Define your mission and goals in life..

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

Finally, Habit 3: Rocks, Pebbles, Sand, Water—And Calculus is which?

A time management specialist was asked to give a presentation on her specialty. She decided to do a demonstration. First she asked her assistants to bring a big bucket and put it on the table in front of the audience. Then she asked for large, grapefruit-sized rocks and filled the bucket with them.
"Is the bucket full," she asked?
"Yes!" said the crowd, but she asked for more to put in anyway. This time her assistants brought in pebbles. She poured the pebbles in the bucket and it held a surprising number in the space between the big rocks.
"Now is the bucket full?" she asked.
"Yes!" "No!" "Yes!" "No!" said various persons in the crowd. Some people were uncertain; some were getting suspicious. The time management specialist asked for more. This time the assistants brought her sand. She poured sand in the bucket and it filled the spaces between the pebbles.
"Now is the bucket full?" she asked.
"No!" they answered. By now, everyone was suspicious. So she asked for water and poured in quite a lot. Now no one could think of anything else that could fit in that bucket.
"What does this process demonstrate?" asked the time management specialist.
One member of the audience spoke up: "No matter how busy you are, you can always fit in one more thing."
"I can see how you might think that was my point, but it is not," said the specialist. "I was trying to show you that if you don't put the big rocks in first, you'll never get them in at all!"

Habit 3: Put First Things First®
Prioritize and do the most important things first.

Best,
Lani

Who Am I ?

2006-03-23T07:09:00.000-06:00

I am your constant companion. I am your greatest helper, or your heaviest burden. I will push you onward, or drag you down to failure. I am at your command. Half of the task you do, you may just as well turn over to me. I will do them quickly and correctly.

I am easily managed, but you must be firm with me. Show me exactly how you want something done. After a few lessons I will do it automatically. I am the servant of all great people; of all great failures as well. Those who are great, I have made great. Those who are failures, I have made failures.

I am not a machine, though I work with all the precision of a machine, plus the intelligence of a person. You may run me for profit, or run me for ruin--it makes no difference to me.

Take me -- train me -- be firm with me, and I will place the world at your feet. Be easy with me, and I will destroy you!

Who Am I ?
And what can this possibly have to do with calculus?

Box Up Sunday!

2006-03-19T12:54:00.000-06:00

This is a clever little game. You've got to get the small blue box inside the large red box. You can only push a box from the inside. The black boxes, if used cleverly, can help you get the blue box inside the red one. But sometimes they're just in the way. I made it to level 4 pretty quickly, but then it starts getting tough. How far can you go? ;-)

Have fun with this!

Seek and ye shall find ...

2006-03-14T16:44:00.000-06:00

The Coin Hunt has officially begun as of 12:30 this afternoon. The race is on! Who will be the first to find the coin? Will the students find their coin before the teachers find theirs? Who will win the pizza party? Which charity will benefit from this year's hunt?

Check the walls of the building as you walk into school in the morning for hints to figuring out the puzzles.

Happy π Day!!
Have fun with it. ;-)

When test time comes, be ready

2006-03-12T15:00:00.000-06:00

It doesn't matter what you do in before the test, what matters is what you do on your own time. This unit was fairly short, section 1 was net distance which is the integral of the velocity function and total distance traveled which is the absolute value of the velocity function then take the integral of that. Section 2 was volumes by slicing, which involves revolving the graph of a function around the x-axis. You can find the volume of that solid by finding the integral from (a,b) of the function A(x). A(x)=(pi)(x)², x represents the radius of the circle, which is the function value. If it is a washer then in place of x, subtract the smaller function from the larger function. In section 3 there was the shell method for solving a problem if a function was revolved around the y-axis. The volume of that solid can be found by taking the integral from (a,b) for 2(pi)*x*f(x). Section 4 was the average value over an interval. The average value of a function is represented by the integral from (a,b) integrate f(x) and multiply it by the constant (1/(b-a)). Section 5 was the hard part of the unit, WORK, and DENSITY. Work = (Force)*(Distance) and Density= mass/volume. The best way to attempt these problems is to break them up in pieces and figure out what each part of the equation is before attempting to find the final solution. For example, if you have a WORK problem, find the FORCE and then find the DISTANCE before combining them in an integral. The problems from this section of the textbook are the hardest we've faced thus far, so good luck.

Sunday Knight

2006-03-12T09:27:00.000-06:00

How far can you go? Play here! ;-)

The Mystery Coin Hunt!

2006-03-09T13:41:00.000-06:00

π Day is around the corner .. it's five days away! Soon, soon, the hunt will be on!

Somewhere on the property of DMCI a coin will be hidden. Hidden so carefully and cleverly that it cannot be discovered by chance or simply by looking for it. On March 14, π Day, the coin's location will be revealed buried in a series of riddles and puzzles. Until it is discovered the coin's location will remain a mystery....

The amount of WORK

2006-03-08T18:35:00.000-06:00

Yes calculus takes a lot amount of work. Imagine our brain cells, constantly moving the data through the brain. Our brain only has a particular volume to it, but it takes a lot of force to move all the data swimming inside it to a particular destination.

*hehe* Just kidding guys.

But really the topic we covered on Tuesday was on WORK. To those who took Physics, you may already have conjured up the formula for work:

WORK = (FORCE) x (DISTANCE)

Applying Calculus…

Overlooking the Problem from Tuesday…

We are given a rectangular prism tank 6 ft long, width of 5 ft, and height of 4 ft. On the top of it is a spout 2 ft. tall. The prism is filled with a liquid whose density is 40 lbs/ft³. Write the integral on amount of force it takes to move the water out of the tank.

When a picture is not provided, it is always good to draw one to understand the problem more.

To start off the problem:
1) Draw a picture, unless it is provided.
2) Write down what you are given.
3) Recall a formula.
For this problem, it’s the WORK formula. Work = F x D

In the problem we are not given the force straight off the bat. We are looking for the force of gravity, which is measured in lbs. We are given that the density in the units lbs/ft³. What do we multiply the density by to leave us off with just lbs.?

Lbs/ft³ x _____ = lbs.

The answer is: multiply it by ft³. In other words we multiply by the volume.

To find the volume of a rectangular prism: (length) (width) (height)

We are to move all of the water out of the tank. Taking an arbitrary slice, the height of the slice will be a very small itty bitty change in y.

We all know that it’s dy.

The length and width is given in the problem 6 ft. and 5 ft.

Good Job. You have found the weight of the substance. Now for distance:

Looking at the prism, imagine an arbitrary slice on the surface. That slice has 2 ft of distance to move out of the tank. Now imagine an arbitrary slice on the very bottom. The height of the prism add the height of the spout gives you 6 ft of distance to cover to get out of the tank. What equation would give you the possible heights?

Answer: 6-y

One thing left to figure out, is the limits of integration. The problem had said the prism is filled with a liquid substance. The height of the prism is 4 ft. So therefore, the limit of integration is 0 to 4.

Now to write the integral it is:

On Tuesday we were given another similar problem but this time it was a cylinder. The only thing that changed was how to find a volume of a cylinder. I guess this is where our bank of volume formulas comes to play once again. It is always good to have any of our formula sheets around.

So it was another day in calculus. I believe we have only 18 classes to go? Or maybe even less before the AP exam. Not to scare anyone away, its one tough exam. Everything Mr. K says is true. It’s nothing like any of the exams we had ever written. Good practice, a good understanding of the concepts is needed, and a great memory too. Take advantage of all resources available to us. We’ll need it.

Well see you all tomorrow =)

Reply to COMMENT: I find the quizzes a good review of the topics we've already covered. It also sets us ready for the time given in the exam to answer multiple choice questions. It also helps us not to be so dependant on the calculator. So overall, its helps us get ready for that exam on the beginning of May =)

Sunday Gridlock

2006-03-05T01:22:00.000-06:00

In this game you have to move the blocks (vertical blocks move only vertically and horizontal blocks move only horizontally) out of the way so that the blue block can slide out the "door" on the right. Although this game can sometimes get frustrating there is always a way out. Remember Sisyphus!

So far I've made it to level 6, how far can you get?

Solid by Revolution

2006-03-03T08:44:00.000-06:00

In today's class we started with a quiz of course, just like every day. We then went on to practice integrals by solving problems using disks and cylindrical shells. One such problem is what would be the volume of a solid obtained by revolving the region bounded by the lines y=9-x^2 and y=-7 around the line y=-7. The points of intersection of the two functions are at x=-4 and x=4.

Using the disk method is the easier method for this particular problem. Because a disk can be thought of as a cylinder and the equation for a cylinder is V=(pi)hr^2, if we were to think of the radius of the disk as the value of the function +7 and if the height of each disk was an infinitely small amount, we could piece together a function for each individual disk as V=(16-x^2)^2 dx.
Now if we were to take the integral of that function with the limits of integration as the points of intersection of the functions (-4 to 4), it would give us the volume of the solid.

The cylindrical shell method is a little more difficult than that however. To get a cylindrical shell, we would have to evaluate the integral as a function of y not a function of x. Therefore we should rearrange our parabola for y, which is y=sqrt(9-y). So now that we have that we need to think of the equation of the volume of a cylindrical shell, the volume being the circumference multiplied by the height and width. Because the distance between the large and small radii will be so insignificant, we can think of the width as dy. The equation for circumference is 2(pi)r, and because r is the distance between y=-7 and the function value, r can be thought of as (y+7). The height of each individual shell will be the value of x at each y value multiplied by two because it is a parabola and it's simetrical, so h=2sqrt(9-y).
So now if we put the two together, we get the equation V=2(pi)(y+7)2sqrt(9-y) dy
Now we integrate the function with the limits of integration being all the y-values between the line y=-7 and the maximum function value (-7 to 9) and that will give us the volume of the solid.

Closer to the end of class Mr. K gave us a stencil with similar problems on it.
(Sorry about my lack of the proper characters, the system I'm on at the moment has very limited resources.)

It's Coming ...

2006-03-02T23:59:00.000-06:00

π
Watch it grow.

Scribe for Wednesday

2006-03-01T16:59:00.000-06:00

We worked on solving for the volume of a function that is revolved around the y-axis. The integral for the interval [a,b], the inner function is the length times the width times the height.

V= (2(3.141...)x)(f(x))

The height is represented by f(x), the width is dx, the length is 2(pi)x.
In class, one of the examples was f(x)=√(-x+3)

V=2(pi) √(-x^3 +3x²)

When the graph is revolved around the line x=5

V=2(pi) (5-x)√(-x+3)

The second part of class was about the average value of a function. Which can be defined as:

[f(x1)+f(x2)+f(x3)+...+f(xn)]/n

dx= (b-a)/n
n=(b-a)/dx

1/(b-a) f(x)
The Average Value of a Function is defined by the previous expression. This expression can be used to find the mean value or f(c).
The next scribe is Chris

Scribe for Monday

2006-02-28T17:21:00.000-06:00

Given f(x)=0.5x^2-2x+4
And g(x) = 2 + rt 4+4x-x^2
S is the region between f and g, write the interval that gives:
a) The volume generated by rotating s around the line y=5.
b) The Volume generated by rotating s around the line y=1.
c) The volume generated by rotating s around the y-axis.
Rotating two lines, It surface will be looks like a washer.
And the Volume of a cylinder V = pai (R^2-r^2)
a)---> pai S(rt (4+4x-X^2) +2 -5) ^2 - (0.5X^2-2x+4-5)^2 dx
paiS(rt(4+4x-x^2-3)^2-(0.5x^2-2x-1)^2 dx
b)---> pai S(rt (4+4x-x^2)+3)^2-(0.5x^2-2x+5)^2 dx
c)---> PaiS(3^2-y^(1/2)) dy

A roll of paper towels has dimesions
Find the volume of the papper
Volume of Pai r^2 h a cylinder
V = Pai h (R^2-r^2)
= pai h (R+h) (R-h)
S (2pai x) f(x) dx
(circle) (height) (width)

Nest scribe: steven :P

Volumes by slicing

2006-02-27T11:52:00.000-06:00

This is definitely the hardest topic I've encountered, so far. It's a good thing that we started the class with a bit of a couple of easy questions that somewhat headed to the real deal.

First of, getting the derivative and the second derivative. Well, we all know how to get them at this point, we're probably good at it, by now, considering myself! (can you believe that? :D) Then the difference between the net and total distance was tackled. The second question involved integrating the area of a region "S" enclosed by graphs f and g, which is by the way an easy thing to do because this whole stuff is not new to us, right? But (a BIG BUT) when we came to the part where we had to find the volume generated by rotating region "S" around the x-axis... Okay, where am I now?

it wasn't very easy for me to understand this chapter because first and foremost, I couldn't picture out the image we were visualizing. So that's probably one note to greatly put everyone's attention to, VISUALIZE. ;) It will absolutely be hard to go through all these questions if merely knowing what it looks like is not met.

Two things I've learned that I thought were new:
- the slicing task results to a WASHER (well, not always)
- the concept of BIG R and small r. (its not new but subtracting R - r is pretty unfamiliar to me, but then again, I know everyone understands it)

IMPORTANT:
> memorize the volume formulae (it helps, big time!)

What should I do?

2006-02-27T10:36:00.000-06:00

Mr. K you'll probably wondering where am I and howcome I'm not in your class for couple of days now. Well this is what happen last friday February 17, when I was in my gym class an unexpected incident happened. I didn't exactly remember what happened because all I heard after I jumped was a crack and at that moment I know I broke my leg. Mr. K I need help I don't know what to do? and I really don't want to drop this class because eventhough I don't have a good grade in this extremely hard class I learned a lot of neat things and I really enjoy being in your class. Well the problem is that I just got surgery on my leg and it will take 3 to 4 months to recover. Now, I'm sitting here with pain don't know what to do? and my hopes of graduating this year is minimal. Now I got 26 credits just need 2 more and I don't know how to get that. Mr. K I need some of your advise on what should I do in order to graduate. And the blog is the only way I can think of to communicate with you to tell you why I'm not in your class for this couple of days. Well thank you in advance Mr. K.....

Let It Grow Sunday!

2006-02-26T10:15:00.000-06:00

This week's game is called Grow.

Draw each item to the center of the ball to Grow it. If you drag them in the right order you will reach the maximum growth level for each object -- that's the challenge and it's not easy. ;-) Lots of trial and error. The number of different ways to play this game is 479 001 600. Can you find the winning strategy?

The Applied Math class will learn how to figure this out this week. Pre-Cal will learn it in about two more months and the AP Calculus students should remember from the Pre-Cal class. Do you?

Have Fun!

Tell Your Parents the Blog is Multilingual!

2006-02-21T18:32:00.000-06:00

You'll notice that all posts on our blog now have a series of flags automatically added to the bottom. Click on a contry flag to have the blog translated into that country's language. You can choose from:

French, German, Italian, Portugese, Spanish, Japanese, Korean, Chinese and Filipino

If you speak any of these languages, let me know if they work well enough to be understood. And tell your parents all about it! ;-) Encourage them to leave comments on the blog as well.

Chinese? Checkers? Chess?

2006-02-19T13:42:00.000-06:00

Chinese Checkers it is called in England. Kinasjakk (Chinese Chess) in Norway. The truth is that it has nothing to do with neither checkers, chess, nor China.

'The Chinese Checkers game board is in the shape of a six pointed star and is playable with two up to six people at the same time. Each player uses pegs or markers of a different color placed within one of the points of the star. The object is to move all your ten pegs across the board (move one step at the time or jump over adjacent pegs) to occupy the star point directly opposite. The player getting all pegs across first wins.' - More.

You can play it here.

(Thanks again to Think Again!)

Life's not easy

2006-02-13T06:16:00.000-06:00

Never in a million times will I ever find ANTIDIFFERENTIATING easy. So to make the long story short, here's what I prepared for, to "somehow" pass the test. It's practically a list of formulae and other important notes for me to remember.

The INVERSE CHAIN RULE:

The method of SUBSTITUTION:

SUBSTITUTE by choosing u = g (x) and write du = g' (x) dx = (du / dx) = dx. Then subsitute both u and du to the original integral producing a new integral in the form of f(u).
ANTIDIFFERENTIATE in terms of u. F ' (u) = f (u)
RESUBSTITUTE g(x) to obtain the antiderivative in terms of x.

The method of INTEGRATION BY PARTS:

* L I A T E *

Antiderivatives of the inverse of Trigonometric functions:

And the NUMERICAL INTEGRATION

the MIDPOINT RULE
the TRAPEZOID RULE
and the SIMPSON SUM (where twice the value of the MIDPOINT is added to the TRAPEZOID value all over 3)

A Prisoner's Sunday Dilemma

2006-02-12T13:54:00.000-06:00

This week it's a logic puzzle ... we look in on a prisoner with a problem ....

For the last trial, the king used not two, nor three, but nine rooms! The prisoner was told that one room contained a Lady and the other eight were either empty or filled with a tiger. The sign on the Lady's door was true, the signs on room with tigers were false, and empty rooms had signs that were either true or false.

These were the signs:

The lady is in an odd-numbered room.

This room is empty.

Either sign 5 is right or sign 7 is wrong.

Sign 1 is wrong.

Either sign 2 or sign 4 is right.

Sign 3 is wrong.

7. The lady is not in room 1.

This room contains a tiger and room 9 is empty.

This room contains a tiger and 6 is wrong.

The prisoner studied the nine signs for a while and came to the conclusion that the problem was unsolvable. The king admitted his mistake and told the poor prisoner if room eight was empty or not.

The prisoner needed no more help. He deduced where the Lady was. What about you?

Problem source: The Lady or the Tiger and other Logic Puzzles by Raymond Smullyan. (With thanks to Think Again!)

Antiderivatives Review

2006-02-10T14:40:00.000-06:00

You can find a good review with a little practice on antiderivatives here. It covers substitution but not inegration by parts. You can also try your hand at a true or false quis over here and some review exercises there.

You can review and practice your Integration by Parts skills here. (The flash animations are really cool!)

You can review Numeric Integration here and there (More flash and animations!).

Lots of practice antidifferentiating using the arc trig functions can be found here. (You may find the solutions provided a little confusing as we used a slightly different technique to solve these problems. If you read it through carefully though you might learn something new!)

You can find LOTS of practice for using Substitution by taking this quiz. There are 50 questions but you can check your answer as you do each one.

There is a whole lot more you can review and learn with this COW (Calculus On the Web). The stuff you're looking for will be in the Calculus Book II link.

Study Hard! Remember, luck has nothing to do with it. It's all about how much effort you're willing to put in. ;-)

Cheers!

Friday's Scribe post.

2006-02-08T17:45:00.000-06:00

late once again...

Friday, the first day of second sem. Even though classes were shortened we had learned a lot. As usual, question were up on the board for us to answer.

The first question asked us to list all the pythagorean trigonometry identities.
- sin2x + cos2x = 1
- tan2x +1 = sec2x
- 1 + cot2x = cscx

The second question was a question similar to the questions we had for homework.

The third, told us to find the derivatives of the three inverse trig functions...

With Friday's lessons, we have added three new derivative rules and three antiderivative rules in our banks.

That's it from Friday.

Scribe

2006-02-07T15:18:00.000-06:00

Today in class we worked on a few antiderivatives. We learned the Simpson Sum which is the best way to find the integral with an estimate.
Simpson Sum = (Tn* + 2Mn*)/3
*(n is a subscript)

The Definite Integral from 0 to 4 for x² dx.
the antiderivative of xÂ² is x^3/3
Evaluate for when x =4 ansubtractct when x=0.
= 1/3(4)^3 - (1/3)(0)^3
= 64/3
= 21.3333

Using Simpson Sum, you get 21.3333, when n = 2.
The Simpson Sum is a combination of the Trapezoid Sum and twice the Midpoint Sum, then divide by 3.
The Trapezoid Sum is a combination of Left Hand Sum and Right Hand Sum, divided by 2.
The Midpoint Sum can be found by using your calculator and one of the programs on the calculator.
The next scribe is Chris.

Math can be Colorful

2006-01-25T22:22:00.000-06:00

To be Organize means to arrange in coherent form or to have a system. Using different colors were very helpful in dealing with integration by parts. To be successful in this part of the course, we need to know how to organize our signs and understand which is which by organizing what we have. Our example for today proved this theory. We found out that the reason why we have simple errors are because we forget our signs and we work very disorganized. Truly, if your work is organized like Mr. K's it is easier to see what we still need to do and what we've done wrong. Almost all of our time was used for answering this questions for this question though not very hard in nature but very complicated in terms of bookkeeping. I'm pondering to myself if it is necessary to elaborate this question for like I said its not challenging. I now decided not to explain the group question further but rather give tips on what we need to remember in answering this questions

We need to understand what is given and what we need to figure out.
Next is to know what steps we need to take in order to go to where we want to go. Choose the right method in solving. Remember not to use a shotgun to kill a mosquito.
Forget not the signs
Be cautious with regards to finishing up the question. Finding the antiderivative is less than half way the battle.

After the group work we discussed transcendental functions which I found out are functions that is not algebraic or cannot be expressed in algebraic form like exponential functions and trigonometric functions. We really didn't have time to discuss fully the topic we need to so I giving the topic to tomorrow's scribe which is...

Prince

Better Choices

2006-01-25T06:54:00.000-06:00

A new way of anti - differentiating was taught today. Not all products can be integrated by reversing the Chain rule. Sometimes, using Substitution will make it look even worse than what we could've imagined. So with the complexity of some equations, the method of INTEGRATION BY PARTS is introduced.

The rule of integration by parts can be written as:

According to the rule, the integral of a product of two functions may be written as a difference, one term of which is a new integral. In most cases, the new integral is simpler than the original integral. Therefore, this technique requires the importance of designating the factors f(x) and g'(x) of the original integrand in such way that f(x) simplifies under differentiation while the factor g'(x) does not become more complicated under the integration.

As what I've just mentioned, assigning the factors is a crucial aspect in performing this technique. To get better chances of heading to the correct answers, L I A T E must be taken into consideration.

L I A T E as in

L - orgarithmic
I - nverse Trig
A - lgebraic
T - rigonometric
E - xponential

It is also significant to know which method applies to a particular equation. In this way, it saves up time and makes it a lot easier.

So make better choices and always remember the past lessons and techniques that were recently learned... then your off to university... I think... well not really... but it's a good start though... *smiles*

Antiderivatives...

2006-01-24T18:59:00.000-06:00

Another scribe that I can't seem to find the time to do.

Well here it goes.

For the past weeks we have learned to differentiate many equations. Ugly or not so ugly, we can differentiate it. But now we run the rules backwards.

Who would've thought that a derivative and an integral undo what each other does? They are both very different from one another. Unlike addition and subtraction or multiplication and division, its not so obvious that they are inverses of one another. But they are, they undo what the other does.

Keep your bank of derivatives in hand, you'll need it. But this time add a couple of new rules, for antidifferentiating. Don't forget your plus C's, why? Because the derivative of any constant is 0. A big fat ZERO. The difference now when you're antidifferentiating is that you're not only getting ONE particular function but a WHOLE FAMILY of functions.

Advice for the entire chapter:
1) Keep your bank of derivative. You may add new ones...
OR
2) Make a bank of anti-derivatives.
3) + C, never forget that.

___________________________________

Natural Log function and Rational function in particular 1/x

Inverses.

Input ---> Output
its inverse is...
Output ---> Input

CORRECT?

X (domain) ---> Y (range)
its inverse is...
Y(range) ----> X (domain)

CORRECT AGAIN?

So where am I trying to lead this to?

If we look at the rational function 1/x
its domain is the reals and x does not equal to 0.

its anti-derivative should have the same domain.

This stands true, antiderivative of 1/x is ln x+c for x>0.

How about when x<0?

ln(-x)
Using chain rule.
d/dx ln(-x) = (-1)1/(-x)= 1/x

So therefore, the antiderivative of 1/x is ln absolute(x) + C.

Don't forget that absolute sign or the + C as they have great distinctions to your answer.

=D

Sorry for the late post.

The method of substitution

2006-01-23T20:16:00.000-06:00

We did the integration using the chain rule yesterday. Today, we had some questions on the board. Firstly, we used chain role solved first 3 questions. But if only use chain role, we wouldn't able to solve some problems like 4 and 5, so some new stuff for today. That's the method of substitution.
Three steps: Substitution, antidifferential, resubstitute
Exp:S cos (3x) 3 dx
Let u = 3x
du = 3dx
S cos (u)du= sin u+ C
=sin 3x + C
Another one:
S X/(rt (x +7) dx )
Let u = rt (x+7)
du = 1/2 (x+7) ^-1/2 dx
du = 1/ (2 rt (x+7)) dx
2du = 1/( rt (x+7) ) dx

u= rt (x+7)
u^2 = x +7
u^2-7 = x

S(u^2-7)* 2 du = 2 S( u^2-7) du = 2 ((u^3/3 – 7u))
=(2/3 u^3 -14u+c)
=2/3 (rt (x+7))^3 -14 rt (x+9) + C

Next scribe: Ara

Scribe for Friday

2006-01-22T21:20:00.000-06:00

Friday, first we answer some questions in the Iflurtz questionaire for Valentines Day. Then for our class we start by answering this question. S (x+x^2)^2 dx.
To answer this question is that you make it to a simpler equation first before you antideffrientiate it. So it would be S (x^2+2x^2+x^4) dx. Now its easy to antideffrientiate it, so it would become x^3/3+x^4/2+x^5/5 + C.

Then Mr. K showed us this F(x)= f(g(x))
F'(x)= f'(g(x)) * g'(x)

So if were given this: S 3x^2 sin (x^3) dx we can say that it is equals to this - cos (x^3)+ C, because of the relationship above. Its like undoing the chain rule.

However, if the 3 is not there we can put the 3 there but in order to make it equals to 1 we should multiply it with 1/3. And we can use this relationship: S k f(x) dx= K S f(x) dx. So if this is given S x^2 sin (x^3) dx, we can put 3 in front of x^2 so that it would look like its derivative. But we should put 1/3 in front of S too to make it balance. 1/3 S 3x^2 sin(x^3)dx then the answer would become - 1/3 cos (x^3) +C .

And that's what we learn last friday....the next scribe is xun...

Sunday Punting Practice

2006-01-22T09:04:00.000-06:00

Like sokoban the target is to push objects (in this case punt-discs or 'pucks') around a maze to cover various targets. In a punt maze however the pusher slides forward tilt-style until it hits an obstacle, and a puck that gets struck will be punted forward a matching distance.'

'Aim: Use the black cross as a pusher to 'punt' the yellow pucks onto the blue targets.
Movement: Use the arrow buttons provided to move the pusher (black cross). The pusher will run in a straight line until it hits a wall or a yellow puck. If it hits a puck the puck will be punted forward a matching distance.'

Are you ready to play?

(Thanks again to Think Again!)

Why Should I Learn Math?

2006-01-19T20:20:00.000-06:00

This is taken from an article (Math Will Rock Your World) from Business Week. A few snippets:

Y'wanna get a really interesting job working with people on lots of interesting things?

But just look at where the mathematicians are now. They're helping to map out advertising campaigns, they're changing the nature of research in newsrooms and in biology labs, and they're enabling marketers to forge new one-on-one relationships with customers. As this occurs, more of the economy falls into the realm of numbers. Says James R. Schatz, chief of the mathematics research group at the National Security Agency: "There has never been a better time to be a mathematician."

Learn math!

How'd ya like a six figure salary?

...new math grads land with six-figure salaries and rich stock deals. Tom Leighton, an entrepreneur and applied math professor at Massachusetts Institute of Technology, says: "All of my students have standing offers at Yahoo! (YHOO) and Google (GOOG)."

Learn math.

D'ya wanna to work on the biggest most cutting edge issues of our day?

This mathematical modeling of humanity promises to be one of the great undertakings of the 21st century. It will grow in scope to include much of the physical world as mathematicians get their hands on new flows of data .... "We turn the world of content into math, and we turn you into math," says Howard Kaushansky, CEO of Boulder (Colo.)-based Umbria Inc., a company that uses math to analyze marketing trends online.

Learn math.

Y'wanna make one of the most significant contributions to the betterment of humanity?

"The next Jonas Salk will be a mathematician, not a doctor."

Learn math.

What are the implications for k-12 education?

Outfitting students with the right quantitative skills is a crucial test facing school boards and education ministries worldwide. This is especially true in America. The U.S. has long leaned on foreigners to provide math talent in universities and corporate research labs. Even in the post-September 11 world, where it is harder for foreigners to get student visas, an estimated half of the 20,000 math grad students now in the U.S. are foreign-born. A similar pattern holds for many other math-based professions, from computer science to engineering.

The challenge facing the U.S. now is twofold. On one hand, the country must breed more top-notch mathematicians at home, especially as foreigners find greater opportunities abroad. This will require revamping education, engaging more girls and ethnic minorities in math, and boosting the number of students who make it through calculus, the gateway for math-based disciplines. "It's critical to the future of our technological society," says Michael Sipser, head of the mathematics department at Massachusetts Institute of Technology. At the same time, school districts must cultivate greater math savvy among the broader population to prepare it for a business world in which numbers will pop up continuously. This may well involve extending the math curriculum to include more applied subjects such as statistics.

Learn more math!

"But I don't like math. Besides, I don't need it. I'm going into the humanities or business!"

As mathematicians expand their domain into the humanities, they're working with new data, much of it untested. "It's very possible for people to misplace faith in numbers," says Craig Silverstein, director of technology at Google. The antidote at Google and elsewhere is to put mathematicians on teams with specialists from other disciplines, including the social sciences.

Just as mathematicians need to grapple with human quirks and mysteries, managers and entrepreneurs must bone up on mathematics. Midcareer managers can delegate much of this work to their staffers. But they still must understand enough about math to question the assumptions behind the numbers. "Now it's easier for people to bamboozle someone by having analysis based on lots of data and graphs," says Paul C. Pfleiderer, a finance professor at the Stanford Graduate School of Business. "We have to train people in business to spot a bogus argument."

Ya gotta learn more math!

Yes, it's a magnificent time to know math.

'Nuff said.

Integral facts

2006-01-18T01:02:00.000-06:00

Inverse of a derivative

Not an antiderivative for it is conceptually different

The infinite sum of small rectangles in an interval

Early use of integration was by Archimedes by finding area of a circle

Generates practically a number not another function

Riemann sum is a method used to approximate values of integral

Accumulation of quantities such as area under the curve or volume displaced

Leibniz introduced the standard long “s” notation of integral

Learn for a week

2006-01-17T23:58:00.000-06:00

WE finished this chapter in just like - a SNAP! -

I can't believe that I have actually ended this topic with a COMPLETE - let me emphasize more on that - ABSOLUTELY EXECUTED set of homework exercises plus an agreeable understanding of what integrals and derivatives are. For the VERY FIRST TIME...

WHOOHOOO!!!

That feels OH so good! Starting the year with a no-absent, no-late attendance... haha! I wish this would go on 'til the end of the next semester. -- i wish? :D

Quite a big information for such a small chapter. It was basically all about ACCUMULATION FUNCTIONS. These functions are called as such because they are integrals evaluated by accumulating the area under the graph of the integrand function f.

What I've gained throughout this short episode:

review on INTEGRALS and the FUNDAMENTAL THEOREM OF CALCULUS Part I
SECOND FUNDAMENTAL THEOREM OF CALCULUS
ACCUMULATION FUNCTIONS

and best of all

ways on getting the area or the integral of a function in a given interval using the calculator - that is so cool! if I had known that before, I've probably aced some of my past tests... or maybe not... who am I kiddin'? hahaha!!! But honestly, that's like so amazing. After practicing on some of those exercises I came up to the point where I said,

The Calculator's not that dumb after all. (laughs)

Blogging on Blogging on Integrals

2006-01-17T16:34:00.000-06:00

When it comes down to integrals, they are kind of confusing. The fact that integrals and derivatives are inverses confuses me. I think the the hardest part of this unit was composite functions involving integrals. Finding the derivative of an accumulation function requires the chain rule.

a=0
b=2x
f(x)= f(t)= tÂ²

You take f(t)substitutee 2x for t and derive t² and multiply that by the derivative of 2x.

F' (x)= (2x)(2)
F'(x)= 4x

I think that the confusing part of it is remembering that the upper limit or 2x in the previous example, is the inner function and that the outer function is f(t). Having clarity on this makes such a difference when solving for the derivative of F(x).

Sunday Slither!

2006-01-15T12:42:00.000-06:00

Thanks go out to Mrs. Armstrong for pointing to today's game from Think Again!.

The game, from Japan, is called Slither Link.

Rules of Slither Link

    1. Connect adjacent dots with vertical or horizontal lines.

    2. A single loop is formed with no crossing or branches.

    3. Each number indicates how many lines surround it, while empty cells may be surrounded by any number of lines.

Play here.

Composite Functions and Integrals

2006-01-14T17:22:00.000-06:00

Friday, we learned about accumulation functions that are a composite of functions. when finding the derivative of a composite of functions involving integrals, is not as difficult as it may seem.

a= 3
b= x^2
f(x)= √(t^6 +1)
dy/dx = (√((xÂ²)^6 +1)(2x)
The inner function is xÂ² and the outer function is the square root function.
dy/dx = 2x√(x^12 + 1)

Another part of the class was about the area between the curve and the x axis. Integral means signed area, and Area is not the Integral. Due to the fact that many graphs have roots and an area below the x axis as well as above it, to compensatete for this, you have to take the absolute value of the function.
abs(f(x))
If you have two functions, the area between them involvesolvingng for the intersection points, then finding the integral of the function that is higher and then subtract the integral of the lower function.
The next scribe is Sarah.

Remember Your Roots

2006-01-12T23:08:00.000-06:00

Well, do I really have to say this???Its been said over and over, well at least for my blogs... If I asked you what we did today, I'm pretty sure you know what we did... :D right?right? Yes I know what you're thinking "well get on with it already..." heheheh well ok, after saying this a million times, here I go again, to start the class, we had to do questions regarding yesterday's class. We are again faced with problems specifically made so that each problem will have confusing parts so we learn better. The first and second problem was about what we learned yesterday. We had an accumulation function and we needed to complete the table of values for the first problem, sketch the graph of that function, name critical numbers of that function and determine the intervals where that function is increasing. All of this is not new to us because we did this before and all of it is just a review so I will not bother discussing it.

What I would like to talk about is the second part of the fundamental theorem of calculus which states that:

If F(t) is continuous and A(x) is defined as an accumulation function:

the A'(x) = f(x)

To be able to remember this we have to go back to our roots. We learn addition first then subtraction. And we know that they are inverses of each other. Just like addition and subtraction, Differentiation and integration are inverses of each other. Even though its not straight forward because getting the slope of tangent line and calculating bunch of little squares are not really making sense but the because of the first part of the theorem we know it is true. Because we know that the first theorem is true then we understand that if we get the derivative of an integral then the function is just the underlying function; that simple.

Integral or Derivative?

2006-01-11T21:47:00.000-06:00

Todays class(and yesterdays class for the three of us not writing our english) started with Mr. K asking us to solve a lot of integrals, but geometrically, not using the fundamental theorem. The integrals were all on the function f(t)=t with the interval between 0 and a series of different values of b. That was all fine until we came upon integrals with a negative value for b. However Mr. K explained that we solve them exactly how we solve any other integral.
For example: on the interval 0 to 1, the integral is solved F(1) - F(0)
on the interval 0 to -1, the integral is solved F(-1) - F(0)

After that, Mr. K asked us to complete a table of values with values ranging from -4 to 4 for the function

We were then asked to graph it, the graph itself turned out to be the same as the function f(x)=(1/2)x². The function f(x)=(1/2)x² is also the anti-derivative of the identity function(f(x)=x). The lesson is a demonstration that integration and derivation are inverses of each other, just in the way that addition and subtraction are.
The function A(x) is an example of an accumulation function. It is called an accumulation function because as x progresses, the function accumulates area.

The next scribe will be..... Jayson

Just For Fun (or Getting Ready to Think Again!)

2006-01-07T13:48:00.000-06:00

I found this "test" over at Teaching and Developing Online. Try it out .... just for fun. ;-)

Below are four (4) questions and a bonus question. You have to answer them instantly. You can't take your time, answer all of them immediately.

OK?

Let's find out just how clever you really are. No looking at the answers in advance.

Ready? GO!!! (scroll down)

First Question:

You are participating in a race. You overtake the second person.

What position are you in?

Answer:If you answered that you are first, then you are absolutely wrong!

If you overtake the second person and you take his place, you are second!

Try not to mess up in the next question.

To answer the second question, don't take as much time as you took for the first question.

Second Question:

If you overtake the last person, then you are...?

Answer:If you answered that you are second to last, then you are wrong again. Tell me, how can you overtake the LAST Person?

You're not very good at this! Are you?

Third Question:

Very tricky math! Note: This must be done in your head only.

Do NOT use paper and pencil or a calculator. Try it.

Take 1000 and add 40 to it. Now add another 1000. Now add 30.
Add another 1000. Now add 20. Now add another 1000 Now add 10.

What is the total?

Scroll down for answer.

Did you get 5000?

The correct answer is actually 4100.

Don't believe it? Check with your calculator!

Today is definitely not your day.

Maybe you will get the last question right?

Fourth Question:

Mary's father has five daughters: 1. Nana, 2. Nene, 3. Nini, 4. Nono.
What is the name of the fifth daughter?

Answer:Nunu?

NO! Of course not.

Her name is Mary!

Read the question again.

Okay, now the bonus round:

There is a mute person who wants to buy a toothbrush. By imitating the action of brushing one's teeth he successfully expresses himself to the shopkeeper and the purchase is done.

Now if there is a blind man who wishes to buy a pair of sunglasses, how should he express himself?

He just has to open his mouth and ask, so simple.

So simple it is ... ;-)

A Comment From a Former Student

2006-01-04T18:59:00.000-06:00

I know many of you don't read the comments left on each other's posts — you really should — sometimes a real gem is hidden there; like this comment from a former student. I think he wanted all of you to read it so here it is:

As far as I know I have been through pre cal and calculus in post secondary school and it's different. Mr K., as strict or fast paced as he may seem, is teaching all of you very well as he is preparing you for the future. I know he has been a great help in my future studies. He has been able to explain things in different ways unlike my new profs who use the book/text only as a reference and do things like completing squares to find roots; which wasn't taught to me in university but was taught in high school and has made life easier.

Also as much as you may think it is stupid to remember the cosine song I wish I could remember it as it would have gained me an extra 10% in post secondary. Also don't ever forget that a log is an exponent. It may seem stupid or useless but it helped me get that extra 10% needed to get a B+ in math. When you are in doubt a log is still an exponent.

This is written from a past student of Mr. K. and he is doing his best and succeeding in preparing you for post secondary. (I wish I would have tried harder). Also, don't mind my English as that wasn't my focus so it may be poor.

Good luck and assignments vs final mark are very close together.

Happy New Year! ;-)

a chapter to remember

2005-12-22T01:53:00.000-06:00

so much for first impressions...

a moment of uncertainty...
an immediate response...
a stimulus reflexed instantaneously...

predicted... erroneously.
judged... superficially.

OPTIMIZATION
- a vague piece of powerful information
- distinctive from the rest
- yet, so interesting, I must admit

after an intensive work...
tremendous effort...
stupendous energy...
and humungous struggle...

it still hasn't paid off..
but it's worth infinite tries... afterall...

PATIENCE is an unbelievable virtue...

*keep that in mind*

More Applications of the derivative

2005-12-21T21:16:00.000-06:00

More Applications of the derivative

The first derivative test:
If c is a critical number and if f’ changes sign at x=c, then
- f has a local minimum at x=c if f’ is negative to the left of c and positive to the right of c;
- f has a local maximum at x=c if f’ is positive to the left of c and negative to the right of c.
A function f has
- a global maximum value f(c) at the input c if f(x) less than or equal f (c) for every x in the domain of f;
- a global minimum value f(c) at the input c if f(x) larger than or equal f(c) for every x in the domain of f.
The extreme Value Theorem
If a function f is continuous on a closed interval (a, b), then f has a global maximum and a global minimum value on this closed interval (a,b).

The Second Derivative Test
- If f’(c ) = 0 and f’’(c )>0 then f has a local minimum at c;
- If f’(c ) = 0 and f’’(c )<0 then f has a local maximun at c.

Find the vertical asymptotoes and horizontal asymptotes of f if f(x)=1/(x^2-1)
X^2-1 = 0
X=1 and x=-1 are vertical asympototes.

Lim(x-->+oo) =0
And
lim(x-->-00) =0
so, y=o is a horizontal asymptote.

Optimization problems
- read the problem. Identify the given quantities and those we must find
- sketch a diagram
- Using available information, express f as a function of just one variable
- Determine the domain of f and draw its graph
- Find the global extreme of f,
- Convert the result

Mean Value Theorem
If the function is continuous on the closed interval (a, b) and differentiable on the open interval (a,b), then there exists at least one number c in the open interval (a, b) such that
F’(c )= ( f(b)-f(a) )/ (b-a)
Ant derivatives:
Given the function f, determine a function F whose derivative is f. the function F is an antiderivative of f
Function -> antiderivative
X^n-------> (X^(n+1) / (n+1) ) +C
Cos X --->Sin X +C
Sin X----> -Cos X +C
(SecX)^2 -> tan X+C
e^X-------> e^X

Antiderivative

2005-12-21T18:26:00.000-06:00

An application of the Derivative
No one knows what C is
The family of curves are related to each other
Indefinite integral
Differentiate backwards
Equals the parent function
Rethink what each term is the derivative of
In each question remember to add C
Very often the power rule has to be reversed
Always check your answer by deriving it
These problem are different from the usual differentiating and solve for a max or min
In all likelihood, you'll forget to add C
Visualize the problem, draw and label a picture depicting the problem
Each problem is solvable!

Infinity - No Limits

2005-12-20T23:37:00.000-06:00

Introduced its symbol by Wallis, JohN
Named after the last greek letter some believed sO
Frankly others found it derived from a roman numeraL
It is not a real number unlike PI
Negative, it grows less than the number it was froM
If positive, it grows beyond, oui? OuI!
The quantity is endless and must be too big to counT
Yes! It does look like the number eight turned sidewayS

Antiderivatives r us

2005-12-20T17:38:00.000-06:00

Today in class, we did an optimization problem which took half of the class. The rest of class was spent on antiderivative problems. I'll show what the antiderivative problem looked like because it is something new that we haven't done before.

A train is traveling at 66 ft/sec when the brakes are applied. If the train decelerates at 8 ft/sec^2. How far did the train travel between the time the brakes were applied and when it stopped?

Solution:

First you have to start with the acceleration function because that is what the problem gives you.

a (t) = -8
Now find the antiderivative.
v (t) = -8t +C
Solve for when t=0 and the velocity equals 66
v (t) = -8(0) +C = 66
C = 66
v (t) = -8t +66
Now find the antiderivative of the velocity function to find the position function.
s(t) = -4t^2 +66t +C
When t =0, the position is 0, so solve for t=0
s(0) = -4(0)^2 +66(0) +C = 0
C = 0
s(t) = -4t^2 + 66t
Solve for t when the function v(t) = 0
0 = -8t +66
0 = -2(4t -33)
t = 33/4 = 8.25
Now evaluate the position function for when t = 8.25
s(8.25) = -4(8.25)^2 +66(8.25)
s(8.25) = -272.25 + 544.5
= 272.25 ft

The next scribe is Chris

Yule Time

2005-12-19T23:23:00.000-06:00

Need not to worry, I got this day covered. Prince told me last Sunday that Im the scribe. Ok? good! To start off today's class we had a group work about optimization. It is about Topher and Sully. The question goes like this:

Topher and Sully decided to go cut down a Christmas tree together, so they ventured out into the snowy woods. After wandering for close to three hours, the two found the perfect tree but also decided that they were lost. Luckily Sully found a map nailed to a tree which showed that they were four miles due north of a point on a perfectly straight road. Six miles east of that point was the park that the two called home. If Topher and Sully can walk and drag the tree at the rate of two miles per hour through the snow, and three miles per hour on the road, what is the minimum amount of time that they would need to get home?

Constraint
T=d/r

Optimization
T(x) = ((x^2+16)^1/2 ) / 2 +(6-x)/3
T(x) = 1/2 (x^2 +16)^1/2 +2 - x/3
T ' (x) = (1/2)(1/2)(x^2+16)^-1/2 (2x) - 1/3
T ' (x) = (3x - 2(x^2 +16)^1/2)/ (6(x^2+16)^1/2

After finding the derivative we find where there are zeros or where the derivative is undefined. The denominator will never equal zero so we just focus on the numerator.

3x - 2(x^2+16)^1/2 = 0
3x= 2(x^2+16)^1/2
(3x)^2= (2(x^2+16)^1/2)^2
9x^2 = 4(x^2 +16)
4x^2 + 64 -9x^2 = 0
(5^.5x-8)(5^.5x+8) = 0

therefore x = 8 / (5^1/2)
then we do the 1st derivative test
- +
T'<---------------8 / (5^1/2)------------->

by the first derivative test 8 / (5^1/2) is a min

then we plug in the root to the parent function to get 3.894
therefore they need three hours to minimize the time they take to walk

After the group work Mr. K gave us derivative functions which we need to find its parent function.

F(x).................f(x)

x^3.................x^4/4 + C
-cos(x)............-sinx + C
sinx..................-cosx + C
sec^2x............. tanx + C
1/x....................lnx + C
e^x....................e^x + C
2^x...................2^x/ln2 + C
x^1/2................2/3x^3/2 + C

Well thats the class...

Remember
This December,
That love weighs more than gold

the next scribe is steve

Applications of Derivatives Acrostics

2005-12-19T15:31:00.000-06:00

Here's the new set of acrostics for you.

Blogging Prompt
Your task is to create an acrostic "poem" that demonstrates an understanding of applications of derivatives related to any one of these concepts:

DERIVATIVE
OPTIMIZATION
ANTIDERIVATIVE
MEAN VALUE
INFINITY
APPLICATION

As an extra challenge (worth an additional bonus mark) try to make a Double Acrostic, that is, each line should begin and end with a letter of the word you are working with.

Remember, this is a bit of a race. Your answers have to be posted to the blog in the comments to this post. If someone has already used a word or phrase in their acrostic you cannot use the same word or phrase. i.e. It gets harder to do the longer you wait. ;-)

Here is an example of an acrostic that Mrs. Armstrong wrote:

Always in 2 dimensions
Region between the boundaries
Entire surface is calculated
Answer is in units²

Be creative and have fun with this!!

Sunday 3x the Funday!

2005-12-18T09:25:00.000-06:00

A triple header this weekend.

Mr. Zhong Kui will make you laugh. I think his "problems" are the easiest ones to solve.

Rat is another "escape" puzzle. Every time you do something wrong he squeaks.

No. 5 is a set of three puzzle/adventures to get a little boy out of trouble.

Have Fun!

Antiderivative

2005-12-15T18:57:00.000-06:00

In today’s class, we went to @_@group work.
Will’s pet Bob~^.^~
Will the wallaby has been looking for the perfect pet for a very long time and he has finally found it – Bob the boa constrictor. Now Willy has to make a close, rectangular cage for Bob but it has to be 4000 cubic feet in volume and the length has to be 5times that of the height of the cage. The material to make the cage costs $0.25 per square foot.
1) What are the dimensions of the least expensive cage?
2) How much does it cost?
Given V=4000
L=5h
V =L * W * h
4000 = 5h w h
4000 = 5h^2 w
W= 800/(h^2)

Closed rectangular cage
S. A = 2 L h + 2 L h + 2 h w
S.A = 2 (5h) w + 2(5h) h + 2 h w
S.A = 10 h w + 10 h^2 + 2 h w
S.A = 12 h w +10 h^2

Cost =1/4 surface area
Cost = 1/4 (12 h w + 10 h^2)
C (h) = 3h (800/h^2) + 5/2 h^2
C (h) = 24000 h^-1 + 5/2 h^2
C’ (h) = -24000 h^-2 + 5h
= -5 h^ -2 (480-h^3)
- - +
--------------1--------------------1---------------------
2 roots: 0 3root480

By the first derivative test, cost is Mimi zed when h = 3 root 480 feet

L= 5 h= 39.15 feet
W = 800/h^2 = 13.04 feet
Cost = 3h (800/h^2) + 5/2 h^2 = $ 331.05

Antiderivatives
Given the function f, determine a function F whose derivative is f. That is, F’ = f. If F’= f, then the function F is called an antiderivative of f.

A function f has an antiderivative it will have a whole family of them and each member of the family can be obtained from one of them by the addition of an appropriate constant.

If f’(X) = 2x, find different possibilities for f, the parent function.
1) f (x) = x^2
2) f (x) = x^2 + e
3) f (x) = x^2 + 10^100
4) f(x) = x^2 + 1 …

Next scrible Prince. +_+

M V T (Mean Value Theorem)

2005-12-14T17:43:00.000-06:00

Now that the plague of Related Rates problems has ended (i don't know with the others but it was for me... laughs!) , a fresh new topic has emerged from its shell for us students to focus on for the next couple of days. I found Optimization problems much easier than related rates... or is it too early to judge?

This morning was the start of the Solving sessions in groups with the new topic Optimization. We had to work on Scuba Steve's Shark Cages. Given the total perimeter of the rectangle 450 feet, we had to divide the whole area into two equal parts so that the sharks wouldn't kill each other. The question is, what is the maximum area of each section of the cage that Scuba Steve can build?

First, know your constraint and optimization equations. From the constraint, solve for one variable, whichever is easier, the length (L) or the width (W). Use this to solve for the Area of the rectangle. Then, get the derivative of that function. After that, do the first derivative test and there you have it, a good start on Optimization problems. ;)

We also started on a new one today, the M V T. Most Valuable Tlayer? (laughs!) Nope, it's the MEAN VALUE THEOREM. M V T states that if a function is continuous on the closed interval [a , b] and differentiable on (a , b), then there exists at least one number c in the open interval (a , b) such that:

f ' ( c ) = f ( b ) - f ( a )

b - a

So given a function and its interval, get the values for intervals a and b. Then use the MVT to get the slope of that interval. Next, get the derivative of that function and equate it to its slope. So at X = c , the slope at c is then equal to the slope of the secant line in the interval a , b.

Sounds easy? Let's just wait and see. :D
Meanwhile, the next scribe is Xun. ;)

OPTIm-EYES-a-SHUN

2005-12-13T19:52:00.000-06:00

Is it just me or just the word Optimization tells us to really use our eyes, OPTI, EYES. So therefore we should use more PICTURES. Shun, means to avoid. How about avoid just reading the problem. Don't just read it, do more with it.

=D

Hello to everyone.

Today’s class we started off with going over questions from homework. Exercises 5.4: 1-9. The problems Mr. K went over were #4, #5, and #9. Hope no one else had any problems with the homework.

TIPS for OPTIMIzation:
(optimi-sm: to look on the best side of things)
1) Remember your geometric formulas.
2) Math is the science of patterns. Always look for patterns.
3) If you ever get stuck with your geometric figure, it’s never wrong to just drop a Cartesian plane, it might even help you see something you didn’t see before.

After going over the three questions from homework. Mr. K put up three optimization problems on the board. The first one being the classic optimization problem.

1) A sheet of cardboard 3 ft by 4 ft will be made into a box by cutting equal size square from each corner and folding up the edges. What will be the dimensions of the box with largest volume?

In each optimization problem, there will be constraint equation we will be working with. The classic optimization problem up there, the constraints were the dimensions of the box and what they did with it (cutting equal size square from each corner). Some will give you a function as a constraint. Read the problems carefully. Write down what is given to you and draw a picture to understand of what is being asked of you to find.

For homework today is the rest of 5.4: 9-18. Hope everyone does ok with the rest of the problems.

The homework question for today is:

A cylindrical can is to hold 20 pi meters cubed. The material for the top and bottom costs $10 per meter square and material for the side costs $8 per meter square. Find the radius and height of the most economic can.

Good luck.

Sunday Jumping Funday!

2005-12-11T00:00:00.000-06:00

'The goal of the puzzle is to switch the the pegs on the left with the pegs on the right by moving one peg at a time.

Move pegs by clicking and dragging them to open slots. A peg may only be moved to an open slot directly in front of it or by jumping over a peg to an open slot on the other side of it. You may not move backwards. The game ends when you win or get stuck.'

Play the game here. Can you win the 8 peg game? ;-)

(Thanks again to Think Again!)

Not much happening

2005-12-08T22:50:00.000-06:00

In todays class, we didn't learn anything new really until closer to the end of the class because Mr. K wanted to keep the talking to a minimum in todays class. So, he gave us a few questions to do, and he gave us a good deal of the class to do them as well. After wrapping up the questions and going over them with us, he began to talk a little bit about more of the applications of the dirivative. We didn't really get very far into it so I do suppose that he will continue this conversation in class tomorow. I wish that I had more material to write on but alas, perhaps next time.

C.O.O.L.

2005-12-05T23:54:00.000-06:00

You know what's cool? Here it is, I never knew Mr. K never cared what are marks were. It was something I never thought he'd say. I understood what he was saying that our marks will only get us IN university and after that well it doesnt matter anymore.What matters the most is that if you have a deepp enough foundation to move forward. Once we're in there if we learned nothing then there is only one way for us, it is downwards. In other words, marks are not important it is better to have a low mark with a high understanding of the concepts rather than to have a high mark but really just know how to do it not why we did the problems.

Convey and Describe

2005-12-05T23:03:00.000-06:00

Our class today was not about learning new stuff, it was about understanding the deeper concepts of what we learned before. Basically it was a day for enhancing our knowledge about the language of calculus.

We learned about the Extreme Value Theorem which states that if a function is continuous and is in a closed interval it has to( Mr. K repeatedly said it has to meaning its important) have a global max and min either in the critical points or the endpoints.

We also learned the 1st Derivative Test which is used to determine if a root of the derivative is a max or a min in the parent function. If the derivative is negative on the left and positive on the right of the root the then it is a local min if the left is positive and right is negative then it is a local max.

Critical points are on the parent function( max or min or where it is undefined)

Critical numbers are the zeros or root in the derivative function

The first example we had is to find the critical points of the given function.
f(x) = x^2 -4x +5
f'(x) = 2x-4
= 2(x-2)

the critical number is x = 2

to get the critical point we get the value of 2 in the parent function. Which is 1
therefore the critical point is (2,1)

we describe this by saying that by the derivative test that there is a min at 2 and the value is 1

Again all we talked about today wasn't something new. All we are learning now are concepts of what we discussed before. The point is we need to know how to convey and describe what is being asked and not just push some numbers in the paper.

The next scribe is.....
CJ

wild crazy stuff!

2005-12-05T20:23:00.000-06:00

In this chapter I have experienced one of the easiest and one of the hardest topics ever that I have encountered in Calculus.

Getting the derivatives of polynomials are not much of a tedious task, although a lot of times, I miss seeing my mistakes so I hardly perfect them. :) Memorizing the derivatives of such trigonometric functions is quite tough but it does not really blow my mind. Chain rule? well... I still forget steps when doing it, but at least I know how to do it.

But if I see a number with a long paragraph beside it... my heart beats faster.. then I FREAK OUT! Yes Mr. K, YES! I am still lost when it comes to related rates. I tried answering 4.6 the best way that I can but I am definitely, positively and undoubtedly sure that only 1 out of 10 numbers will I get correctly. Not that I am being pessimistic, it's just that, for me, IT IS REALLY, REALLY HARD. But I'm still trying, even until now. I keep on reading and reading until the whole problem finally gets stuck in my head.

So, that's about it. I hope we all pass guys...

(...Lord please enlighten our wonderful teacher, that he may be easy on us tomorrow in our test...)

- just kiddin'! I know that the harder the problems we take, the better knowledge we gain. :D

GOODLUCK everyone! ;D

Blogging On Differentiation rule

2005-12-05T20:03:00.000-06:00

In this Chapter, I found the most difficult stuff was relates rates problems.
We need to read the question and find a formula, or even probably two relate to the information that question told us. We need to know everything that might relate to the question, and solve the problem.
Exp: Water is flowing into a cone-shaped tank at the rate of 5 cubic inches per second. If the cone has an attitude of 4 inches and a base radius of 3 inches, how fast is the water level rising when the water is 2 inches deep?
The problem is given us the rate of change volume, dv/dt =5, r = 3, and asked to determine the rate change of height when h is 2, dh/dt when h=2.
We know the formula for conical volume is: V = 1/3 pai (r)^2 h
By similar triangle we have r/3=h/4,
So, r=3/4 h
V=1/3 pai r^2 h
V=1/3 pai (3/4h)^2 h
V= 1/3 pai 9/16 h^2 h
V= 3/16 pai h^3
Defferentiate implicitly:
dv/dt = 3/16*pai* 3h^2 (dh/dt)
5 = 3/16 pai 3(2)^2 (dh/dt)
20/9pai = dh/dt
The water level is rising at a rate of 20/9pai inches per second when the water is 2 inches deep.

Blogging on Related Rates

2005-12-05T17:54:00.000-06:00

Instead of reflecting, I'm going to list some information that might help you understand differentiating functions.

Constant Function f (x) = c f'(x) = 0
Constant Multiple Rule d/dx [c*f(x)] = c*f'(x)
Sum and Difference Rule d/dx[f(x) + g(x)] = f'(x) + g'(x) you can replace the + signs with a - sign
Power Rule d/dx(x^n) = nx^(n-1)
Derivative of e^x d/dx(e^x) = e^x
Exponential Function d/dx(a^x) = (ln a)*a^x
Product Rule d/dx[f (x)*g (x)] = f (x)*g'(x) + f'(x)*g (x)
Quotient Rule d/dx [f(x)/g(x)] = [g(x)*f'(x) - f(x)*g'(x)]/[g(x)]^2
Derivatives of Trigonometric Functions
d/dx (sin x) = cos x
d/dx (cos x) = -sin x
d/dx (tan x) = sec^2 x
d/dx (csc x) = -csc x cot x
d/dx (sec x) = sec x tan x
d/dx (cot x) = -csc^2 x
The Chain Rule [fog]' (x) = f'(g(x))*g'(x)
d/dx (ln x) = 1/x
Local linearization of f at x = a f (x) ~ f(a) + f'(a)(x-a)

All of these derivative rules are probably going to be on the test, so it is a good idea to learn how to use them. I excluded related rates because that depends on the question, and there is no definite formula for all related rates problems. These formulas and derivatives cover most of the unit and they are good to know.

Cubeoban Sunday

2005-12-03T22:37:00.000-06:00

The objective of Cubeoban is to push/pull all the blocks to their corresponding lights. Do this by clicking on the blocks and drag them in the direction you want to push them. Play it here.

Level 1 was so easy that even I could do it. Level 2 (the image), started my thinking.

(Thanks again to Think Again!)

Friday's Pre-test

2005-12-03T09:26:00.000-06:00

Like other unit before the test we had a Pre-test, in order for us to get ready for the test on Tuesday. The test is consist of 5 practice multiple type of questions and one Open response question. In today's pre-test we got this question:

The sides of this rectangle increase in such a way that dz/dt=1 and dx/dt=3dy/dt. At the instant when x=4 and y=3, what is the value of dx/dt?

First we create a relationship in about this problem. and we got this z^2=x^2+y^2
then we find the derivative of that relationship.
2z dz/dt=2x dx/dt + 2y dy/dt. The two are reduces z dz/dt= x dx/dt + y dy/dt.
By the pythagorean theorem we found that z=5. So after that we just plug in the values.
(5)(1)=4(3)dy/dt+ 3 dy/dt
=15 dy/dt / 5
= 1/3 dy/dt
Since dx/dt= 3 dy/dt we got the value of dx/dt= 1.

Now our pre-test was done all we have to do is study for our test on Tuesday. Good Luck guys.... The next scribe is Steve....

Tangent Line Approximations and Newton's Method

2005-11-30T20:32:00.000-06:00

I haven't exactly been helpful at all. I started something and I slowed down. I guess that's what happens sometimes. I don't know wether to say sorry or not. But here's one of the missing scribes.
Wednesday, November 30, 2005

We learned half of the last section of Chapter 4.
First area it covers was on Linear Approximations.
If we were given a function.
For example f(x)= square root(x).
And we were to find an approximation value of f(26).
What do we do then?
In this particular example, we find the closest perfect square.
Which happens to be 25.
What if we were given other kind of functions?
You try to find a coordinate that you know, and is close to the value you are finding.

Now we have a set of coordinates (25, 5).
Since 26 is close to 25, we can use a tangent line to find f(26).
Differentiate f(x).
f(x)= x^(1/2)f'(x)= 1/2(x)^(-1/2)
We then find the slope at our given coordinate.
f'(25)=1/2(25)^(-1/2)f'(25)=0.1 or 1/10
Now we have all the "ingredients" to put into our slope-point formula.
y-y1=m(x-x1)

y-5=0.1(x-25)

So basically, that equation up there is all we need to find f(26).
All we have to do now is solve for y when x=26.
In the end we get, y= 5.1 or f(26)=5.1

Let me sum up the steps of Linear Aprroximation.
1) Identify the function.
2) Identify what value you have to find.
3) Think of a set of KNOWN coordinate (x, y), close to the value you have to find.
4) Differentiate the function.
5) Find the slope of the tangent line at the KNOWN coordinate.
6) You have the slope and set of coordinate points, plug it into the slope-point formula.
7) Let x=the value you have to find, and solve for y.
8) *optional, but recommended. Do a check. You never know you might have made a mistake somewhere along the way.

That day we were also introcuded to Newton's Method.
Newton's Method is a method of finding zeroes of a function. You are given a function, an interval or point where the xero is close to. Zeroes of Tangent lines are then used to determine if its the zero of the function as well.
At the start we are given f(x) and (Xo, f(Xo)). From the given you can find f'(x) and the derivative at the given point.
y- f(Xo) = f'(Xo)(X-Xo) <----- slope-point formula
Since we are finding a root of the tangent line, we let y=0. From this equation we are trying the find the x value of the root. So we re-arrange the equation.
0- f(Xo) = f'(Xo)(X-Xo)
-f(Xo) / f'(Xo) = x - Xo
Xo - [f(Xo) / f'(Xo)] = x

x would then be the first x value.

When using Newton's Method it may take multiple times till you finally get close to a good approximation of where the zero of the function is. That's why doing this method by hand is very very very very painful. Thank You for our stupid calculators. In our calculators we have a program stored called "NEWT". It is used for this kind of method. We can also do this method manually, as showed during next class.

Hope this was some info. Hopefully is useful for someday reference. Sorry it was very very very very very very late.

Newton's Method

2005-11-30T14:59:00.000-06:00

One application of derivatives it to use the equation of a tangent line to approximate values close to a known value on a curve. i.e. finding √35 on the function f(x) = √x.

Another use of derivatives is to find a tangent line whose root is the same as the root of a more complicated funtion. Every approimation we make gets us closer to the actual root of the more complicated curve. This is known as Newton's Method and it's named after the man who first developed the technique, me! No, just kidding, Sir Isaac Newton developed the technique. Watch how it works here and here.

Here is a flash tutorial that explains the process we discussed in class today. You can watch Newton's Methis in action using this example. And then try these exercises to see how well you can apply what you've learned; detailed solutions are provided.

Tangent Line Approximation

2005-11-29T12:02:00.000-06:00

Stan the Clause - another one of the usual related rates problem Mr. K always gives every morning. As to what I did, I never preassumed that a wide set of concepts are to be considered in answering these kinds of problems. Claiming to that perception, I twisted my mind a little bit to come up with not "The Answer" but something just "close" to it. And to my dismay, I got way far from it. (laughs!)

A lot of mathematical ideas come into use when these kinds of questions are given:
- Find the rate at which the angle of elevation from the naughty boy's house to the flying can is changing in radians per second when the angle of elevation is pi over 12 -

To make life easier, visualize the whole problem. Try to illustrate. Know all the given facts or numbers. Then, give it your best shot! ;) It is just like solving an everyday math question, only it is more complex. Students might have to apply all the stuff they have learned throughout their mathlife. Regarding this problem, I was not able to get this one completely because I never thought this would include trigonometric concepts and all. I was stuck when I came to that point because I have slightly forgotten some stuff about it. So now I think I might have to work on refreshing myself with trigo.

In the remaining minutes of the period, we learned a new lesson that is the Tangent Line Approximation. This is basically applied in (as to what Mr. K used as an example) finding the value of a square root function with a non-perfect square number.

Given f(x) = √37 find the value

First, formulate its derivative equation - f ' (x) = 1 / (2 √x)
Then, obtain a perfect square number closest to the given number, in this example the closest is 36.
Substitute the x variables in the derivative equation by 36 and by doing that, the slope at that point will result. - f ' (x) = 1 / (2√36) = 1 / 12
Use this slope to form a derivative function for the given number 37 using the point-slope form. Y - Y1 = m (X - X1)

-> f(X) - 6 = 1 / 12 (37 - 36)

So, f (37) = 1 / 12 + 6 = 73 / 12

And using the calculator, the √37 is 6.08333 that is, almost as close to 73 / 12

Derivative Acrostics

2005-11-28T22:03:00.000-06:00

Kudos to Mrs. Armstrong for turning me on to the idea of acrostics in math.

Blogging Prompt
Your task is to create an acrostic "poem" that demonstrates an understanding of calculus related to any one of these concepts:

DERIVATIVE
POWER RULE
PRODUCT
QUOTIENT
CHAIN RULE
TANGENT LINE
NEWTON'S METHOD

As an extra challenge (worth an additional bonus mark) try to make a Double Acrostic, that is, each line should begin and end with a letter of the word you are working with.

Remember, this is a bit of a race. Your answers have to be posted to the blog in the comments to this post. If someone has already used a word or phrase in their acrostic you cannot use the same word or phrase. i.e. It gets harder to do the longer you wait. ;-)

Here is an example of an acrostic that Mrs. Armstrong wrote:

Always in 2 dimensions
Region between the boundaries
Entire surface is calculated
Answer is in units²

Be creative and have fun with this!!

Sunday Connected-Slide Funday!

2005-11-27T13:48:00.000-06:00

ConSlide puzzles are a new type of sliding block puzzles invented by M. Oskar van Deventer... The pieces of a ConSlide puzzle move like regular sliding block puzzles, but some of the pieces have sections connected by bars of various heights. This means pieces can pass over and under one another as long as the bars and posts don't run into one another. The goal of each of these puzzles is to move the red block to the upper left corner. Play it here.

(Once again, thanks to Think Again!)

Friday's class...

2005-11-27T11:20:00.000-06:00

Last Friday we started our class by having our second review quiz in preparing for the Test in May. We're given 8 minutes to answer 4 multiple type questions. Then after that Mr. K showed us the rubric for the Prince of Calculand Project. He discuss it to us for a few minutes. After that Mr. K talked about the Editor's Initiative and he asks our thoughts about it.

After all the discussions we had about the project and the Editor's Initiative, Mr. K showed us and taught us how to answer this related problem that we have been working for a few days.

A spotlight at a school dance is fastened to a wall 8 m above the floor. A girl 1.75 m tall moves away from the wall at a speed of 0.75 m/s.
a. At what rate is the length of her shadow increasing?
b. At what speed is the tip of her shadow moving?

First in solving in this problem like other kind of problems is drawing a diagram on what is given to you.

Then next is analyze the drawing. In our diagram there are two similar triangles and based on that we could create a relationship between the two.
b/1.75= a+b/8 (a+b=c as shown in the diagram)

8b=1.75a+1.75b
6.25b=1.75a

Then find the derivative of that,
6.25 db/dt= 1.75da/dt
Since we know da/dt= 0.75 m/s, juss plug it in and we could figure out db/dt and justplug it in.

db/dt=1.75(0.75)/6.25
=0.21 m/s
So the length of the girl's shadow is increasing 0.21 m/s.

Then in order to know what speed is the tip of her shadow. We could use this relationship c=a+b and find the derivative of that,
dc/dt=da/dt+db/dt and plug in the values to get dc/dt
dc/dt=(0.75)+(0.21)
dc/dt=0.96 m/s

The tip of the girl's shadow is moving 0.96 m/s.

That's all we did last Friday....The next scribe is Ara..I'm willing to accept any constructive criticism so don't be afraid to leave a comment. Thanks!

The Editor's Initiative

2005-11-26T21:19:00.000-06:00

Instead of posting a pre-test reflective comment on your progress in this class you may undertake The Editors' Initiative. Here's how it works:

  Step 1: Scan through the previously posted Scribe Posts on the blog. Find one that has one or more errors.

  Step 2: Discuss the error(s) and what you think the correction(s) should be with me. If I agree with your editorial proposal go to Step 3.

  Step 3: Discuss the editorial change with the author of the post. The author will chose to proceed in one of the following two ways.

3a	3b
The Editor is briefly allowed administrative privileges on the blog. They will edit the post to make any necessary corrections. They then sign the post at the bottom: Edited by: [name] on [date]	The author will edit the post in consultation with the editor who will vet the author's changes until they are correct. The author then signs the post at the bottom: Consulted editor [editor's name] on [date]

Students may chose to make more than one edit. Each additional edit will earn them a bonus mark on the next test. Your mark on the previous test determines the maximum number of edits/bonus marks available to you.

Mark on Last Test / Max Edits Allowed
> 90 / 1

80-89 / 2 (1 bonus mark)

70-79 / 3 (2 bonus marks)

60-69 / 4 (3 bonus marks)

50-59 / 5 (4 bonus marks)

40-49 / 6 (5 bonus marks)

30-39 / 7 (6 bonus marks)

20-29 / 8 (7 bonus marks)

10-19 / 9 (8 bonus marks)

0-9 / 10 (9 bonus marks)

You may also assume the role of Content Consultant to earn marks as outlined above. Here's how it works:

  Step 1: Scan through the previously posted Scribe Posts on the blog. Find one that doesn't provide enough detail or leaves out too much information. Decide what additional content should be added.

  Step 2: Discuss the new content you think should be added with me. If I agree with your editorial proposal go to Step 3.

  Step 3: Discuss the editorial change with the author of the post. Together, you will chose to proceed in one of the following two ways.

3a	3b
The Content Consultant will add a new post to the blog inserted at the appropriate time and date. They then sign the post at the bottom: Additional Content by: [name] on [date]	The author will edit the post to include the additional content provided by the consultant. Additional content will appear under a heading "Additional Content". The author then signs the post at the bottom: Additional Content Provided by [consultant's name] on [date]

Students may chose to make several additional content contributions for bonus marks according to the table above.

As we discussed in class, you can edit your own scribe posts only if the whole class agrees with your proposal and we are convinced that you tried to do your best work the first time around. When it's your turn to be scribe try to write a post that is so excellent no will be able to edit it. ;-)