<body><script type="text/javascript"> function setAttributeOnload(object, attribute, val) { if(window.addEventListener) { window.addEventListener('load', function(){ object[attribute] = val; }, false); } else { window.attachEvent('onload', function(){ object[attribute] = val; }); } } </script> <div id="navbar-iframe-container"></div> <script type="text/javascript" src="https://apis.google.com/js/plusone.js"></script> <script type="text/javascript"> gapi.load("gapi.iframes:gapi.iframes.style.bubble", function() { if (gapi.iframes && gapi.iframes.getContext) { gapi.iframes.getContext().openChild({ url: 'https://www.blogger.com/navbar.g?targetBlogID\x3d14085554\x26blogName\x3dAP+Calculus+AB\x26publishMode\x3dPUBLISH_MODE_BLOGSPOT\x26navbarType\x3dBLUE\x26layoutType\x3dCLASSIC\x26searchRoot\x3dhttp://apcalc.blogspot.com/search\x26blogLocale\x3den_US\x26v\x3d2\x26homepageUrl\x3dhttp://apcalc.blogspot.com/\x26vt\x3d-7891817501038621673', where: document.getElementById("navbar-iframe-container"), id: "navbar-iframe" }); } }); </script>

# AP Calculus AB

An interactive log for students and parents in my AP Calculus class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.

## Sunday, April 30, 2006

### Your second wind--

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?

## Friday, April 28, 2006

### From ZZZZZ's to A's

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

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??

## Wednesday, April 26, 2006

### Your self portrait!

How is your self portrait coming?

## Monday, April 24, 2006

### Visualizing excellence in calculus!

"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

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

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

## Sunday, April 23, 2006

### Steve's Blog Assignment

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

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

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

## Thursday, April 20, 2006

### Sarah's Blog Assignment

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.

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
or
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.

## Wednesday, April 19, 2006

### The Main Thing

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?

## Thursday, April 13, 2006

### endings-- beginnings--

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

"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*

## Sunday, April 09, 2006

### A New Feed Window

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

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

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!)

## Wednesday, April 05, 2006

### Hot Coffee, give it a few minutes

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

## Sunday, April 02, 2006

### Roboclaw Sunday!

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??

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