December 29, 2006
His story is spellbinding. There's a really important message for you in the last two minutes ... but you won't really understand it unless you watch the whole thing.
Click on the picture. (18 min. 48 sec.)
December 28, 2006
Rather than simply receiving financial support, winners of the TED Prize are granted something extraordinary: something which children dream about, but which adults assume is merely the stuff of fairy-tales.
They are granted three WISHES to change the world.
They may wish for anything. And TED will seek to make their wishes come true.
Bono won the TED prize in 2005.
Rock star and activist Bono accepts the 2005 TEDPrize with a riveting talk about our moral obligation (and economic incentive) to help lift Africa out of poverty. He unveils his TEDPrize wishes by challenging the TED community to help build a social movement of more than one million American activists for Africa; to tell people one billion times about the ONE campaign; and to connect every hospital, health clinic, and school in one African country, Ethiopia, to the Internet.
Click on the picture. (28 min. 37 sec.)
For more: Read the update on Bono's wishes.
December 27, 2006
Click on the picture. (3 min. 40 sec.)
December 26, 2006
The pictures are just stunning.
Click on the picture. (18 min. 42 sec.)
December 25, 2006
Watch how he displays and talks about statistics. He'll make you laugh and he'll make you think.
December 22, 2006
Thanks to everyone who helped put this together. I found this to be a great experience for me as a teacher I hope it was also valuable to you as a student.
Anyway, here it is, version 1.0 fresh out of beta. ;-)
Flickr Assignment Rubric
It is paramount that the picture be in tune with the purpose of the assignment. It should show, first of all, the student's understanding of how the photo is related to mathematics. The hot spots are important too, because that's essentially your way of teaching other people. Creativity is a factor, because keeping one's interest in the photo contributes to the learning process. Finally, the picture quality should be kept in mind too. If we can't see the picture, it's going to be hard achieving all the other requirements.
The picture must be tagged properly with the course tag and assignment tag. If tags are misspelled or no tags are present the photo cannot be graded and will receive a grade of ZERO. Not tagging your photo properly and accurately is analogous to not handing in your work or not putting your name on it.
| Classification ||Mathematical Content (50%)||Hot Spots (35%)||Photograph (15%)|
| Level 4 ||Packed with mathematical concepts/facts. (Minimum 7 concepts/facts.)||All hot spots accessible; i.e. "smaller" hot spots are "on top" of larger ones, they do not obscure each other. All hot spots are actually labels and relate to parts of the photo (not on blank space with filled in notes). One or more hot spots include a link to a relevant supporting resource on the internet. Minimum 7 hot spots.||In focus or appropriately focused for effect. The subject of the picture occurs "naturally," it is not a contrived shot. Really makes the viewer "see" math in a place they hadn't realized it existed. (Example: trigonometry)|
| Level 3 ||Significant number of concepts/facts included. (Minimum 5 concepts/facts.)||All hot spots accessible. Most hot spots are actually labels and relate to parts of the photo. Not more than one hot spot on blank space. One or more hot spots may include a link to a relevant supporting resource on the internet. Minimum 5 hot spots.||In focus or appropriately focused for effect. The subject of the photo has been "set up" or contrived yet still illustrates math found in "the real world." (Example: derivative)|
| Level 2 ||Some effort to include content evident. (Minimum 3 concepts/facts.)||Most hot spots accessible. Most hot spots are actually labels and relate to parts of the photo. More than one hot spot is on "blank" space. May or may not include links to relevant supporting resource on the internet. Minimum 3 hot spots.||In focus or appropriately focused for effect. Although it is a "real world" picture, objects have been used to "draw" the math. An obviously contrived shot. (Example: trigonometry)|
| Level 1 ||Very scarce content related to assignment.||Less than three hot spots are visible or have information related to the theme of the assignment.||It is evident that little effort went into finding and shooting a picture that reflects the theme of the assignment.|
| Level 0 ||Content unrelated to theme of assignment.||No hot spots or mostly unrelated to the theme of the assignment.||Out of focus and/or otherwise difficult to look at.|
Creativity (up to 5% bonus)
The maximum possible mark for this assignment is 105%. You can earn up to 5% bonus marks for being creative in the way you approach this assignment. This is not a rigidly defined category and is open to interpretation. You can earn this bonus if your work can be described in one or more of these ways:
- unique and creative way of looking at the world, not something you'd usually think of;
- original and expressive;
- fresh and unusual;
- a truly original approach.
December 21, 2006
December 20, 2006
I think the problem for me was the arithmetic part. I would get the right idea, but I would write the equation or something like that wrong. The optimization problems were alright i guess. I liked the mean value theorem. It was the easiest to understand. Antiderivatives were tricky. I found it alright. Mr. K explained it a WHOLE lot better than in the textbook. I'm not a textbook kinda person. I either get confused or can't get interested in it. The holidays played a part in my laziness this past unit. School just sort of drags on because you can't wait and thats BAD because thats when things get crazy! I had like three other tests to study for. It's like they all plan to make it like that- to get you all busy, worried, and crazy. And it doesn't help when people, like my bonkers family, getting even more crazy over stuff they should have did a long time ago. Kind of sounds familiar, doesn't it... XD. Good luck to you all and don't forget C.
December 19, 2006
(1) Review what we've done so far.
(2) Take notes
(3) Do the assigned problems
(4) Don't stress out =)
That's all for my bob!
Multiple Choice Part
1) The Graph of the function f(x)= 2x5/3 - 5x2/3 is increasing on which of the following intervals?
- We get this by differentiating f into f'(x) = (10/3)x2/3 - (10/3)x -1/3.
- Then we can factor it into f'(x) = (10/3)x -1/3(x-1) by taking out the lowest common factor to find its roots in order to make a number line
- We now know that there is a root at 0, and 1, so where ever it is positive, the function f is increasing.
2) Let f(x) = x5 - 3x2 + 4. For how many inputs c between a=-2 and b=2 is it true that f(b) - f(a) / b - a = f'(c)
The answer is 2.
- First we'd find f(b) and f(a), which equals 24, and -40 respectively.
- Then plug it into the equation to get f'(c) to equal 16.
- Then differentiate f to get f'(x) = 5x4 - 6x.
- And since we're looking at the point f'(c), and we know what the point is, we get 16 = 5x4 - 6x.
- Then we'd just solve for the roots and we get the answer for how many we would have.
- I'd just punch in the equation the calculator and look at it over the [-2,2] interval and look at the zeros.
3) The table below gives some values of the derivative of a function g.
Based on this information it appears that on the interval covered by the table...
The answer is that g has a point of inflection.
- From the information, we know that g will be increasing everywhere, since it g' is positive throughout the interval.
- But we notice it g' starts to decrease after increasing, so at the highest point before decreasing there is a horizontal tangent line which means that there is a point of inflection because there is a change of sign from a positive to negative rate of change.
4) Suppose f is a continuous and differentiable function on the interval [0,1] and g(x) = f(3x). The table below gives some values of f
What is the approximate value of g'(0.1)?
The answer is approx. 3.84.
- For this we use the chain rule law.
- We know that g(x) = f(3x).
- So then we find g'(x) = f'(3x)(3) by the chain rule.
- After just plug in the values to get g'((.1)) = f'(3(.1))(3).
- Which turns into g'((.1)) = f'(.3)(3), then solve using the symmetric difference quotient.
5) If f(x) = ln(x) - k√(x) has a local minimum at x = 4 then the value of k is:
The answer is 1.
- If x = 4 is a minimum on the graph of f, then f'(4) = 0 must be true.
- So differiantiate f into f'(x) = 1/x - (1/2)k(x)-1/2
- Then plug in the 4 f'(4) = 1/4 - (1/2)k(4)-1/2
- Now solve for k.
Long Answer Part
1) Let f be a function given by f(x)= (3x - 2) / ( √(2x2 + 1) )
a) Find the domain of f.
The answer is (-∞,∞)
When deciding domain, we need look only at the denominator. And in this function the denominator will never equal 0, meaning it's never undefined which means the function won't have vertical asymptotes. From that we know that the domain includes all the real numbers.
b) On the graph below, sketch the graph of f.
- We can sketch this by first looking where the function has roots. Roots occur when the function equals 0, or undefined. Look at both the numerator and denominator for roots. In this function the denominator can never equal 0 because of the square root function, which means there is no vertical asymptote. But in the numerator we notice that there is a root at 2/3.
- Then we can look at where we have y-intercepts, which occur when x=0. By plugging the 0 in for x into our function, it is easily located.
- Next, we can look at if theres any horizontal asymptote. And there is since the highest degree of power in the numerator is equal to the degree in the denominator. Horizontal asymptotes are discovered by the limits of infinity concepts. From this information we can just take the leading coefficient in the numerator which is 3, and divide it to the leading coefficient in the denominator which is √(2), so there is one at 3/√(2). And we can't forget about the -∞ side, to find that there is another horizontal asymptote at -3/√(2).
- Then we need to find out where the function is increasing and decreasing, and its concavity. For this we need the derivative. But the derivative was already given in part (d). So from finding the root to be -3/4, we make a number line to see that it is, decreasing to the left of it, and increasing to the right, which means there's a minimum at that point.
c) Write an equation for each horizontal asymptote of the graph of f.
The answer is lim x->∞ f(x)= 3/√(2) and lim x->-∞ f(x)= -3/√(2).
- Applying the infinity concept to our function. The -2 and 1, in the numerator and denominator respectively are insignificant to infinity, so we disregard it.
- So the denominator can be rewritten as √(2) * √(x2), which is √(2)*(x).
d) Find the range of f. [ Use f'(x) = 4x + 3 / (2x2 + 1)3/2 to justify your answer]
The answer is [ -17/4√(17/8), 3√(2) ).
- Since there's a minimum at x = -3/4, we can just plug in that value in our function to get the minimum output.
- Our maximum output is approaching 3/√(2) but never actually touching because of the asymptote there, found in part c).
Click here (part 1, part 2)to look at the paper with the correct circle answers and other multiple choices.
And that was it for the whole day. *Whew!* I'm a bit still unsure about parts of the long answer part, sooo if you're confused, I am too =). I tried my best to answer these problems in full detail as much to my understanding. Our group did pretty poorly on the test afterall. I'm scared for the test *gulp..* Anyway.. next scribe for Thursday will be Danny. Later days!
December 18, 2006
EDIT * nevermind I'm not ready. after that pre-test...boo it's 11:30 pm right now STUDY TIME.
I really enjoyed this unit. We did a lot of problem solving and group work for this unit. I found the problem solving the fun part in this unit. It was also fun to hear and learn with my fellow classmates. We had some disagreements and we had some clueless moments. We were there to learn as a group and we were there to help clarify anything that we found challenging. At this point in the course, i have realized that the way we think is different. At the beginning of the course, we were learning the basic idea about derivatives. Here we are now learning how to find anti-derivatives and we are learning how to solve optimization equations. In summary, i have not had much difficulty with this unit. I think it is understanding what steps to take to solve an equation/question. An example, solving word problems involving optimization.
Today, we continued our discussion on Antiderivatives. Here we go...
Last Friday, we started our discussion and we came up with the first antiderivative rule.
An antiderivative is basically just DIFFERENTIATING the function backwards. The rule for finding the FIRST derivative from the PARENT function has been reversed.
For example: Find the antiderivative of f'(x) = x.
Applying the rules for the antiderivative, we came up with the answer. If your wondering where the "C" came from, good for you! C is an unknown constant in the parent function. We don't have any evidences that a constant exists within the parent function.
Note: Remember that the derivative of any constant is 0.
For this example, we were given a point in the graph, (1,2). Using this information, we can identify the unknown constant, "C", by substituting the values into the parent function.
Then, we discussed the other antiderivative rules.Note: Remember that the derivative of cosX is -sinX. Therefore the "antiderivative" of "sinX" is the negative of "cosX".
Note: "Anything you can do, you can Undo..." (Mr.K) Every single derivative rule we have learned has an equivalent antiderivative rule!
How do we know if the antiderivative is correct? We can DIFFERENTIATE the antiderivative we came up with and see if the answer is the given question. It is the same as checking a set of factors by distributing the factors and see if the answer is the same as the given function.
f(x) = 2x^2 - 4x
Note: "Any high school level functions CAN be differentiated or antidifferentiated..." (Mr.K)
Then, we reviewed the "Definite Integral".
If you're wondering why antiderivatives have a stretched "s" like a definite integral, congratulations! The format of the antiderivative itself is formally known as an "Indefinite Integral". It's the opposite of a definite integral because it has NO limits/intervals (the number at the 2 ends of the "s").
Here are some examples we discussed:
If your wondering why number 3 has no answer, THERE IS NO ANSWER TO IT for now. No one knows the antiderivative for this function. If anyone answers this, that lucky person will be famous and his/her name will be written in the history books! Cool eh? XD
Lastly, we discussed this concept:
Correction: "no inflection points [f"(x) not equal 0]"
We explored the concept of this function. We found the first and second derivatives. Then, we used the first derivative to find the critical number, -b/2. Using this critical number, we can explore the graph of the function. Using three values, b>0, b=0, b<0,>
Whew... that took a while... XD. Anyways, I hope I covered all the things we did today.
Reminders: Pre-test tomorrow! Put up your BOB's before Wednesday.
Oh and BTW, Manny is scribe for tomorrow. XD
December 16, 2006
In order to solve these kind of problems, you'll need three things:
-the values of the closed interval plugged into the equation
-the Mean Value Theorem
HMWK is Exc. 5.6 Questions 1-14
Also, if you haven't already done so, comment on Christian's story http://pandaproject-apcalc06.blogspot.com/2006/11/pedro-pi-rate-panda-part-1.html for the Panda Project.
And, think about your integral pictures...
Last but not least, JANN is tomorrow's scribe =)
December 14, 2006
topher and sully decided to go cut down a Christman 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 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?
then we did examples on the board
linger your next
December 13, 2006
Today we started off with 2 groupwork sheets based on optimization problems. The first sheet was pretty straight forward and I believe every group got it right. =)
Then came the second sheet!!! It was much more difficult than the first and wasn't as straight forward.
GROUP WORK SHEET #2: Willy's Pet Bob
2. Willy the Wallaby has been looking for the perfect pet for a very long time and he has finally found it - Bob the boa consticter. Now Willy has to make a closed, rectangular cage for Bob but it has to be 4000 cubic feet in volume and the length has to be 5 times 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?
NOTE: Click on pic. and it will be easier to read.... =)
Then we went over the "CLASSIC OPTIMIZING QUESTION" that we didn't get to do yesterday...
You are standing at the edge of a slow moving river which is one mile wide and wish to return to your campground on the opposite side of the river. You can swim at 2mph and walk at 3 mph. You must first swim across the river to any point on the opposite side. From there you will walk to the campground which is one mile from the point directly across the river from where you start your swim. What route will take the least amount of time?
NOTE: THE RED BLOTCHY NUMBER IN THE NAVY BLUE BOX IS #3...
Homework: finish up 5.4 Excericises
The next scribe is Charlene ;]
December 12, 2006
We started the class of with questions on the board like always...
An open rectangular box with a square base is made from 48 ft squared of material. What dimensions will result in a box with largest possible volume?
A sheet of cardboard 3 ft by 4 ft will ne amde into a box by cutting equal sized squares from each corner and folding the four edges. What will be the dimensions of the box with the largest volume?
A cylindrical can is to hold 20π m cubed. The material for the toop and bottom costs $10/m squared and the material for the sides cost $8/m squared. Find the radius, r, and height, h, of the most economical can.
Mr.K was trying to be ambitous like he always is and tried really hard to squeeze in the "classic" optimizing question but we were saved by the bell.
Homework is ex 5.4 questions 10-18
Test is on Friday! So study!!!
Anyways I was so hungry I decided to draw this..=)
Next scribe : Charlene =)
December 11, 2006
To start things off, I would like to remind you guys that Mr.K gave us an extension for our derivative photos. You have until the end of the week to make it great.
Next, the white board questions! I like to draw weird cartoon things. If find it easier to read things when there's something drawing your attention to it. So, I drew. XD. I also use sticky notes...and I don't like typing much either -_-;; I prefer paper and pen...yes, anyway, moving on..
1. Evaluate these limits:
2. Determine the total number of max and min points of the function whose derivative is:
3. Build a rectangular pen with 3 parallel partitions using 500 feet of fencing. What dimensions will maximize the area of the pen?
4. An open rectangular box with square base is made from 48 ft^2 of material. What dimensinos will result in a box with largest possible volume?
CONSTRAINT: 48 ft^2, surface area
*NOTE: We'll work on this problem tomorrow!!
5. Find the point (x,y) on the graph f(x)=root of x closest to the point (4,o)
Next scribe is ANH!!!!
December 10, 2006
We started off today’s class by going over Friday’s Quiz. I think the first question was quite straightforward so I won’t get into that.
2) Use linear approximation to find the value of (1.001)^100. I had trouble understanding this question, but after Mr.K explained it, it made much more sense. The important thing to recognixe is that you have a value very close to one that would be easy to solve for.
1) Find a function to represent the question. We also find the derivative of the function, so that we can find the equation of the line later.
2) Choose a simple number close to the value we’re approximating. In this case, 1 is the easiest value. By finding the f(1) and f ‘(1), we can use the point-slope formula to find the tangent line.
3)Now we can plug in the value we’re approximating, 1.001, and we get an answer of 1.1, which is a good approximation of the actual value, 1.1051.
3) Using 3 iterations of Newton’s Method, by hand, find the root of f(x)= x^5-x+1.
Start with x= -1.
Newton’s method uses progressively more accurate tangent lines to approximate the root of a function. First of all we need to have f(x) and f ‘(x) so that we can find the tangent line.
The second step shows the familiar point-slope formula. Because the root is where y = 0, we can get rid of y, and then rearrange the equation to solve for the next x-value.
The first iteration is done by finding f(-1) and f ‘(-1). Then plugging these into the rearranged formula, we have the next x-value. We repeat the process using this next value instead of x=-1, and so on until reaching however accurate a solution we want.
From here we moved on to the lesson itself, all about finding horizontal and vertical asymptotes and using derivatives to find out what's happening around the asymptotes.
1) Draw a graph of f(x)= 1/(x-3)
Because the limit of the function as x approaches 3 is one divided by zero (undefined), we know that there is a vertical asymptote at x=3. If the limit were zero divided by zero, it would be indeterminate (impossible to know what is happening).
By finding where the function is increasing/decreasing and concave up/down, we can determine what the graph is doing on either side of the asymptote. We know that there are no roots because the numerator always exists. So we get a graph like this:
2) Draw a graph of g(x)= 1/(x^2+1)
Once again, the numerator always exists so there are no roots. And there are no vertical asymptotes because the denominator will always exist. By finding the limit as x approaches infinity, we can determine that there is a horizontal asymptote at y=0.
Technically, we should also find the limit as x approaches negative infinity, in case the graph is doing something else on the other side of the x-axis, but this is a nice happy function so we don’t bother.
From here it’s more or less the same as the previous question. Finding g’ and g”, we learn key information about the graph that allows us to draw it. For example, we know there’s a max at (0,1) because g’ changes from positive to negative at x+0.
This graph is a very well-known one in mathmatics, called the 'Witch of Agnesi". To find out why, look here: http://mathworld.wolfram.com/WitchofAgnesi.html
3) Next we looked at (2x^2)/(x^2-5x+6).
If we try to find the limit as x approaches infinity, we will get ∞/∞, which is indeterminate. In this case, to find the horizontal asymptote we divide each term by the highest power of x in the function. There are three possible outcomes when we do this:
1) If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at y=0.
2) If the degree of the numerator is greater than the degree of the denominator, the horizontal asymptote does not exist.
3) If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the leading coefficient of the numerator divided by the leading coefficient of the denominator.
It is probably easier to memorize these three rules, but if you think you will forget them over time, it is probably better to remember that dividing by the highest power of x will allow you to find the horizontal asymptote. That's what Sherlock Holmes would do, anyway. :)
In this problem, the third case applies. So we get an asymptote at 2/1.
If the numerator had been 2x^3, the second case would apply and the horizontal asymptote wouldn’t exist.
And if the x^2 in the denominator had been x^3, then the horizontal asymptote would have been at y=0.
So that’s about it. Just remember...
By finding the limit as x approaches infinity you can find any horizontal asymptotes.
If the limit as x approaches some value a approaches infinity, there's a vertical asymptote.
And Jupiter has 63 known moons.
Next scribe shall be… Anh.