A window through the walls of our classroom. This is an interactive learning ecology 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.
In this episode of Student Voices three Advanced Placement Calculus students, Chris, Craig, and Graeme, talk about a wiki assignment they did to prepare for the exam. Then the conversation transitions to a discussion of the many things they learned while doing their Developing Expert Voices project. It ends with a challenge, the result of which will be featured in a future podcast.
Let Chris, Craig, and Graeme know what you thought about the podcast by leaving a comment here on this post.
I have to say I feel more confident going into this test then I did going into the other test earlier today. I get how these problems work most of the time, just using the unit logic and the idea of taking one of the parameters and chopping it up into little bits. I grok the average value of a function idea. I waver a bit on the solids but really they are not terribly hard. I think I will do fairly well tomorrow. G'night.
Here is my scribe! Supposed to be spinning solids but I think it made my head spin more than anything. Got 4 out of the 7 wrong according to the sheet but I worked hard and if I had time I would go through them with a fine tooth comb but cannot do that today (Plus number 7 was just mean).
Well I'm looking forward to this test! I think that this one is definitely aceable. I feel fairly comfortable with all the work that we have done except for maybe that brutal sheet of antiderivatives but we were assured that things like those wouldn't appear on the test so it's all good.
Hello everyone! We started off todays class with a quick talk about the software out there to solve integrals and how we are better. A program called Maxima can find the answer to solvable integrals but they will be messy answers usually and we can do better! So we did as shown below... (Sorry for the terseness of the slides, it's not my style but I really don't have the time for all that tonight as it got busy real fast)
Hello everyone! Feeling mighty good about this upcoming test! I do so enjoy math when I get the concepts. That workshop we had today really re-affirmed this. Hope the tests will be as easy as that. Well I'm off to do some chemistry (on that note, Mr.K that whole thing on half-lives last year applies really well to first order reactions in chem).
Sorry for the late bob, boot sector viruses suck. Well I'm feeling pretty good for this unit, even the pretest felt pretty good. Just waiting for that related rates problem today! Figure I can get an 80 atleast.
Well, I gotta say that this unit has been tough for me. Have to chalk that up to only getting to about 10 to 15 done in the homework, but it took me a long time and you just lose your will after a while. Well have started getting back on the horse the last couple of days and have finished up a portion of the homework I haven't done. Coupled with some hardcore studying tonight, tomorrow at my spare and lunch I figure I can pull an 80. I really do get the straight up rules for the derivatives, just identifying what I have to do in question is still my problem. Well good luck every1.
Hello everyone and welcome to my scribe! Enjoy (the angled words and other formatting came out a bit funky, but it's still easily legible except for maybe that second slide)
The next lucky recipient of the scribe will be Robert since he'll be back tomorrow, I think, if he isn't it's Craig, if Craig somehow doesn't get to class, it's Van, if not Van then it will be Sandy. Think I've covered the bases there.
Well I have to say I seem to get this unit fairly well. Enough so that I feel confident that I will get a good mark on the test. Especially after that pretest, felt pretty good throughout that. Must say this has been far easier than grade 12 so far. I wish you all luck on the test!
Alright a pretest scribe! I guess I have some explaining to do.... about the questions on the pretest!
Question 1) If f(x) = ln(sqr(x)) then the average rate of change on the interval [3,7] is... then 5 possible answers. Using Roberts method you punch the equation into the calculator and use that handy-dandy slope program and out pops the answer or you can do it manually by hand and come up with A) 0.106
Question 2) Suppose that the number of bacteria in a certain organism grows over time and the number, N(t), of bacteria (measured in thousands) at any time is given by:
N(t) = t(2 + cos(t))^(4/3) + 3t
At approximately what time ,t, in the first ten days is the colony growing the fastest?
First and foremost I must remind you all that IN THIS CLASS CALCULATORS SHOULD BE IN RADIANS! It is truly *very* annoying to miss a mark because of your calculator...
Now entered into your calculator you turn on that oh so useful derivative-of-whatever-you-have-in-your-Y1-slot graph and you look for the peak! (Then use that maximum function to give you a value)
The answer is.... 5.18!
Question 3) Limit as x approaches one of:
(2x^2 + x - 3)/(3x^2 -x -2)
is equal to.... ?.
Well first you try putting one in right away and find your dividing by 0! Algebraic massage required... These both factor readily and two of the terms reduce. Then 1 can be entered in and you end up with 5/5 or 1... which is the answer!
Question 4) The graph of the second derivative of a function f is shown at the right. Which of the following statements are true (I realize that there is no graph there but there but there is on this slide)
I) The graph of f has an inflection point at x = -1. II) The f graph is concave down on the interval (-1,3). III) The graph of the derivative function f prime is increasing at x = 1.
Then you find which of those statements is true. I) An inflection point on a second derivative graph would be a zero and there happens to be one where they say that there is an inflection point therefore it is true! II) The second derivative measures concavity, since it is negative on the interval that they say then the statement is true! III) The second derivative is a measure of the slope of the first derivative. Since the value is negative the slope of that point is negative so it is decreasing at that point making this statement false.
So the answer is I and II are true!
Question 5) Suppose a function f is defined so that is has derivatives: f'(x) = x^2(x-1) and f''(x) = x(3x-2) Over what intervals is f both increasing and concave up? Well for it to be increasing the first derivative must be positive. Finding the roots of that we find them to be x = 0 and 1. Then do a sign check by entering a value before, in between and after to find when it is positive. The process is repeated for the second derivative because where that is positive the graph is concave up. We find the overlapping positive areas to be when x > 1!
Question 6) a) Condider the following table of data. Estimate f'(5.2) as accurately as possible. Well f' would be the slope at that point! Since they are on equal intervals and there are values before and after 5.2 we are able to use the symmetric difference quotient to get a more accurate result. It is found by finding the slope of that point with the points before and after and averaging it which results in you getting -9/4.
b) Write and equation for the slope of the tangeant line at that point. Well... you are given a point.... you are given a slope... so let's use point slope form shall we? and you end up with y - 8.8 = -2.25 (x - 5.2)
c) Use your answer in part (b) to approximate f'(5.26) Since the two points are so close together we can just plug that value into the equation we made to get a fairy accurate answer... which ends up being 8.665.
d) What is the sign of f''(5.2)? Explain. Well this one I got wrong. I drew a graph where this point really looked like a point of inflection and that's what I based my answer on. WRONG! Mr.K ends up saying something along the lines of "Using the algebra to prove a point is far better then the graphical approach and will be expected of you on the exam". Since f' is decreasing to the left and right of this point we can deduce that this point is also decreasing meaning the graph is concave down meaning that f'' is negative at that point! (Though I still think it could be a point of inflection, but I do see the logic in what was said).
Well my first bob of the year... I must say I have been finding the base concepts of this unit fairly easy. But my problem is with the recognition of those base concepts. They are as straight forward as can be in the text but I know that Mr.K is never that straight forward. I just have to work on identification of what is given in the question and use the knowledge that I'm comfortable with to get the questions. If I can do that this test will be a cake walk.
ok so the red graph is the original, the green graph is the first derivative and the teal graph is the second derivative. The purple graph is unrelated.
Well there's the slideshow, didn't upload like it was on the smart software and I tried in 5 different formats... If you want some more practice I attempted to make an applet in JAVA. I must say it's one of the ugliest computer programs I've ever made and it was a heck of a time making it work online (even though it is still sketchy even now seems to work all the time on ei6 and from time to time completely kill firefox). For any computer programmer people out there who look at my code I know it could have been done better with arrays but I remembered that JAVA arrays were kinda funky compared to VB so I didn't bother re-reading that part of my book and just avoided it. Here is the program (hopefully) the code is under it for anyone else who wants to compile and use it (but I doubt it). I'll try to make a smoother, better looking, more complete(more angles) and more stable program some other time. I didn't this time because there are just too many little things that went wrong. Spent an hour debugging and the bug was a single equals sign. ugh.
final String answer1 = "1/2"; final String answer2 = "sqr(2)/2"; final String answer3 = "sqr(3)/2"; final String answer4 = "1/sqr(3)"; final String answer5 = "1"; final String answer6 = "sqr(3)"; final String squareRoot = "sqr(";
final String question1 = "What is the sine of pi/6?"; final String question2 = "What is the sine of pi/4?"; final String question3 = "What is the sine of pi/3?";
final String question6 = "What is the cosine of pi/6?"; final String question5 = "What is the cosine of pi/4?"; final String question4 = "What is the cosine of pi/3?";
final String question7 = "What is the tangeant of pi/6?"; final String question8 = "What is the tangeant of pi/4?"; final String question9 = "What is the tangeant of pi/3?";
String[] answers; String[] questions;
public void init(){ mTracker = new MediaTracker(this);
setLayout(null);
submit = new Button("Submit"); submit.setBounds(20,100,100,40); submit.addActionListener(new ButtonSubmitListener());
Today started off with an in depth study of our homework. The first question we looked at was well... number 1...
The point of doing number one was to show how to find a good domain in a problem that does not give any concrete numbers to work with. The example was talking about the number of daylight hours per day. Mr.K then dived right into the in depth analysis of the problem. We discussed the day with the most daylight (Summer solstice), the day with the least light (Winter solstice) and the days which the amount of daylight is equal to the amount of night (cannot remember what they were called). With this information in hand and the assumption that we are in Saskatchewan and at the equator to simplify this we delved into finding a suitable domain and range. Since the cycle comes full circle from summer solstice to summer solstice it was decided that a year would be a fair domain. For range we estimated that the daylight would shift by three hours off dead even at the solstices (at this point we also realized that this is a sine or cosine graph pretty much straight out of last years questions). So we found the domain and range and the shape of the graph which satisfied the question. But Mr.K is not satisfied with teaching us merely the rudiments of what we need to know so to the weather network statistics page we went and we observed all the different patterns Winnipeg goes through.
The second question that we moved onto goes as follows:
An open box is made by cutting squares of side x from teh four corners of a sheet of cardboard that is 8.5 by 11 inches and then folding up the sides.
a) Express the volume of the box as a function of x.
No without crunching numbers terribly much we came up with the domain. We knew that it could not be 4.25 inches or more because we'd lose the third dimension therefor losing the volume and you cannot cut 0 or less because you cannot cut negative distances and if you don't cut at all you don't have a third dimension and therefor no volume.
From there we continued to make x useful and make it the size of our cut. From there we deduced that the 8.5 inch side must be 8-2x because two sides will be cut from it and same with the 11 inch side as shown in the diagram below.
Then using what we know about volume, we found the function that related x to the volume. All we had to do is multiply the L x W x H. So our function ended up as f(x) = x(8.5-2x)(11-2x). (Though the slides show it as 8 - 2x).
Then Mr.K got rid of a wasp.
We then moved onto the golf ball question which ended up looking very similar to many questions we have answered in the past years and we quickly caught the drift of it. The only thing that was a little bit tricky was the fact you had to look back in your work for the answer (view slide 2).
The last question that we handled from our homework was as follows:
A wire 6 meters long is cut into twelve pieces. The pieces are welded together to form the frame of a rectangular box with a square base. a) Define a function that relates the height of the box to the length of one edge of the base. Well first off we knew that all the sides of the base and top must be the same because it is a square and the four sides connecting the squares must be the same length as well, so our equation for the perimeter became 4H + 8L = 6. b) Define a function that relates the total surface area of the box to the length of one edge of the base. Well we quickly came up with a formula for the surface area 2L^2 + 4LH = SA. But we had to express it in terms of one length of the base. So finagling with our first equation we isolated H and substituted it in and took it as solved (slide 3).
From this point on we were just drawing graphs... The only one that was tricky was the graph of f(x) = sqr(-x-2). Within it you have to factor out the -1 in both terms or else when you do your flip then translation it will be wrong.
Remember to do your homework! If I can this weekend I'll update this to try out a JAVA applet I'll create (found a couple stumbling blocks). Have a good weekend!
