Friday, June 26, 2009
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!
Farewell, Auf Wiedersehen, Adieu, and all those good bye things. ;-)
Tuesday, June 16, 2009
Friday, June 12, 2009
Without any further ado, here are the results of our class's survey. Please share your thoughts by commenting (anonymously if you wish) below .....
The questions in this section were ranked using this 5 point scale:
|Strongly Disagree||Disagree||Neutral||Agree||Strongly Agree|
The bold numbers after each item are the average ratings given by the entire class.
1. The teacher was enthusiastic about teaching the course. 4.89
2. The teacher made students feel welcome in seeking help in/outside of class. 4.33
3. My interest in math has increased because of this course. 4.11
4. Students were encouraged to ask questions and were given meaningful answers. 4.56
5. The teacher enhanced the class through the use of humour. 4.33
6. Course materials were well understood and explained clearly by the teacher. 4.44
7. Graded materials fairly represented student understanding and effort. 4.11
8. The teacher showed a genuine interest in individual students. 4.33
9. I have learned something that I consider valuable. 4.78
10. The teacher normally came to class well prepared. 4.33
Overall Impression of the Course
The questions in this section were ranked using this 5 point scale:
|Very Poor||Poor||Average||Good||Very Good|
1. Compared with other high school courses I have taken, I would say this course was: 4.56
2. Compared with other high school teachers I have had, I would say this teacher is: 4.78
3. As an overall rating, I would say this teacher is: 4.78
1. Course difficulty, compared to other high school courses:
2. Course workload, compared to other high school courses:
3. Hours per week required outside of class (n = 8):
|0 to 2|
|2 to 3|
|3 to 5|
|5 to 7|
4. Expected grade in the course:
AP Exam Preparation
The bold numbers after each item are the average ratings given by the entire class.
How prepared were you to write this exam? 76.5%
How much effort did you put into preparing for this exam? (n = 8) 81.9%
How good a job did your teacher do preparing you for this exam? 95.0%
Did you have enough preparation using your calculator?
Did you have enough preparation without using your calculator?
Was your teacher too hard or too easy on you?
[Ed. Note: Numbers in parentheses indicate the number of students, over 1, that gave the same answer.]
What was your best learning experience in this course?
|There is more than one way of doing things.|
Instead of memorizing, each student gained the ability to find patterns.
The creative projects.
Developing Expert Voices project (3)
The AP Exam.
Any time spent on the SMARTboard. (4)
All group work in class.
The Final Take Home Exam.
Learning techniques to help make learning easier in the future.
Train of thought.
Being able to "think" in calculus.
What was your worst learning experience in this course?
|Difficulty at the beginning of the course.|
The constant changing of teachers at the beginning. (2)
Our substitute. (2)
Everything at the beginning.
When Mr. K. wasn't here.
What changes would you suggest to improve the way this course is taught?
|More students going up to SMARTboard to solve questions.|
I felt satisfied with what happened in this course.
The teacher could be more prepared for class.
Getting marked tests back. (2)
Make sure to get everyone in class involved.
Really spend a lot of time on concept building.
Continue to ensure the concept of a derivative is firmly understood.
Better room ventilation.
More wiki assignments.
Make everyone goes up to the SMARTboard at least once.
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 ....
Wednesday, May 27, 2009
Monday, May 11, 2009
As you know, everyone is working on their DEV's however, due to May-Madness, which includes the arrival of AP exams including Bio, Calculus, and Chem, our DEV's progress has hit a snag. Although I only speak on behalf on Team FLJ, I can undoubtedly say that all other groups have been hit with scheduling crisis's.
I myself have been kept busy with Career Internship, work, and projects from other classes. Unfortunately, all the due dates and AP exams are clustered up in the same region, leaving little or no room for flexibility. Other members of my team have also been busy with their own studying, individual class projects and others with myself included took part in this year's Fashion Show.
- Calc exam = May 6
- Fashion Show = May 7
- Bio Exam = May 11
- Chem exam = May 12
- English project = May 15
- Career Internship Portfolio = May 20
Tuesday, May 5, 2009
Today was, THE LAST DAY before the exam. I still can't get that thought out of my head. It's so close. It seemed like only yesterday that we first started the class.
BEFORE we officially started the class, we were reminded of the really important rules we should follow to avoid automatic failure. As well as that, we were given the following tips since our exam would be running from 8 AM [be there by 7:45] until noon.
Make sure to:
- Sleep reasonably early, preferably before midnight-- don't DARE to do last minute cramming, it's way too late to do that, but sleep EARLY to wake up EARLY.
- EAT a really healthy meal before [not immediately before] your bedtime and have a really hearty breakfast. Eggs are good brain food, perhaps pack a granola bar, dipped in chocolate. Get Twizzlers, if you wish. HAHA
As for the actual exam itself, it will run approx. 4 hours in four sections.
- Multiple choice, calculator [50 mins]
- Multiple choice, non-calculator
- Free response, calculator [45 mins, 3 questions]
- Free response, non-calculator [45 mins, 3 questions]
NEVER recopy something that is already defined in the question. Also, don't feel the need to show off your algebra skills. This is already expected of you. Plug things into your calculator for efficiency. Don't do more than you're asked to do.
Finally, we started on the actual review section of the class. It was sort of a race, and whoever had the answer buzzed in with the phrase "Got it."
There was an exquisite blend of both calculator and non-calculator questions. We had fun with it, and I don't know about everyone else, but I somewhat forgot why we were doing it. We were all about business, but simultaneously having a good time. Especially when playing Jenga with the pile of Twizzler packs on my desk.
The questions were pretty straight forward multiple choice, and by now, we should know what we're doing given the circumstances of each question. We might not be as speedy as Benchmen, but I think we've proven that we could get there. I believe in you guys and I just don't feel like explaining slides right now. haha. I will though. I think.
I'm going to follow the advice given to us, and rest up, get nourished and do some light reading and test myself and see if I know what I'm doing.
But for now, while I'm on "standby" mode, enjoy this vid. It's not precisely funny, but it was part of our school's production of The Wiz. AWESOME JOB and the rest of the videos are on the channel.
The sound isn't really that good, and I think the footage is from the first matinee show. But still I miss the days where I eased on down the road. xD
...and here is one last kicker for you guys. This song is called "Believe In Yourself". I guess it's a really nice inspirational message before something like an exam comes up, don't you think?
Monday, May 4, 2009
We started with:
This was easily found with the use of a calculator. But if not given use to a calculator, it's good to know how to solve this question.
We found the antiderivative of the given function over the interval of 0 to pi and replaced all the values of x with the values of the interval to find the difference in area. Now if no calculator is given, it's simple, just simply simplify it to the simplest form. Then repeat for the other question.
Our next question had to deal with the same functions as the last, but we have to find the integral equation of the volume of the shape over the same interval of 0 to pi, but when it's generated over the x-axis, and then the we found the volume when generated over the y-axis.
Now we cut this shape into tiny cylinders and find the volume of each cylinder, given by V = (pi)r2h but because we are dealing with 2 different functions, and finding the difference in the volumes, each graph has its own radius. The volume is then simply stated as V = pi(R2 - r2)h, we use an infinite amount of cross sections, and because of this the height is quiet small, its so small that we call it dx. Now we throw the functions' radius' into the equation and voila.
When revolving around the y-axis, it's the same deal, but with the use of a different equation. Note: I'm not sure what it is but I'll edit it in when I find out.
Our next question involved trapezoid sums.
The question shows a table with 9 sets of information when x is various values. We need to find the difference of area between the 2 given functions. Using a trapezoidal approximation, we find the upper limits and lower limits of f(x) and find the differences between these limits of g(x) and find the average. Quite simple.
Next question deals with finding the area of the region of a graph.
Now because the graph has a root between the given interval, we split the region into 2 seperate intervals. One finding the area of the region between the root at -1.100457 and -1.5 (I'm not sure why we're finding the area seemingly backwards on the interval) and we add this to the area of the function in the interval -1.100457 to 1.5
This is all we did in today's class. Review questions are in today's slides. Our exam is in 2 days and good luck to everyone. Next scribe will be .:. J + ME .:. or better known as Jamie.
Friday, May 1, 2009
Today we reviewed Techniques of integration.
The first one we looked at was the U-Substitution technique. This can be used when you are antidifferentiating a function that is composed of a factor that is a composite function and the other factor is the derivative of the inner function of the composite function.
I did this question on the smartboard. First I chose a part of the function to be substituted for U. I saw that the denominator and the exponent of e was root x, so I chose that one.
So I let u equal root x and differentiated u as shown in the image above. I also solved for solved for x using the substitution, just in case I needed it. So I substituted all values necessary and I got an antiderivative. Great we found an antiderivative, how do we know if it correct? Just differentiated the antiderivative that you found and if you get the function that was to be antidifferentiated, then you got the right antiderivative.
This question is similar to the previous one but this one is an integral. What is the difference you ask? An antiderivative will give you a function, an integral will give you a number. Even though they are different, they are also related.
To answer this question, just antidifferentiate the function and substitute the limits into the antiderivative to find the answer. This is the definition of the second part of the Fundamental Theorem of Calculus.
Oh no! There is an antiderivative that we've never seen before. What is the antiderivative of cosecant, let alone the antiderivative of cosecant squared. I can't remember how we thought of this but Mr K got us to differentiate Cotangent. Using the Quotient Rule because cotangent can be written as cosine divided by sine as shown in the image above using abbreviations. Don't try to memorize the antiderivative/derivative relationship. Just understand that you can easily build it when needed.
Mr. K began talking about the antiderivative of x to the power of some number times e to the x. He said that it was a telescopic function which means as you evaluate it, the function will get longer, but things can be reduced or factored in such a way that it will shrink like a telescope.
I was playing around with some examples and I discovered the pattern. It is such a cool antiderivative. I'll give an example:
I got to this part by using integration by parts. So f = x³, my f' = 3x², g' = e^x, and g = e^x. Notice that the second term is still an antiderivative that we cannot find easily, but it is a factor of 2 functions, so let's use integration by parts again. So we'll get:
Once again we cant antidifferentiate one of the functions so let's integrate by parts once more:
Now we can finally evaluate that antiderivative. Once you do that you'll notice that there is a common factor of e^x. So let's factor that out and you might see something completely amazing. I'll show you:
Okay look at the polynomial in the brackets. When have you seen those specific terms before? If you think and look carefully the consecutive terms are derivatives of the previous term. I tried this again with another power of x and I got the same pattern. So when you are asked to find the antiderivative of x to the power of some number times e^x, the antiderivative will be e^x times the power of x in the function being antidifferentiated minus the derivative of the previous function until you reach a constant. Don't forget the plus c.
tn-1 means the previous term.
Sorry if my algebra was bad on the image above. But if you try it yourself, it really is a beautiful thing :P
The next few slides on the slide show posted by Mr K is for homework. Study for the exam on Wednesday. Next Scribe will be Francis.
Here is the video of the Scribe:
These guys have very funny podcasts.
That's all folks.
Thursday, April 30, 2009
Sorry that this is very late, some urgent things came up and I wasn't able to get home until 1, so I'll just give a quick run down of what happened since everyone was in class.
So as you can see, to solve it, we found that the hypotenuse of the triangle at the beginning is 30 root two. we then found the remaining area near the end of the distance to be x since we do't know how long that distance was. Because triangle is a triangle with two sides equal, the top of it would be 30, thus the 170 - x (it's 170 because 200 - 30)
So that makes the second part of the equation for part a. The next slope we found one of the sides to be 45 feet. So then we found the hypotenuse to be root x squared + 2025. And now we can build our equation.
The RDT triangle is an example of how to change from rates, to distances, to times.
Unfortunately, this is as far as we got as we ran out of time.
Next scribe will be Ben
Wednesday, April 29, 2009
And here's the youtube video. Its the nom nom bunny we loved =).
And the next scribe is ...hmm..there's so many to choose from! Well..since it looks like its going to rain...and rain starts with the letter "r"...*Stares at scribe list*. I choose Rence, Lawrence!
Tuesday, April 28, 2009
Anyways, todays class was a relatively simple one, as for reasons unknown to me, Mr. K wasn't there. Instead we had a sub, who's instructions to us were to work on the questions we had been given yesterday (I think? I wasn't all to clear about which questions he was talking about.) So the majority of the class did that. Myself, Lawrence, and Francis used that opportunity to do a bit more work on our DEV and go over some questions from the text that we had been struggling with.
Overall the class was mostly geared towards review and practice questions in preparation for our ever looming AP Calculus Exam (May 6th D:)
So I think thats everything, and its good too, because now I must get ready to go to work :] See everyone tomorrow!
SO theres one swear in it, I didnt notice it until just now. Theres also some mild drug references, although they're in good nature. Basically, make sure theres no young children around and it should be fine :]
Edit: Oh yeah, Kristina's scribbbbeeeeee Sorry for the lateness. I forgot ^^;
Monday, April 27, 2009
As always we did a mini exam to help us review for the exam:
Find the derivative of both function, which V1=cos(t) & V2=-2e^(-2t)
It's in terms of V because the X1 and X2 are the postion fungction the derivative of postion is velocity which is V.
Then you let V1=V2, which is cos(t)=-2e^(-2t). You can graph both is the calacror or find the interstion point of the two functions. OR you can let the whole thing equals to 0, cos(t)+2e^(-2t)=0
then find the root of the function. The answer should be D)three
In this parblem you applie the derivative test rules. Graph the function, then find where the function have it's roots over the interval [-1,]3]. Then at the three roots find the point that is going from negative to postive, because it's a local minimum. Your answer should be E)2.507
From the graph, take a slice and look at it as a disk with a washer in it. Then , cos(x)=x which becomes 0=cost(x)-x. use the area formular the pluge everthing in to find the function . If you done everthing correctly you should end up with C)1.520
will this question camed up the third time, I think everyone is good with it. The answer is D)8647 gallons.
Next scribe is Justus.
Friday, April 24, 2009
This was a gimme question. Very straight forward. We just have to apply the average value formula:
This was a washer question. Ah good times right? "I don't see the hole!" In washer questions we just subtract big radius squared from little radius squared and integratate. Don't forget your pi.
After simplifying we realize that the function is just a line. So we know I is true since the limit from both sides will be equal since this is a line. II is also true because f(a) is defined. III however is not true because x cannot equal a. So there's a hole in our line.
There was a little typo with the questions here. But it's all better now. The table gives us numerical values so that was a hint that we numerically find the derivative at 1.2. We do that by finding the slope from the left and right of 1.2 by using the f(x+h)-f(x)/h formula.
So we first start by cutting the graph up into 4 intervals. The ending time was 24 and they want it into 4 intervals so 24/4=6. So our intervals are 6 hours apart. To find a midpoint Riemann sum we use the middle value of the interval as the length of the rectangle and of course width is the change in time.
K that's it folks. I think this is my last scribe post =(. The exams getting closer so don't forget to do your three questions a day. Hope you guys are enjoying you're long weekend. Next scribe will be Yi Nan.
Thursday, April 23, 2009
Wednesday, April 22, 2009
We had another exam practice session today. The question that we did was a very famous one from past AP exams. It was THE AMUSEMENT PARK QUESTION.
Here is how to get the answer for part a.
This part of the question was quite simple, but there was a part to this question that may mess up some people. Let me show you how to answer it than I'll talk about that tricky part. To get the answer integrate the entering function for 9 to 17 with respect to t. You do that because since the entering function is a rate (derivative) function and if you integrate a derivative you'll get the total change of the parent function which in this case is the total number of people that entered the amusement park. If you do that you should get 6004 people. Never forget the units. You will know what unit should be part of the answer if you understand the concepts and/or technique used to find the answer.
One tricky thing that people had trouble with was understanding what the question was asking for. Some people included the Leaving Function. If you do that you are solving for how many people were in the amusement park, not how many people entered it. Those are 2 totally different things since there are entering and leaving functions.
Part a) involved the process from part a) plus a little simple multiplication. Since there are 2 costs for tickets at 2 different time intervals, you will need to do 2 integrations. So integrate the entering function for the first time interval (from 9 to 17) to get the number of people and multiply by the cost to get the amount of money made for that time interval. Do that again for the second interval and add the results together and you should get the answer in the image above. Again, don't forget the units.
This part of the question involved an accumulation function. As you can see this accumulation function represents the total number of people in the amusement park over a time interval from 9:00AM to x o'clock because the function involves the integration of the difference of the Entering and the Exiting functions. The question is asking for the derivative of the accumulation function. So if you differentiate the function you will get the integrand of the accumulation function but respect to the variable limit instead because an accumulation function is a composite of functions. If you differentiate a composite of functions, you must apply the chain rule. The differentiation of the accumulation function above results in the differences of the Entering and Leaving functions which is the change in the number of people in the park. If you did the math correctly your answer should be the answer in the image above. For the Free Response questions on the AP exam, a word answer is required. The word answer should be very specific but not long because this is math class not english class.
The last part of this amusement park question is an optimization question where you are looking for what time is it where there is a maximum number of people during the open hours of the park. So what you have to do is differentiate the accumulation function from part c) and find where the resulting function is 0. A function has a maximum or a minimum where ever the derivative has a root or is undefined. When the critical number is found, do a line analysis of the derivative to find where the parent function is increasing or decreasing. If the function is increasing on the left and decreasing on the right of the critical number, the critical number found is a maximum. The time when there is a maximum number of people is about 15.7948 hours after midnight(as in 12:00 am of the current day).
The averages of past AP exams are found on the next slide and as you can see as a class (average) we are just above average which is good because that means we could be expecting a 3-4 on the exam.
Thats my scribe and sorry if it was horrible. Very busy since the APs are coming in about 2-3 weeks.
Remember, try to do 3 AP Free Response questions a night. Get constructively modifying the wiki questions.
The next scribe will be Joyce.
Here's the Youtube video:
Tuesday, April 21, 2009
- Exam Review Intro
- Related Rates: Area, Perimeter, and Diagonal of a Rectangle
- Related Rates: Shadows Including Similar Triangles
- Related Rates: Distance and Velocity and Spongebob!
Exam Review Intro
We do our exam review in three ways:
- Doing mini-exams pre-test style
- Doing questions in class related to rusty topics
- Doing old exam free-response questions and study how each question evolved throughout the years and do 3 questions a night
Related Rates: Area, Perimeter, and Diagonal of a Rectangle
The length (L), the width (W), and the rates of which the lengths (dL/dt) and widths (dW/dt) are changing.
By convention, we're going to designate an increasing rate as a positive rate and a decreasing rate as a negative rate.
a) We know the formula for the area of a rectangle as A = L * W, where A is area, L is length, and W is width. Since we're looking for the rates of change (That's the definition of a derivative!) we differentiate the formula, with respect to time.
Since L * W is a product, we use The Product Rule to differentiate.
Since area, length, and width are all with respect to time--meaning that area, length, and width are functions of time--we must use The Chain Rule to differentiate.
Answer: dA/dt = 32 cm/s^2; increasing
b) We know the formula for the perimeter of a rectangle as P = 2L + 2W. Differentiate the formula using The Chain Rule, since P, L, and W are with respect to time--a function within a function! Then plug and chug.
Answer: dP/dt = -2 cm/s; decreasing
c) We know that the diagonal (D), the length (L), and the width (W) are related in The Pythagorean Theorem: the square of two sides (in this case, L and W) equals the square of the hypotenuse (D), so D^2 = L^2 + W^2. Differentiate the formula using The Power Rule and The Chain Rule.
Answer: dD/dt = -33 cm/13 s; decreasing
Related Rates: Shadows Including Similar Triangles
We know the height of the lamppost (L = 16), the height of the man (M = 6), the rate at which the man walks toward the streetlight (db/dt = -5), and the length of the man's shadow from the base of the lamppost (b = 10).
Since the man is walking toward the lamppost, by convention, the rate is negative.
b) Using similar triangles, we can see that the ratio of L to M equals the ratio of s to b+s. We simplify our proportions and differentiate to determine the rates of change (That's the definition of a derivative!). To differentiate, we use The Product Rule and The Chain Rule--Refer to Related Rates: Area, Perimeter, and Diagonal of a Rectangle for reference. We plug in the numbers to obtain ds/dt.
Answer: -3 ft/sec
a) Note this question is underlined in blue. The exam would never ask you to do part B because you need to do part B anyways to answer part A. The rate of the tip of the man's shadow is dP/dt. To obtain dP/dt, we differentiate P = b + s.
Answer: -8 ft/sec
Related Rates: Distance and Velocity and Spongebob!
- Next scribe is bench.
- Wiki constructive modification due Sunday midnight.
- Developing Expert Voices projects due soon.
- AP calculus exam is in two weeks.
- Three AP calculus exam free-response per night.
- Homework: Olympic Spongebob!
- We will be reviewing applications of derivatives (related rates and optimization), applications of integrals (density and volume), and techniques of antidifferentiating (integration by parts).
I really need to stop with these late night scribe posts.
For the record, Joseph is scribe, and the list REALLY needs updating.
In this question, the answer should be straight forward. If you approach X from the left (low), then the largest integer value is 1. This is demonstrated in the first line. If you approach from the right (high), the largest integer value is 2. These are also known as roof and floor values. Because the right approach (floor) is not equal to the low approach (roof) then the value cannot exist, meaning the answer is (E).
For this question, we need to absolutely remember what the definition of a derivative is. Because this is it! Written right there. 1 is the value of f(x), which is also equal to sin(π/2). You will never actually see this, but you need to recognize it. Once you see it, this question becomes easy. The derivative of sin(π/2) is 0, because at sin(π/2) you are at the top of the curve and the tangent line (derivative) will be perfectly horizontal. Thus, the value must be zero, or (C).
Just keep deriving. You’ll get there eventually. (C)
A painfully simple solution to this question. To find a line, you need to have the slope and a point. You are given the point. Therefore, logically you should find the slope. Derive the equation of your curve, and you know that equation also equals the line with y = 1 and x = –1 (the equation for any line is y – y1 = m(x – x1). Solve for m with algebra. Once done that, do some rearranging with algebra, but the answer should become obvious as being (A).
Speed is the scalar version of velocity, which has vector (direction). When velocity is negative, you are moving backwards. Because speed is scalar and has no direction, it is the absolute value of velocity, so in terms of speed even if you are moving “backwards”, your speed will be positive because you are moving period. So if you re plot the absolute value of the graph, you will see the greatest value is at t = 8 which is answer (E).
If at t = 3 our object is at the origin, and its velocity is positive, it means the object is moving away from the origin in the positive direction for 3 units (the triangle formed by the graph from 3 to 6 is base 3 and height 2, so (1/2)2(3) = 3). So, we want to find where the object moves in the negative direction for 3 units, Since the value of the triangle created by the graph from 6 to 7 is base 1 and height 2, the area and total distance changed is 1 in the negative direction, meaning we havent gone backwards enough to be back at the origin. If we find the area from 6 to 8 then the base is and the height is 4, so by t = 8 you’ve moved 4 units negatively, which is past the origin, but we can logically deduce that at some point between t = 7 and t =8 we had moved 3 units negatively and thus returned to the origin. So, logically, the answer is (E).
Question (7) Free Response
For (a), you know S(0) = 6 so use that to solve for C. Quick math determines that C = 6 because e ^(0k) = 1 so 1C = 6. You also know that consumption doubles every 5 years, so S(5) = 2(6) and 2(6) = 6e^(5k) so e^5k = 2. Take the ln of both sides, so ln2 = 5k, and solve for k.
In (b), you want the integral from 3 to 13 (remember that your function counts years from 1980 so 1983 – 1980 = 3 as your starting value, and you want a 10 year period so 10+3 = 13 as your other limit) of the function. Calculator please. Really, you dont want to integrate by hand, but be my guest.
(d): “The integral of S(t)dt from 3 to 7 gives the total amount of cola consumed in the United States in billions of gallons per year during the time period from January 1, 1983 to January 1, 1987.”
Remember guys to start making your significant edits to the wiki solutions manual, and to at least try 3 exam questions a night. This is crunch time!
I dont really have a new funny/interesting youtube video, so here’s Intergalactic.
Monday, April 20, 2009
Is keeping first things first important as you come down this last stretch toward to AP exam? And if it is, how are you at making sure you tackle those 3 suggested review problems every night?
I’m hoping that in my sharing one of my experiences, you can find some bit to help guarantee your success on your test day! In early 2003, I too faced an important test date and I too appreciated the opportunity and felt I had to find a way to achieve. Achieving this goal was too important to me and my test if I passed would lead to a national teacher certification (only 40% of test takers passed). 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!!! Urgent “stuff” kept happening and I was always responding to that. Does that sound familiar?
I was overwhelmed!!!! I sat down and broke my test review into manageable chunks (I had 2 months until the test) 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 before but was always interrupted by the phone or my students on instant message with questions about our studies.
Then I looked at my other “stuff”, and categorized it: vital, important, or nice. Then I took the vital “stuff” and categorized them again: vital, important, or nice. And I let go the 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 students’ 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 goal. 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 all of us to some tips that are really helping you manage to put first things first?
Sunday, April 19, 2009
WARNING: CONTAINS SOME FOUL LANGUAGE
Almost forgot my youtube contribution. Thanks for reviving it Jamie :). And yeah, my reaction to the upcoming exam would probably be something like the Troll 2 scene XD.
And as usual, if you're too lazy to check the slides, the next scribe is Paul. Wiki questions are due at midnight tonight also, so get moving everyone.
Friday, April 17, 2009
Thursday, April 16, 2009
So Differential equations. For the most part, I found this unit quite easy conceptually. I understood most of what I was supposed to do and when and things, but I would always make stupid mistakes which cost me the whole question. I think this is mostly a matter of needing to do more questions so that I get out of the habit of overlooking those little bits. That said I don't think I did so well on the test D:
For the most part all I can really say about the test is I blanked. I was feeling pretty confident in the concepts and applications of them, and then as soon as I sat down to write the test, it was like someone opened a window in my brain, and whoosh, it all flew out. bleh, to say the least.
Anyways I'm looking forward to reviewing further, and covering all the loose ends I have with the course, as at this point in time it feels like I have alot ;P
But chyeah, Justus out for now.
Night all :]
I thought the test was quite easy. I completed the test with about 20 minutes to spare so I decided to do all of the free response questions. Yes, I did omit one of them.
How did everyone feel about the test?
The exam is coming in about a month. I feel pretty good but as a review I would like to go over the accumulation functions as well as the related rates problems. The related rates problems are giving me the hardest time in this course.
How does everyone feel about the exam coming up? Is there anything that you would like to go over?
OK I went through the Chapter yesterday, everything seems to be easy and also hard.
I'm having trouble with most of the word problem question.
In this unit you always have to think through the whole question and find the giving data.
It's really easy to miss some parts.
OK I have to o back to review now, good luck everyone!!!
The other thing that I had trouble with, was the basic math. Simple precal. I seem to be over thinking things. Not really anymore, since I've been practicing, so I feel really confident about this test. That part at least.
Newton's law. Again, the only thing was the basic math, but now, the bigger problem to me is how to set up the function. But seriously.. I've never felt better about any other test. Let's hope this feeling stays this way and hope that Euler's doesn't get out and take me down.
PS. I LOVE LEIBNIZ NOTATION!
The Eulers Method I will study up a bit on because I don't quite remember it... At all.
Newtons Method of Cooling I believe I had the easiest time with (being that it was taught last) and that I seemed to catch on how to do it. I felt comfortable doing the questions and they just seemed tedious, and sometimes mind wrapping.
Wednesday, April 15, 2009
We went into the velocity and acceleration problems next. It's much easier to follow the question by writing down everything that's given in the question, and from the initial information, finding hidden information by doing a bit of work.
Newton's Law of Cooling was not bad in class, but looking back I can't remember much about it. I still have trouble on when to use certain equations and using the calculator to solve certain problems, so I'm still sort of worried. I just wish we had a class today for review or maybe a formula sheet, I suppose this would help by refreshing my memory on the different equations. Good luck on the test everyone.
K so studying for the next 2 hours. I really wanna do good for this last chapter :(
Tuesday, April 14, 2009
Anyhow, I guess I'll explain the pretest. If you can explain something, your understanding of it can be deeper.
In this question, an equation is given for acceleration. As well as that, initial values are given to help determine the equations for velocity and position.
The objective of the question is to find an equation that follow the conditions of the initial values and to help find the correct equation, it's important to remember, that acceleration is the second derivative of position. Ergo, the method of finding the position under these circumstances is antidifferentiating; once to get the velocity and another time, to get the position. C is the correct answer.
Here, a derivative is given and it's asking for the AVERAGE rate of change. As soon as this is indicated, so is the pattern. The general formula of the average rate of change is 1/b-a multiplied by the integral of the function from a to b, with respect to x. From there, it's as easy as "plug and chug" to get the answer which happens to be A.
This is the multiple choice question that confused me. Here, it really helps to write down the information that the question is giving you and think of other ways to skin a cat. One of the things the class learned was why Leibniz notation is very much appreciated. The question tells us that the slope of the tangent line [derivative] of y is equal to ey.
From there, we separate the variables and antidifferentiate, of course, not forgetting the "plus C" for the constant. In order to determine what value the constant is, we plug in the initial value to find the constant and it results in finding a general differentiable equation-- the answer for this question; letter c.
This is the last of the multiple choice questions. Another question that is easier to solve in Leibniz notation. This may look like a mini monster of a question, but one of the first things to do is to algebraically massage the equation to separate the x's on one side and the y's on the other.
Then, anti-differentiate both sides and find y. Again, to determine the constant, plug in the initial values given in the question and plug in that constant in order to determine a specific function, rather than settling with a family of functions caused by that "plus C".
Finally, there was the three part open response question.
For the first part, I was ashamed of myself... I spent ten more minutes on this part of the question than I should have. All of the needed information was already given in the question. All I needed was to plug the values in.
In part B, I got the right idea but I didn't get the right variables for Leibniz notation and I felt like i was going in circles. But I HAD IT!!! This question is following a very similar pattern to those questions from the multiple choice. Isolate variables, integrate each side to get the parent function... or the family of parent functions and solve for the constant to get the explicit equation.
Another way of checking if the equation is correct is to plug in the initial values and see if they are equal to each other.
Finally, since we've found the parent function in part B, all we have to do for this question is plug in the time it takes and determine the velocity.
There's the pretest for you. A quick reminder that Mr. K mentioned that he will not be here tomorrow and on our test day.. which is the next day after tomorrow. There isn't going to be a class tomorrow so.... enjoy.. your hour? haha.
The rest of the slides are question that are probably homework/review questions in preparation for the test.
Also.. our assignment.. Claim your wiki-questions soon!
Next scribe is....Kristina.
ZOMG>>>!! I almost forgot about our youtube tradition [that was broken...]
This one is really.. just.. incompetent.. but I love this movie.. it's been so long.. since I've seen this.