AP Calculus 2008: Without Bound

So Long ...

noreply@blogger.com (dkuropatwa) — Tue, 28 Jul 2009 22:03:25 PDT

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

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

Just to have a laugh or learn some math,

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

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

So long everybody!

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

I Found This...

noreply@blogger.com (Benofschool) — Tue, 16 Jun 2009 16:39:47 PDT

This video reminds me of Mr.K's talks about technology and how it is evolving constantly:

Class Survey

noreply@blogger.com (dkuropatwa) — Fri, 12 Jun 2009 10:27:23 PDT

The exam is over and we did a little survey in class. The results are below; 9 students participated. If you'd like to add another comment on what you see here email me or leave a comment below this post.

Without any further ado, here are the results of our class's survey. Please share your thoughts by commenting (anonymously if you wish) below .....

Classroom Environment
The questions in this section were ranked using this 5 point scale:

Strongly Disagree	Disagree	Neutral	Agree	Strongly Agree
1	2	3	4	5

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	2	3	4	5

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

Course Characteristics

1. Course difficulty, compared to other high school courses:

Very Easy
0%

Easy
11.1%

Average
0%

Difficult
44.4%

Very Difficult
44.4%

2. Course workload, compared to other high school courses:

Very Easy
0%

Easy
0%

Average
66.7%

Difficult
33.3%

Very Difficult
0%

3. Hours per week required outside of class (n = 8):

0 to 2
0%

2 to 3
0%

3 to 5
37.5%

5 to 7
50.0%

over 7
12.5%

4. Expected grade in the course:

F
0%

D
11.1%

C
33.3%

B
44.4%

A
11.1%

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?

Yes
88.9%

No
11.1%

Did you have enough preparation without using your calculator?

Yes
77.8%

No
22.2%

Was your teacher too hard or too easy on you?

Too Hard
0%

Too Easy
11.1%

Just Right
88.9%

Specific Feedback
[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)
Work ethic.
The AP Exam.
Any time spent on the SMARTboard. (4)
All group work in class.
This blog.
The Final Take Home Exam.
"Commercial" assignments.
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)
Workshops.
Everything at the beginning.
Riemann sums.
Related rates.
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.
More workshops.
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 ....

Baby Madness

noreply@blogger.com (Rence) — Mon, 01 Jun 2009 19:08:02 PDT

I thought that this was just absolutely HILARIOUS.

DEV Status Report

noreply@blogger.com (Rence) — Mon, 11 May 2009 18:13:50 PDT

Hey Mr. K,

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

Due to this cluster of due dates, I on behalf of Team FLJ, am requesting a DEV extension. I will not keep this request exclusive to us, and I'm letting other groups also request an extension via Comment.

Twizzler MANIA!

noreply@blogger.com (.:. J + ME .:.) — Tue, 05 May 2009 17:13:07 PDT

As we all know, tomorrow MORNING, the bomb goes off. The AP is here! HOLY GOODNESS. Of course, anxiety is kicking me in the head right now. I'm slowly snapping back to reality.

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

These tips given, need MUCH emphasis, no matter how much we'd love to defy them.

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]

**during the exam, remember to maximize your time.
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?

Today's Slides: May 5

noreply@blogger.com (dkuropatwa) — Tue, 05 May 2009 09:29:29 PDT

Here they are ...

AP Calculus AB May 5, 2009

View more presentations from dkuropatwa.

Exam Review: Applications of Integrals

noreply@blogger.com (Francis) — Mon, 04 May 2009 22:50:15 PDT

We started off class by talking a little bit on how sushi is made. It's a roll, but when sliced (hondamitsubushi!) its made into cross-sections. By knowing the area of these cross sections, we can find the total volume of the initial roll if we have the length. This class was pretty much dedicated to applications of integrals for our exam review.

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.

Today's Slides: May 4

noreply@blogger.com (dkuropatwa) — Mon, 04 May 2009 13:15:56 PDT

Here they are ...

AP Calculus AB May 4, 2009

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3rd Last Class Before the Exam

noreply@blogger.com (Benofschool) — Sun, 03 May 2009 08:10:32 PDT

WOW! Just 2 more classes until the big exam. That was very fast.

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.

Todays Slides: May 1

noreply@blogger.com (dkuropatwa) — Fri, 01 May 2009 10:49:52 PDT

Here they are ...

AP Calculus AB May 1, 2009

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The Scuba Steve Dillema

noreply@blogger.com (Rence) — Fri, 01 May 2009 06:35:26 PDT

I knew I forgot to do something yesterday. And that was check to blog -__-"

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.

Anyways, we worked on another question, this one being about Scuba Steve. There was a lot of information, so the question had to be read carefully to fully understand it.

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.

*Red is what Paul stated. Blue = edited work.
The RDT triangle is an example of how to change from rates, to distances, to times.

The next part is done by benchmen (Yes, he's a plural. That's not weird), and he found the derivative of the function, and set it to zero to find x.

Unfortunately, this is as far as we got as we ran out of time.
Next scribe will be Ben

The Truth about Bruce Lee and Chuck Norris

Some real fighting, none of that ninja fighting bullet dodging special effects haha.

Today's Slides: April 30

noreply@blogger.com (dkuropatwa) — Thu, 30 Apr 2009 11:09:17 PDT

Here they are ...

AP Calculus AB April 30, 2009

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Today's Slides: April 29

noreply@blogger.com (dkuropatwa) — Wed, 29 Apr 2009 14:18:54 PDT

Here they are ...

AP Calculus AB April 29, 2009

View more presentations from dkuropatwa.

The Return of Optimization!

noreply@blogger.com (kristina) — Wed, 29 Apr 2009 19:24:39 PDT

Optimization Review

View more presentations from k_ina.

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!

Almost Final Scribe Posteyness

noreply@blogger.com (Skyline) — Tue, 28 Apr 2009 22:13:10 PDT

Alrighty guys, as chosen by yinan I'm the scribe for todays class, which is great, because its a short scribe, and I have to go to work in about half an hour, so it works out perfect.

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 ^^;

apr.27th/09

noreply@blogger.com (yíＮＡЙ) — Mon, 27 Apr 2009 18:55:03 PDT

In today's class Mr.K went through the our wiki, to show us some tips to compare the difference between two person's edits of the question. All you have to do is to quick the page history, then choose the two that you wants and qiuck compare. The green parts are the parts that are added, and the red is the parts that are been deleated. Also the deline for DEV project is extened, due to the high purssre of the upcoming AP Exam.

As always we did a mini exam to help us review for the exam:
Q1/
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

Q2/
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

Q3/
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

Q4/
will this question camed up the third time, I think everyone is good with it. The answer is D)8647 gallons.

The end..............
Next scribe is Justus.

Today's Slides: April 27

noreply@blogger.com (dkuropatwa) — Mon, 27 Apr 2009 13:32:58 PDT

Here they are ...

AP Calculus AB April 27, 2009

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Mini Exam 3

noreply@blogger.com (Joyce) — Fri, 24 Apr 2009 13:47:31 PDT

Thought I'd do my scribe early to make room for studying =(.

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.

Today's Slides: April 23

noreply@blogger.com (dkuropatwa) — Thu, 23 Apr 2009 12:55:39 PDT

Here they are ...

AP Calculus AB April 23, 2009

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The Famous Amusement Park Question

noreply@blogger.com (Benofschool) — Thu, 23 Apr 2009 05:46:31 PDT

Hi everyone, I'm Benchmen and I'll be your scribe for today. I'm sorry if this scribe is boring because I have to finish this quickly so I can finish off a project for another class.

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.

Good Night

Here's the Youtube video:

Today's Slides: April 22

noreply@blogger.com (dkuropatwa) — Wed, 22 Apr 2009 13:28:13 PDT

Here they are ...

AP Calculus AB April 22, 2009

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Related Rates Exam Review

noreply@blogger.com (zeph) — Tue, 21 Apr 2009 20:37:59 PDT

SUMMARY
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

Today's class we decided to do #2, reviewing related rates.

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!

HOMEWORK!

HOUSEKEEPING

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).

Today's Slides: April 21

noreply@blogger.com (dkuropatwa) — Tue, 21 Apr 2009 10:43:02 PDT

Here they are ...

AP Calculus AB April 21, 2009

View more presentations from dkuropatwa.

Mini Test: The Calculus Strikes Back

noreply@blogger.com (Not Paul) — Tue, 21 Apr 2009 01:30:35 PDT

I really need to stop with these late night scribe posts.

For the record, Joseph is scribe, and the list REALLY needs updating.

Question (1)

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).

Question (2)

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).

Question (3)

Just keep deriving. You’ll get there eventually. (C)

Question (4)

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).

Question (5)

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).

Question (6)

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.