Related Rates Exam Review

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:

1. Doing mini-exams pre-test style
2. Doing questions in class related to rusty topics
3. 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).
  

AP Exam Practice Quiz 1 and Intro to the Intermediate Value Theorem of Integrals

Introducing the first of a series of AP exam practice quizzes, the intermediate value theorem of integrals, the day when the AP Calculus 2008: Without Bound blog is unblocked at Daniel McIntyre Collegiate Institute, the series of YouTube videos in tandem with Flickr pics, and kristina's powerpoint presentations!

AP EXAM PRACTICE QUIZ 1


Let h be a function defined for all x does not equal 0 such that h(4) = -3 and the derivative of h is given by h'(x) = (x^2-2)/x for all x does not equal 0.


a) Find all values of x for which the graph of h has a horizontal tangent, and determine whether h has a local maximum, a local minimum, or neither at each of these values.

To determine if there are local extrema at the critical numbers, we must first determine the critical numbers. Remember that there are critical numbers when a function has an asymptote or is undefined. h' = 0 when x = sqrt(2) and -sqrt(2). h' is undefined at x = 0 since 0 isn't in the domain (as stated in the question). So critical numbers are x = sqrt(2), -sqrt(2), and 0.

We use the first derivative test on h' to see where h' is positive or negative. Why? Because when h' is positive, h is increasing; when h' is negative, h' is decreasing. So a change in sign in h' would indicate a slope of zero at that point and that's where there are local extrema. According to the line analysis, we see that to the left of -sqrt(2), h' is negative; between -sqrt(2) and 0, h' is positive; between 0 and sqrt(2), h' is negative; and to the right of sqrt(2), h' is positive. Wherever h' changes sign from negative to positive, h has a local minimum; wherever h' changes sign from positive to negative, h has a local maximum. By the first derivative test (line analysis), there are local minimums at x = -sqrt(2) and x = sqrt(2). We don't look at 0 because it's not part of the domain of the function.


b) On what intervals, if any, is the graph of h concave up?

Rememberize these rules (from chapter 5 of your textbook):

If the second derivative is positive, the first derivative is increasing, and the parent function is concave up.

If the second derivative is negative, the first derivative is decreasing, and the parent function is concave down.

So wherever h" is positive, h is concave up.

We determine h" by differentiating h using the quotient rule.

We see that h" is positive, so h is concave up everywhere.


c) Write an equation for the line tangent to the graph of h at x = 4.

Pull out the point-slope formula: y-y1=m(x-x1)

The question gave us the x-coordinate: x = 4.
The question gave us the y-coordinate: y= -3.
m is the slope at x = 4, so plug x = 4 into h' which spits out 7/2.

Plug those numbers into the equation. BING! BANG! BOOM! We're done part c.

y+3=(7/2)(x-4)


d) Does the line tangent to the graph of h at x = 4 lie above or below the graph of h for x > 4?

If we draw a line tangent to the graph at x = 4, the line is below the graph at x > 4, because h is concave up everywhere.


INTERMEDIATE VALUE THEOREM OF INTEGRALS

Similar to the intermediate (or mean) value theorem of derivatives (that in a closed interval between a and b there exists a point on a continuous function which equals the average value), there exists a point on a continuous function which equals the average integral of the function.


We have b = 3. We have a = 0. We have f(x) = 1-2x. Plug the numbers into the equation. BING! BANG! BOOM! We get the average integral. But why does it work?

In this graph, pivot the yellow area found between c and b to the white area found between c and a. Notice that they fit together like a jigsaw puzzle to yield one big yellow rectangle. We can imagine that one big rectangle having the same area as the area under the graph, which we can see as the average value of the function.

HOUSEKEEPING

• Next scribe is benofschool.
• I'll be away on Monday to write the grade 12 English pilot exam.
• Don't forget to check Graeme's comment on Rence's post as it will prepare you for the exam!
  
• Don't forget to give Jamie your $1.25 donations so she can make her Betty Crocker style cheesecake!
• Pi Approximation Day is coming! Are we organizing the annual Coin Hunt for Pi Approximation Day too?

Post Test for Applications of Derivatives

Another day and another test.
Time for another round of thoughts and reflections on the last test.

I'll start it off. This test was quite tough being that it was the first test back from the long winter holidays. The multiple choice questions were much more challenging compared to the free response questions. The last few multiple choice questions were a tad easier.
For the free response, the optimization problem was extremely easy (I was happy when I saw the question). The other free response question took a bit longer to answer. I was not 100% sure if the given function contained a cusp or a curve. I was unsure because of the resolution and pixels of the picture because it was taken from a TI Calculator. So I asked Mr. K for what it was and all I hear was curve, so I took that word for it and continued to answer the question with the function having no cusp.

That was my reflection. I want to know how everyone else thought about the test. Was it easy? Hard? Muddy? Clear? Was it Hot? Was it Cold? Is that a Yes or a No? Come in and answer, don't stay out. What were your ups? What were your downs?

BOB, Applications on Derivatives

After break, everything just BROKE. No, I'm joking, but I did forget a lot of stuff about this unit. It's awesome how I'm one of the last BOB's so I can read other peoples BOB's and have a sort of mini review on everything, especially Benofschool's BOB. I was chatting with him earlier, and I asked him "whats up?" and you know what he says? He says "I'm helping people study" Wow, can that kid get an award or something? Thanks bud!

Applications of derivatives, where should I start? How can I start? I don't remember a lot, but I did read other people BOB's! I know about the 1st and 2nd derivative tests, and what they're used for, pretty much, but I don't know if I will be able to graph them, and I have a feeling that's going to be a big chunk of the test. Optimization problems seem quite blurry to me, but I did look at some questions on the internet, such as examples, so hopefully I can just understand them by reading a couple more. The mean value theorm was easy enough to remember, atleast the theorm part of it, but if there are any questions on it, I'm not so sure I can make it through alive! That's pretty much it for BOB and I. Good luck everyone.

Bobbing for 2009

So, it seems just by glancing over everyone's bobs, that the break really took a toll on their "remembrance" of the material we did prior to the break. I hate to admit it, but I'm pretty much in the same boat.

For me, it seems that the pre-test, was basically a bunch of questions, that I knew what to use to figure them out, but couldn't remember how to do it at all. Like I'd see one, and be like, "Oh! you need to use mean -value theorem to solve this. Wait. How do you do that again?"

As you could probably imagine that became terribly frustrating quite quickly.

In terms of the unit itself, I found it somewhat high on the difficulty scale, not really super mega hard, but no where near as easy as integrals. The easiest part was definitely the bit with taking the limit of a rational function as it approaches infinity, where half the terms like disappear and things. That was a fun bit! :D

Besides that though, I'm not enjoying these units. :[ Difficulty wise anyways ;p

Hmm, so thats all I can really think of for this time, so I'll see everyone tomorrow. Good luck!

ciao

OMB. I hope I'm not in a pickle. I'd settle for a cucumber or some Polski Ogorki.

O M B...OH MY BOB. I was...minimally productive over break. I was anxious, stressed out, and away for too long. My brain will show that I guess. Maybe...but hopefully not. I haven't...forgotten EVERYTHING, but I've fogged up the concepts a bit over break. I guess the weather was quite frightful in a way. Really. I spent a whole week just sitting and thinking...having that panicking thought of how I was going to end up in the future. Those paranoid thoughts just rushed through. Then I tried to... well...start...eating differently. I'll leave it at that and just BOB.

I understand the main part of the uses of the 1st and 2nd deriv. tests and have an average understanding of what they're for. 1st derivative test is to find critical points including Local/Global Maxima (Max or Mins) of the parent fcn. Sorry. Rough, but I'm rushing. People are telling me to sleep. 2nd deriv is for finding concavity of the function. I'm blazing fast typing this and forgive the typos that I don't spot. I almost spelt "fungtion"

A disaster that I can't seem to prevent is optimization problems. I don't think I can do it by myself yet. Not without someone to hold my hand across the street. I can go as far as distinguish what needs to be maximized or minimized and what can be used, but there's always a part where I get compuzzled and somehow not put the pieces together correctly.

As for the mean value theorem. It quite simple, but there's something in my gut that's telling me that I don't understand it 100%. Maybe I do get it, it's just that I'm probably doubting myself again. Yeah I get it...I think. I apologize for the bipolar uncertainty. And the fact that I said "bipolar" and "uncertainty" together makes me wonder if those two words contradict each other, as if I could have just stopped at one word.

My insides are shuddering vigorously. To delay or not delay? I mean I don't know if I'm more comfortable doing this test tomorrow or if we did it before break. I just don't think I'm ready. But sometimes we have to dive without knowing. But after diving I just hope I come back up.

G33K LUV everyone.
<3.1415.......

The New Years BOB

Unfortunately, like everyone else, I have forgotten a majority of what we learned and it remains faint in my mind. It's there, but I can't access it. All I can think of is food, eggnog, present's and fireworks. Really.

The first and second derivative test however seems to stay with me, and I still kinda understand it. I always had problems with the optimization questions and always hit the wall hard. I have scraps of pieces of the Mean Value theorum, but after looking at a few scribe posts, it still doesn't ring bells. I must've lost too many brain cells over the break.

I honestly feel very uncomfortable going into this test (I didn't even know it was tomorrow until I looked up the blog just now), due to the fact that I did so horribly on my pre-test and guessed most of it. Woe is me.

~Rence, Out.

BOBing for Derivatives

Sadly I too have forgotten a huge chunk of Calculus during the winter break, and I probably will do poorly on the test tomorrow, especially since I didn't do well on the pre-test... ;o;

From what I still remember I'm ok with the 1st and 2nd derivative test, but when it comes to the optimization problems it would be a train wreck X_X as for the rest in chapter 5 (excluding the anti-derivatives) its more of a room that hasn't been cleaned and touched in for years. Hopefully that everyone will do well on the test tomorrow! Good Luck!

BOB for Chapter Five

Darn, that break really messed me up. Before the break I felt like I was on a roll with the questions in the chapter. Now that I look at the questions one last time before the test, I feel like I'm almost new to the subject. What I do remember from "Last Year" was the Mean Value Theorem, some Optimization Problems, and How to use the First and Second Derivatives.

The Mean Value Theorem says that if a function is differentiable and continuous within a closed interval, the secant line connecting the 2 endpoints of the function will have a slope equal to, at least, one other tangent line of another point on that function within the interval.

Optimization Problems are word problems involving situations where you want to have the most or least of something. Some examples include maximizing number products sold while minimizing production costs, or maximizing the area of a shape inscribed another shape with limited dimensions.

The main point of this chapter is to learn the many uses of the First and Second Derivatives.
The First Derivative can be used to find many things about the parent function, but the main use is to find the critical point(s) of the parent function that may exist and on what interval(s) is the parent function is increasing or decreasing. Critical Points include Local/Global Maxima (Max or Mins).
The main use of the Second Derivative is to find on which interval(s) is the parent function concave up or down and where the parent function has an inflection point (change in concavity).

Once I saw the word Antiderivatives again after the break, everything that we learned "Last Year" flew back into my head. *That was the coolest flashback I ever had.
To antidifferentiate a function is to find the parent function if the first derivative is given, the first derivative if the second derivative is known, etc.

Overall I think I'm going to do okay. It may take me a while to complete it though.
Good luck and see you all tomorrow.

BOB: Applications of Derivatives

Hmm, well after doing the Pre-Test, I can say that I'm kinda in good shape for the test. The only problems I really had with it was the long answer question, specifically the last two parts of it. Although, from what Benchmen told me, I had the last part kinda right. >__>

As for this unit overall, the major problem that I had, and I'm assuming the same goes for the rest of the class, were the Optimization Problems. Just when I thought the Rates of Change questions were bad, these types of questions had to come into the big picture. Really, although I'm pretty confident I can do the square/rectangle ones and NOT the inscribed stuff since those are really hard for me -__-;

Yes...as for the rest of the unit. I can say that I really enjoy finding critical points for some reason, plus I was able to learn how to use chain rule on the natural logarithm because of critical points questions, so I feel quite fondly of them :(

Mhm, so good luck on the test everyone and if it ends up not being tomorrow don't shout at me again like last time haha XD

Post Holidays Day 2

Since there were a few people who were taking their Provincial English Exam today, all we did in class was Page 293 (Supplementary Problems) ODD.

Might as well add the fact that the test is tomorrow (According to Benchmen, don't blame me if I'm wrong again XD). Yes, so don't forget to BOB for those of you who haven't yet! :D

Next scribe will be Shelly.

BOB

Well, I think I didn't do well on this unit that i need to review more. It's mostly about 1st and 2ND derivative test, i think I'm fine with that part or maybe not*_*. The Limits part was confusing, i didn't really remember much from that part. I feel like i didn't remember a thing from the optimization problems, i think i need to review more on that part. Well for the rest i still need so time to study.
Bye for now!

BOB for Chapter 5

I knew i should've bobbed loooong time ago. The holiday break has really made me forget a lot of things but I've reviewed the past chapters and i'm regaining my memory. My muddiest point for this chapter is the optimization problems. It seems so simple with all the steps we were given but it just isn't. No topic has ever made me so frustrated. So the pre test we had yesterday was so so for me. I got the multiple choice ones but the long answer was harder. After Benchi explained that all i had to do was apply the 1st derivative test it became clear what I had to do from there.

KK we've postponed this test long enough and tomorrow is when we'll finally have to do it =(. Hopefully i'll do good. *prays*

First day back.

Today in class, we didn't do much. Our main focus though, was the pre-test. It wasn't worth any marks but it was valuable. The answers can be found on today's class slides. The pre-test was actually quite hard, but maybe that's because it was our first day back, and we just forgot about calculus over the holidays. I know I'm not alone here.

We found out that our DEV will have a final due date on May 3rd, which means we have 4 months to complete it, which is plenty of time, so don't stress, but don't procrastinate either. We have to copy and not create questions similar to what we will be doing by the end of this week for our DEV. 4 questions for a soloist, 5 questions if it's you and a partner, and 6 questions if you got a group of three.

Mr. K also showed us a video reply to our calculus ad "What is a derivative?" which was made by someone else, it was pretty cool and pretty flashy. Our next integrals commercial will have a time cap of 60 secs, or one whole minute, and will be due on February 1st.

Time to study everyone! Exam week is the last week of January, but from the looks of things our class doesn't have to worry much, because it'll be pretty lightweight on us calculus heavyweights.

That was pretty much what we went over in today's calculus class. That next scribe will now be Kristina.

Today's Slides: January 5

Here they are ...

AP Calculus January 5, 2009

View SlideShare presentation or Upload your own. (tags: apcalc2008 math)

Determining the Properties of the Parent Function Using the First and Second Derivatives

OVERVIEW:

1. Calculus Commercials: How Many Ways Can You Define the Derivative?
2. Mathematical and Computer Science Definitions of "AND" and "OR" Notes (slide 10)
   
3. Determine the Properties of the Parent Function Using the First and Second Derivatives (slide 2 to 9)
   

*************************************************************************************

1. Calculus Commercials: How Many Ways Can You Define the Derivative?

A continuation of last class' lesson, but first, we watched calculus commercials, which can be accessed here:

http://apcalc2008.blogspot.com/2008/12/gauntlet-is-thrown-down.html

The class agreed that the time restriction of 30 seconds to define the derivative was challenging, but there are our finished products! Note that people have commented on our projects.

*************************************************************************************

2. Mathematical and Computer Science Definitions of "AND" and "OR"

Let
A = go to the store
B = wear the hat

One day, you aren't sure if you should go to the store or wear the hat. So you think and think and think, but in terms of mathematical and computer science definitions, you notice that their definition of AND and OR are different from the AND and OR of English. The OR used in math is the inclusive OR. The OR used in English is the exclusive OR.

Here is a list of all the possibilities:

1. You go to the store and wear the hat. (A = true; B = true)
   
2. You go to the store but not wear the hat. (A = true; B = false)
   
3. You don't go to the store but wear the hat. (A = false; B = true)
   
4. You don't go to the store and you don't wear the hat. (A = false; B = false)
   

A U B = satisfy the conditions of A OR B
A ^ B = satisfy the conditions of A AND B


*************************************************************************************

3. Determine the Properties of the Parent Function Using the First and Second Derivatives

Using the derivative rules, we found the roots of the function.

Remember, k is a constant, not a variable!

We let k equal a negative number, like -1 000 000, in f' and discovered that to the left of 1/k, f is increasing, so f' there is positive.

We let k equal a positive number, like 1 000 000, in f' and discovered that to the right of 1/k, f is decreasing, so the f' is negative.

Since f is increasing to the left of 1/k and decreasing to the right of 1/k, we can imagine that at 1/k, there is a local max.

Remember, k is a constant, not a variable!

In the previous slide, we evaluated f' when 1/k>0.

In this slide, we evaluated f' when 1/k<0. style="font-style: italic;">

Since f is decreasing to the left of 1/k and increasing to the right of 1/k, we can imagine that at 1/k, there is a local min.

We have now obtained as much info as we can from the f'. On to f''!

Using the derivative rules, we determined f''.

We determined the roots of f'', so we can determine the sign of f'' on either side of the roots; this info can be used to determine f's concavity to either side of the roots.

We let k equal a negative number, like -1 000 000, in f'' and discovered that to the left of 2/k, f'' is positive, so f is concave up.

We let k equal a positive number, like 1 000 000, in f'' and discovered that to the right of 1/k, f'' is negative, so f is concave down.



Remember, k is a constant, not a variable!

In the previous slide, we evaluated f'' when 2/k>0.

In this slide, we evaluated f'' when 2/k<0. style="font-style: italic;">


Synthesizing all that we know, there is that list of properties f has.

• We know where f is increasing or decreasing at certain intervals.
• We know f's concavity.
• We know where f changes sign.
• We know f's local extema.

*************************************************************************************

END NOTES:

• Homework: the rest of 5.5 Antiderivatives
• Looking forward to the rest of the presentations tomorrow!
• Ending our scribe for 2008 is Francis.
• Reminder tomorrow's class will range from 10 to 25 minutes, with the Holiday Inn Gym Riot, starting at 10:30 AM, hosted by the DMCI Student Council. We will be called down to the gym between 10:15 AM to 10:30 AM. See you there! =)

Today's Slides: December 18

Here they are ...

AP Calculus December 18, 2008

View SlideShare presentation or Upload your own. (tags: applicationsofderivatives calculus)

The Absence of Evidence

In Absense of Spring by flickr user abstract splotcHes'
Anyone else feelin' a little chilly?

We started off the class with a pretty funny clip which featured Calculus and the Numa, Numa song. Basically the person in the video showed two pieces of paper, one a function and one an anti-derivative.

Following that, we got into the stuffing of the class. We reviewed derivatives and what they are in order to grasp the concept of an anti-derivative. We were told to remember something when we do anti-derivatives, which is "If you can do something, you can undo it." For anti-derivatives, this means that any derivative can be anti-differentiated.

A key point is to know that an anti-derivative is a family of functions, however, if you have a given point, you can find one particular function within the family of functions.

An Indefinite Integral is not like a definite integral. It picks our a family and if a point is known, it picks a particular one. This sounds a lot like an Anti-Derivative.

Moving along, we are told that for every derivative rule, there is an anti-derivative rule, as this is automatic when a derivative rule is made.

So, knowing that, can we find the parent function with first and second derivatives?

Let's use an example.

If you know the derivative of sine, you know the anti-derivative of cosine.
Therefore if you know the derivative of sine, you know the anti-derivative of sine.

In words: The derivative of sinx is cos x, so the anti-derivative of cosx is sin x + C where C is an arbitrary constant. This basically means that C can be any value, because the function can exist in a number of places in the "family of functions."

The derivative of cos x is negative sin x, so the anti-derivative of sinx is negative cosx + C.

After this, Mr. K gave us a question so that we can apply this.

Consider the function (not the way it's supposed to be written but sitmo won't let me write it properly.) * "k" is a parameter and can be any value.*Using the first and second derivatives, describe the features of the family of functions generated for different values of "k".

Bench did the honours of doing this question.

Using the quotient rule and the chain rule...and factoring out...and reducing...and then we let it equal 0, so that we can find x.

o = 1 - kx
kx = 1

*Remember that critical numbers are at zero and undefined points, however, for this function there are no undefined points.
Therefore, this function will have no discontinuities (cusps, corners, magic tricks, hat tricks, pen tricks, kick flips etc.)

So by the first derivative test, when x = 1/k, the function is at a max.

Okay, it's 12 AM. I need to study for Physics... Ugh.

Next scribe will be Zeph!

Rence ~ Out

Today's Slides: December 17

Here they are ...

AP Calculus December 17, 2008

View SlideShare presentation or Upload your own. (tags: apcalc2008 math)

(Di)Lemma's, Proofs, and Other Things.

By now it should be no mystery to everyone that I, (Hi I'm Justus, glad to meet you) am the scribe for today. Now I regret to inform you that this scribe post might now be wholly as spectacular or colorful as the one I last left you with, and this is due to the fact that well, I just don't have that much time on my hands today ;p As a result you get the scribe post you may or may not be about to read. I apologize in advance for any trouble this may cause.

So without further ado, onward with the show!

Where to begin...where to begin...

This is indeed a hard question, since we really didn't start with anything in particular, but instead jumped all over the place across a few topics we all felt would be good to talk about. Well I'll just start with what I have first in my notes here.

All right, so to kick things off we started with some more proofs (Or rather, logical explanations) for the differentiation rules (Which Jamie so graciously typed up for us in a lovely format.)

The first (according to my notes) was;



Now Mr.K explained to us that it is quite possible to go through the algebraic explanation of this differentiation rule, but also said that it is much simpler, so just remember the graphs of sine(x) and cosine(x)


Sine Graph



Cosine Graph



Looking at the sine graph as the graph of some function we want to differentiate, we may pick out the important points/bits of information found within the graph itself. By doing a quick glance you can find the local minimums and maximums, as well as the inflection points for the graph. These will be used to construct the derivative graph of sine(x).

Our experience with the graphs of functions and the graphs of their derivatives tells us that the red points are the local minima/maxima and that the blue points are inflection points. We know that because the red dots are local extremum, they will become roots for the derivative graph of f. We also know that the blue point(s) is a maximum because the slope at the inflection point is positive.




At this stage it should be easy to see, that connecting these dots gives you a derivative graph identical to cos(x). Therefore;



We also dealt with the derivative of cos(x) being -Sin(x) following basically the exact same steps. Those can be seen in the following image.






Moving along nicely then :]

The next topic of discussion, is kind of a twisty windy, jump off into several lemmas, resulting in an organization dilemma, and me having to sort out some thoughts kinda topic. For the most part, it involves this;



and the question, how would you find the derivative of that logarithm? This is indeed a tough question, as we didn't (at that point in time) have any differentiation rules for logarithms of any base (which in reality, is what we were trying to find at the time.) However, by using the change of base law, and some knowledge about differentiation rules with did know at the time, Mr.K showed us the way.

LEMMA TIME

(something about lemmas imagery here)

Change of Base Law, how it works, and what it looks like on the inside.

So, say you have a logarithm base "a", that equals "n", that you would like to change to base "c."



You must first start by re-writing the logarithm as a power (In this case, a to the n equals b.)



The next step is to take the log of both sides giving you the following;



After you have completed that, you would use the power law to bring down the "n" exponent on the left side leaving you with something you can work with.



After that, all you have to do, is isolate N. Looking back at the beginning we see that in the very definition of what N was, lies the logarithm we wanted to change the base of. What we're left with, is N (aka. our logarithm) equals a new logarithm, now with a different (hopefully more useful) base.



I hope that made sense ^_^;, just in case, here's all the steps together, so you can see it as one fluid thing, instead of a bunch of seperate steps.

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2.

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5.

Back To Our Regularly Scheduled Programming.

So we have this logarithmic function



which we would like to find the derivative of. The first step would be to apply our newly a wholly understood change of base law to end up with this;



From here, we will pull out 1/ln2 so that we have;



the reason for this lies in the fact that although ln2 is a logarithm, it is still just a number, and therefore, a constant. By pulling it out of the whole thing, we get the nice and easy to differentiate, lnx by way of the constant multiple rule. For those of you who may be having trouble remembering exactly what that is, it basically means that;

"The derivative of a constant times a function is the constant times the derivative of the function."

As a result of this differentiation rule, all that remains is to find the derivative of lnx which we know to be 1/x

*Note* We did a Lemma for the differentiation rule for lnx but when I looked at my notes I couldnt really makes heads or tails of it. I'll need to check it over with Mr.K and get that up here once I understand it. Sorry guys D:

The result, is the derivative, which in this case, happens to be



Now if we think back to the beginning, the primary goal was to see if we could find a rule for all logarithmic functions, and to this end I believe we achieved our goal. By simply substituting a variable (lets say a) into the spot of 2, you get the rule for differentiating logarithms.



With that stuff out of the way we moved onto the proof for d/dx of tan x, which happens to go something like this.




After converting tan into something more friendly, you simply apply the quotient rule.







Tadaa! If you wanted to find the derivative of say Cotangent, you would do the exact same thing. Turn cotan into Cos x over sin x, and apply the quotient rule from there.

Another neat little tidbit of information for you. According to Mr.K, the derivative of all "co" trig functions is negative. Just thought you'd like to know ;p

Now, after alllll that stuff behind us, we got back to antiderivatives. Remember that an antiderivative lives on the idea that anything that can be done, can be undone. However, with derivatives, because of how we figure them out, there is a small kink. Any constant in a parent function, is lost when you take the derivative of that function. Essentially, you may know a functions derivative, but you may NOT know its vertical positioning by the same method. In effect, when you differentiate a funtion, you are finding a family of functions, not just one. Now there is a way to figure out what the constant was, if you have a point from the graph you'd like to single out. These are called intial value problems. Lets see an example.




Because we know an antiderivative gets rid of the constant, we need to add that back in. We do so with the value "c", seen above.

Know this is called the inital value theorem because we have a value at the beginning (initial) of the problem. In this case its the value of f at 1



From this point we substitute the initial value into the antiderivative and solve for c.





And thats really all there is to it :] In case that didnt make much sense, here's the slide that explains it better then me I think D:



Okay guys, I think that about wraps it all up for tonight. Not gonna go into a big spiel this time, its late, and I'm tired, and I have basketball tomorrow, and the day after, and the day after lol.

SO with that said, lawrence shall be scribe for the next day (which is technically today.) Alright? alright. Night all :]