Scribe Post for 3/1/13

Alrighty folks, so that was an action packed Calculus class this past Fri. so I'm gonna dive right in!

Bluester began the period with an exciting look at what's next!  (Un)fortunately, we are almost done with new material, something that should be a relief but also tragic and somewhat nostalgic for us hardcore AP Calculus nerds. All we've got left is Unit 5 and 6 (review) which means a lot more extremely stressful IW's and our last finely crafted opportunity day! In a quick blast to the past, I hope you all got your Unit 4 SuperCorrections in to OB on Friday! This coming Tuesday holds the daunting, equally-weighted follow-up test, which is a chance to shine or realize you never understood the unit! Then the Bluester Cluester showed us a divine sneak peek at the AP Calculus Bible, a sneaky book of everything we need to get a 5!  It had things like:
-Limits
-Asymptotes
-Continuity
-Derivatives as functions
-Concept of derivatives
-Derivative at points
-Slope of curves at a point
-Local linearity
-Applications of derivatives
-Computation of derivatives
-Definite integrals

and LOTS more!!!!!

If you didn't remember a lot of those things, its time to review! Luckily, we are going to have a lot of time to do so and we will be using the wonderful Khan Academy to aid us! Bluester's goal is for everyone to get at least a 3, and it IS possible! Khan Academy is really cool.  We took a quick look at this link: https://www.khanacademy.org/math/calculus/differential-calculus/differential-calculus/e
It was really cool and our answers and progress could all be monitored just as if Bluester was Big Brother! We will be using Khan Academy more so here's the link to the site itself, for your personal gain: https://www.khanacademy.org/math/calculus/
Then, we saw a new sneaky bullet point on the list that was completely NEW!!!
"Antiderivatives by substitution of variables"
<- that's today's class, SURPRISE!
But for now, O'Brien decided to deviously delay the lesson and return to Caroline's scribe post!
We set up a GeoGebra sketch. First, we made slider "a" and set it equal to 1.
Then, we created f(x) = x^2
Finally, we created A, a point (x,0) somewhere on the x-axis.  Then we took the integral, using the GeoGebra integral template to fill in function (f) and boundaries (a, x(A)). At this point, it looked a little like this.

 Finally, we plotted (x(A), b) as point B, and turned on trace for B.  The traced line looked like a function! In fact, it looked a lot like a cubic function, which Becca intuitively pointed out to be the function:

Well well well. That function was actually AN antiderivative of the earlier function f(x).  It wasn't THE antiderivative because antiderivatives can be plentiful due to that constant "C," the tricky reason why many of us screwed up that antiderivative question on the Unit 4 Test.  So yeah, this was pretty cool.

BUT DOES IT WORK WITH OTHER FUNCTIONS????????

We changed f(x) to equal sin(x). With trace, things looked a little like this:

Wowzer! It worked again, somebody correctly answered that the trend we were seeing was the function:

We checked that it was an antiderivative, and it was. Another interesting to note was that in the function, the 1 in the cos(1) really was the variable a, and it was a constant translation for the function as we moved slider "a."

Alrighty, next up on "Guess That Antiderivative Function" was ln(x):

Yippee! Three for three! The antiderivative of ln(x) was the function:

Some interesting things to take into consideration. First, Bluester realized that we needed to change the function to ln(t) instead of ln(x) because x was a variable with different representation.  Second, it worked! The function was an antiderivative function. We continued to prove that the lower limit of the integral (represented with a 1) can be changed with the slider for that variable "a." When changed it always alters the constant of the function.  

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Here OB began to shift directions.  He showed that due to the FUNDAMENTAL THEOREM OF CALCULUS, the following is true:

But how do we know that xln(x) has a derivative of just plain ln(x)?

Well this is how:

In short, OB says "you gotta know your derivatives to do antiderivatives!"

Then OB showed us why there is a constant in the first place! It's just like trying to find integrals when you evaluate a function's antiderivative at each end of an interval and find the difference.  The constant must be subtracted at the end.  Here's a more elaborate explanation from OB:

You can see that -(2ln(2)-2) is the antiderivative evaluated at 2, the lower end of the interval in this case.  That's an exciting new way to find constant C! We moved on.

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The wily Bluester Cluester decided to set up a new learning springboard:

Does it matter whether, in the integral, policy is f(x) or f(u)? dx or du?

YES, it DOES!

We proved this by testing the functionality of three different looking integrals.

You can see that for part a, the antiderivative of the original function was taken to get an answer.  On part b, we used f(u) and du instead of f(x) and dx.  The process was rather similar until we substituted u's back for x's again at the end and found that the answer was different! Finally, on part c, with a little hybrid equation, we had to substitute in x's at an earlier step, but got an entirely new answer.  Overall, Bluester made it clear that IT MATTERS!!!!

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Now lets use the function:

What's the antiderivative? Ummmmm....

While everyone sits in discombobulation, crafty Connor immediately goes to Wolfram Alpha!

The answer is this:

What the heck? Where did THAT come from? Suddenly, Alex Wilder has a moment of true revelation and says, "It all makes sense!"

Taking the derivative of that craziness proves the insanity! See:

Amazing! This leads O'Brien into a frenzy, it is finally time to learn the main idea of the day,  "Integration by Substitution." 

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We started out with the same function tan(x). 

Then we took some new steps:

1. Discovered we could rewrite tan(x) as sin(x)/cos(x).

2. Allowed u (the substitution variable) to equal cos(x).

3. This meant the derivative of u was equal to -sin(x).

4. -sin(x)dx ended up being equal to du, so we can sub in du if we can find one of those.

5. We found one of those by adding a -1/-1 fufoo!

6. We arrived at an easy function to work with, we anti-differentiated.

7. We substituted x back in for u and got the crazy answer that before could only be done on Wolfram!

Here is a picture of our work:

Well that was new and exciting! Bluester wanted us to practice so we decided to tackle a few questions from the IW (that you should all be working on about now). 

First up, was #32!

We started with a really nasty looking thing but using the steps above came up with a much easier function to anti-differentiate. Yay! The work is below:

Next up, we moved on to #34. Again, things looked dire, but we used Bluester's step by step procedure and prevailed handsomely! I popped the question to OB, "Don't things have to just be perfect for this type of thing to work out?" OB responded, "Yes, it's rare that things will be as nice as they are in these book problems, so this technique might have little real world practical use." But let's be real. You're a student in AP Calculus. When will any of this stuff have "real world practical applications?"

Finally, we moved onto the last problem which OB scrawled out just before announcement time!

The work for #45 is below: 

So that was some good practice using the substitution techniques! Keep working on more to master the skill! Also, if you want some help from an MIT professor, check out this youtube video, suggested to watch on your own time due to length: http://www.youtube.com/watch?v=W7sNkRpcydk

OB also reminded us of some sneaky tricks to help us on substitution problems like these:

1. Remember Fufoos! They can come in common to make things work nicely!

2. Substitute in du, get rid of all x's in the problem before anti-differentiating.

That's it folks!

Announcements...

Half Period Notes:

We didn't even look at my scribe post :( :( :( :( :( :( :( :( 

OB came in and jumped right into the Unit 4 test answers!  

The first question could just be done with fnInt. 

The second question was a trapezoid rule, a lot of people just took an underestimate, but we learned how to do things the right way. (picture and working to come)

The third question we discussed during SuperCorrection days, it was about the pizza!

The fourth question we needed to find the integral (picture and working to come)

The fifth question was a big old free response.  Part a was found using fnInt, (but don't write that!) Part b was as simple as writing down the definite interval and taking the absolute value of that velocity.  For part c, we needed to add it to the starting position and not use absolute value.  

The sixth question was all about mean value theorem and Riemann sums! You had to be able to interpret the intervals and get average acceleration.  Then, you could use things like symmetric difference quotient to get part c! On part d, you had to be able to interpret the unit of the definite integral.  

The seventh question was a function with different regions.  You also had to be able to interpret the addition of constants to an integral.  

The eight question made us find the derivative/antiderivative to find that g prime and f prime were both ln(x). Then, you had to explain that it was a translation.  

The ninth question made us test whether Riemann sums and Trapezoidal sums were underestimates or overestimates of different function graphs. 

The tenth question was just the antiderivative, but because of chain rule, it was D.  

The eleventh question required absolute value signs around the 3-x in ln(3-x).  Part b was more antiderivatives. 

The twelfth question was again an evaluation of an integral with some anti-differentiation.

The thirteenth question was best found with a graph that allowed us to see the absolute value function shifted over.  Then the definite integral can be taken from that.  

The fourteenth question was a test of our knowledge for properties and rules of integrals (like zero rule!)

The fifteenth question made us solve for k, which could be found by getting an antiderivative and then evaluating. 

The sixteenth question tested our knowledge of slope fields, which was tricky for some, also we had to perform separation of variables and anti-differentiation to solve for f(x)

Nobody had any questions, but you better study up for tomorrow's test!

Then we went over question 48 and question 54 from the IW.  

Pictures will be coming with more details of the work on these problems! (as soon as I can offload them from my phone!)

Cole Ellison is next scribe. (The sneaky boy who was supposed to be scribe a while ago and somehow slinked his way out of the devious duty.)

Scribe Post 10/4

Hello Everybody!


UPDATE!!!!

We talked about three different ways to write the derivative:

We also discussed how to type a cube root into your calculator:
cbrt(x) in Geogebra, also you can raise it to the 1/3rd power.

We also checked out the function:

and talked about how there are two different slopes: 1 and -1.  This forms a corner not a cusp.

We also looked at a few equations to prove that non-differentiable (when there is no derivative) at cusps, corners, vertical tangents, and discontinuities.  

We decided to use a function with a cube root to show that if f is differentiable, f is continuous.  (Thus, proving yet another theorem).


p. 124/1, 5, 11, 25, 29, 32, 37, 39, 53, 54
p. 126/1, 2, 4

We began our electrifying class with the 40 minute SuperCorrection test created by the masterful Mr. O'Brien.  It (hopefully) jogged our memory on how to solve interesting and exciting limit problems!

We discussed the very thrilling math meet on Wednesday!  The mathletes came in so very close behind Lincoln Academy.  Many of us made varsity, including Duncan, Crockett, Eddie, and Cal.  Wednesday the 31st is our Halloween Math Meet in Boothbay! We will be in costume!  Think about joining us and talk to O'B!

Then we got down to business. 

THEOREMS AND PROOFS

We first went over question 3 from the IW.  There were two equations:

First Equation

Second Equation

O'B showed us that for these problems, we could use the derivative equation (which should be memorized by now):

Derivative Equation

Then, since  

we can efficiently substitute k in for f(x).  That means, since x is approaching c, we can also substitute k in for f(c).  Doing this, we come up with the final result of this equation:

which ends up equaling 0 when the two k's cancel.  

Then, we tackled part two of question three.  The equation was:

We took the same approach, by using the derivative formula and then substituting in mx+b whenever there was an f(x) or f(c).

In the end we got this:

Then, that simplified as well:

Victory!  Then we went over question 10.  Oh, wait! We were almost starting to discuss when Rebecca had to ask a question about #3...

"Can you cancel instead of factoring m out?"

O'Brien retorts, "its all sort of the same idea." In principle, the limit of x-c/x-c is 1, and the limit properties allow us to extract the m. 

If you still need more help understanding how to utilize the limit properties, check out this link from Scotty's scribe post. 
O'B reminds us all of the handy second form of the derivative equation.  

Derivative Equation Take 2

To tackle question 10, O'Brien wittily decides to alter the equation from:

into

This is because we will be working with h as a component of the derivative equation. The variable "s" was chosen to stand as the function letter because it sounds like sum and there are two functions being added together in the "s" function.

Now we will prove the Sum or Difference theorem when it comes to derivatives! Using the derivative equation above, we changed all the f's to s's.  Then we started to expand everything.

THEOREM:

We determined that since s was the sum of f and g functions, s(x+h) would really be f(x+h) added to g(x+h) and so on and so forth.  We applied these applications to get the following monster:

AHHHH, that is scary looking! Luckily O'Brien knew what he was doing, and was using this as a part of his master plan of genius.
"Split the big thing into two different limit pieces!" -O'B (Using the Limit Properties again!)
And thus, it was cleft in twain:

since each limit piece is the derivative equation itself, we can conclude that

But of course, Rebecca had to ask yet another question:

"What's with the big thing over h, how did it become the two separate limits?"

O'Brien sneakily explained how he first used the distributive property:

then, he used the commutative property to rearrange the order of things:

Finally, limit properties were used again to separate the two limits.  

Hope that clears everything up!
If you want to see a real live math student perform the above actions, check out this Youtube Video! He's a real bro and he does pretty much exactly what we just did, except with j and k instead of f and g.

Next order of business!

"Let's prove the power rule!" -OB 

The power rule (also known as the infamous "Damian Rule") is that clever trick that allows us to quickly find the derivative by bringing down the exponent and multiplying it by the coefficient to get a new coefficient.  Then, you reduce the top exponent by one.  See the theorem below:

                     THEOREM:        

Then, we can apply the Constant Multiple rule to come up with the official Damian rule below:

To be warned: this proof will only concern natural numbers, proving irrationals and other unnatural numbers will come later.  
For now however, to acknowledge that we will prove it, let it be known that:

Starting out we plunked down the beginnings of a proof:

Then, using substitution, we ended up with this:

Then it got tricky, so we put it on the back burner, and referred to question 11 from the IW to help with this long and complicated proof.  

Let's travel to Wolfram Alpha Land!
Question 11 looked kind of like this


11. Next class, we will prove the wonderful Power Rule. To prepare us for this event, please visit 
wolframalpha.com and calculate (a + b) raised to the power of n for some natural number values of n. What do you observe about the 

powers? How are the coefficients related to Pascal’s Triangle?

So we set off to experiment with different natural number values of n.

Our exploration proved fruitful and left us with the following chart (in mid-creation in the shot below):

                                                 

                                                                            and so it goes...

There was something clearly up, though there usually is when O'Brien is this deep in a complicated math exploration.  Then, all of a sudden, SURPRISE! The ball was dropped... it was PASCAL'S TRIANGLE!

Pascal's Triangle of Awesomeness

Remember this crazy thing? We all learned about it in our math childhood.  It's that cool repeating pattern where a number is just the sum of the two numbers top left and top right of it.  This all fits in, because you can see thats whats happening with the coefficients as n gets larger with our Wolfram Alpha equations above! The exponents of a just decrease by one each term and increase by one for b.  Math is truly wonderful.

Then, O'B relates and asks whether our old teachers ever told us about the connection to combinations. Scotty retaliates "Fitz didn't."

Well combinations are pretty cool, basically you can plug them in on your calculator and answer cool probability questions with ease (think SAT?) Here's a FUN link about Combinations and Permutations! If you thoroughly scour that site, you'll see lots of cool connections with Pascal, including the fact that with combination nCr, n represents the number of rows down in the triangle, while r represents how many spots in.
Basically a combination has a few parts, in nCr, n stands for the number of items total, of which you have to choose r items not paying attention to order.
Example: 5 elephants, you have to pick 2, order doesn't matter.  n=5, r=2.  There are 10 combinations of elephants.
This is the factorial explanation, used by Patrick JMT in a video a little ways down the post:

Here's how you do it on a calculator:

n value -> Math -> Prb -> nCr -> r value

Here's me typing it in on my TI-83:

Since we know that Pascal, which can be represented combinations, determines coefficients, and that exponents decrease each term by 1 from n for a, and increase by 1 for each term from 0 for b, we can deduce the following sequence to represent 

etc...

Patrick JMT, in this helpful video, explains the above definition, also known as the BINOMIAL THEOREM.
However, note, he doesn't use the term "C" as combination but rather uses the definition of a combination as 

Anyhoo, we've got this crazy thing: etc... and we ended up sticking that onto our proof right after that step way back when we had 

Luckily, things simplify a lot because stuff cancels and something as confusing as that first term:

 can reduce all the way down to x raised to the nth.  

Yippee, now we can look at something as simple as this:                                  

Yeah, just kidding, thats not simple in the slightest, but O'Brien points out that since we know the big repeating thing ends in a -x to the nth, we can cancel a lot more terms.  Everything cancels with its opposite except for the golden term, 

At this point, a frazzled O'Brien, hair tossed, glasses askew, compares himself to a mad professor.

The scribbles on the board look like the crazed etchings of a delusional scientist.

The proof is crazy, everybody hopes there are ways to look at the power rule which are more intuitive.

A polite insult to Ms. Damian: "Math is too hard for you."

Yes, this proof is difficult, but maybe this guy can help.  It really starts getting interesting about halfway through, as he'll show you a way to get the same answer by step by step substitution. Thanks calculussuccess!

WHAT DO I ACTUALLY NEED TO KNOW? <- you are probably asking yourself this

You need to know two different forms, recognize the derivative, and should know that all these cool results that we are proving come from these definitions.

Work on these problems still! They are tough! He's not gonna post the answers for a little while because he is maniacal, devious, tricky, and trying to help!

Scribe for Next Class: The MAGNIFICENT Duncan Hall!!!!!!