1/23 Scribe Post (FTS)

January 23, 2013: Another semester, another scribe post. Sunny. Blue sky. Butt cold. MOTHER O' PEARL IT'S COLD! Luckily we're inside, studying calculus. With our nice laptops. Speaking of laptops, for those of you that didn't know, the MLTI Laptop Program expires next year. Survey's are being sent out to homes asking for student and parent input regarding the pros and cons of having laptops. Hopefully the school will work something out for next year. But if not . . . SUCKS FOR YOU, JUNIORS!
 

Instructions for Class:

1) Sit somewhere you've never sat before with someone new

2) Take out exam Sections A and B 

     Today, O'Brien provided us with an opportunity to thoroughly look over the midterm. We started by looking at the Free Response Questions (FRQ). He drew our attention to how the question was from Form B, which is the test that people who are "sick" on exam day take.

*Some initial questions to ask: What do the x and y axes represent? (Height and position) Is the skateboard component significant? (No) What type of function is this? (Quadratic)

Below is the scoring guide. Because we don't actually have access to our tests anymore, I thought it would be pointless to write up all of the work (especially because it's available on the web). Instead, I thought it would be more beneficial to highlight some useful points that O'Brien went over.

Things You Should Be Able To Do With Your Calculator:
1) Make a graph
2) Solve equations (Graphically or with Solver!)
3) nDeriv at a single point
4) fnInt

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UPDATE:
So...what the f*^% IS "fnInt"?

     We know by the integral symbol (the long, slinky 's') that we need to take the antiderivative of tan(x), evaluate this antiderivative at π/4, and subtract from this the antiderivative evaluated at 0. But we also all know that taking antiderivatives can be kind of difficult. I mean, what the heck is the antiderivative of tan(x)? Sure, O'Brien just taught us a little bit about substitution of variables (brief lesson below) and sure, some antiderivatives aren't that bad. BUT if they give us the information above and ask us to solve, we don't really have to do anything.
     There is a special function on your calculator that will take the antiderivative of the given function, evaluate the antiderivative at the upper boundary(in this case, π/4), and subtract from this value the antiderivative evaluated at the lower boundary (0).
1) Press "Math" key.
2) Press "9".
3a) If you have a TI-84 Plus Silver Edition, just put in the lower limit, the upper limit, the function, and then X. Press enter, and BAM. The answer.
3b) If you have a slightly older calculator, follow steps 1 and 2. Then type your equation, a comma (the button underneath the sine function key), "X", comma,  lower limit, comma, upper limit, comma, and then enter. BAM. Answer = 0.347 (three decimal places!)

Easy as π.
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Both parts a and b involved simple substitutions of given information. Here is how to find the derivative of g(x) for part c:

For the second part of part c, you have to prove that g(x) does not meet Requirement iii. You cannot simply make a table of values because it doesn't "use calculus" (#stupid). To get the marks, you have to use calculus justification. We are talking about the function g(x) always INCREASING . . . well, if g(x) is always increasing, then it's derivative will always be positive. DERIVATIVES!!! We already took the derivative of g(x). We need to see if it is negative between x=0 and x=4. We can get the critical points of our derivative by solving for 0 and by looking at the end points.

 

If we evaluate g'(x) at values of x between the critical points, we can see where (if ever) g'(x) is negative. 

An acceptable answer is something like g'(x)<0 for 0<x<4/3, so g does not satisfy (iii). O'Brien thought that g'(1) < 0 would also work. Although it may seem like this answer isn't really any different than using a table of values, the fact that we have g'(x) (the mighty derivative!) shows that we know calculus. Although part d seems complicated, you just have to realize that you aren't going to use logs because the variable is not in the exponent but rather is in the base. After realizing this, you just have to use the chain rule and substitution. The last thing we talked about was being cautious about the phrases "the function" and "it" - they are too ambiguous and could be referring to the original equation, the derivative of the original, the second derivative, etc. Instead, be specific. Use f(x) or f'(x) or f"(x).

*Some initial observations: The radius will stay at a constant 5 inches. Therefore, the Volume of the Coffee is actually 25πh. This is a related rates problem!

Again, here is the scoring guide:

Part a is pretty straightforward. It's all about translating the given, worded information.
Part b has a few EASY points that you should always look for. Just Remember:

SEPARATE THE F*#KING VARIABLES!!!

and

 ADD C AFTER TAKING THE F*#KING ANTIDERIVATIVE!!!!!!!!!!!!!!!!! 

and

YOU CAN ALWAYS SOLVE FOR C. THEY WILL ALWAYS GIVE YOU AN INITIAL CONDITION, IF THEY DON'T, IT'S PROBABLY (0,0).

Note: For part b, you can only earn up to two points if you forget to put + C after taking the antiderivative. If you forget to separate variables, you are not able to earn any points. Yah. That's how important they are. Lastly, Part c is very easy - just set h'(t) to 0 and solve.

Strategies for Approaching FRQ's:

1) Do not let the wording prevent you from finishing.

~O'Brien introduced today's lesson by recognizing that the math for FRQ's isn't actually that difficult – in reality, the wording of the problem is what creates confusion. It can often be to your benefit to take a minute to read the entire question and think about it before you start writing. Once you understand the question, you may find that you won't even need the whole fifteen minutes to work on the problem. 

2) CUT THE BULLSHIT.

 ~He told us that each question is out of nine marks and each mark is allocated for specific things. Sometimes you don't even need that much writing. Furthermore, you actually CAN say too much -- if your answer is too ambiguous, the reader is not able to give any points. Just answer the question.

3) Go back and check your work.

~FRQ's ≠ O'Brien. In other words, there is no partial credit with FRQ's. So make sure you get as many points as you can. That being said . . . 

4) If you get stuck on a problem, move on. 

~If you can get five out of the nine possible marks on each FRQ, you will more than likely get a 5 on the exam. That being said, sometimes you will get (in the words of Damianator) a "Killer Problem" where you many only get three marks. Therefore, you want to aim to get as many marks as possible on every problem (obviously).

5) Make sure you used something that you learned in this course.

~Answers that do not use calculus will not be given credit. For example, when asked if f(x) is always increasing, you cannot use a table with values for x and f(x). You must show that f'(x) is always greater than 0. Try to differentiate yourself from the "shmo in Missouri".

6) Show as much work as possible with the knowledge that anything related to calculus will be most helpful. 

~This tip was given in response to Dunks' question: "If you hit the points and you show no work . . . are you okay?"-- CLEARLY a question that a junior would ask . . .

 ------------------- 
Questions On Multiple Choice
Note: I scribed all of the ones we went over in class. However, I put #14 first because this is where I discuss Integration By Substitution, a new skill. If you did well on Multiple Choice or you frankly don't give a damn, just read the first bit.
 
14) For this question, you could potentially just take the derivative of all of the answers. But this is kind of time consuming. I approached this problem by thinking about what the antiderivative of cos(x). I knew that it was sin(x). I also knew that the original equation would be
The derivative of this would be:

 

ARGH. So close to what I need it to be, which is:

But!!! If I divided the original function by 3, I could do it. But wait...wouldn't I have to then use quotient rule? NO! BECAUSE OF THE CONSTANT MULTIPLE RULE! (Note: The only reason I can use this rule is because 1/3 is a constant number. If it were a variable, I would have to find another way...). So the answer is:

Another way I can solve this problem is by integration by substitution.
In this, we take the original problem and make it "look easier".....

EXTRANEOUS TO LIFE:
5) Use Chain Rule to find derivative:

 

10)






There are a few important things about inverse functions. To obtain the graph of a function's inverse, reflect the function across the line y=x. If you know the slope of the line tangent to the function at a point, you can find the slope of the line tangent to the function's inverse by taking the negative reciprocal. And finally, the point (x,y) on a function becomes (y,x) on its inverse. This being said, we can figure out this problem quite easily. We know that a point on g(x) is (2,1). Therefore, a point on f(x) is (1,2). Take the derivative of f(x) @ (1,2)and find that f'(1)=4. This means that g'(2) = 1/4.

11) For x ≥ 0, the horizontal line y = 2 is an asymptote for the graph of the function f. Which of the following statements must be true?
O'Brien clarified that this problem was mainly looking at end behavior. He told us that asymptotes can actually be crossed - it just matters the end behavior of the function does not touch the asymptote. Because of this fact, we know that B is not true. f(0) does not have to be 2 -- it could be anywhere. We don't know if f(2) is undefined because the graph can cross the asymptote. And lastly, D implies a vertical asymptote, where as the graph approaches x=2, the y-value does not touch the vertical line x=2. All that is left is E.



16) In this question, we are given the equation of the tangent line (2x+3). We can easily take the antiderivative and solve for C.

25) O'Brien hinted that he wanted us to look over this one in class. We are given function g that is twice-differentiable. g'(x) > 0 and g"(x) > 0 for all real numbers x, such that g(4)=12 and g(5)=18. If g"(x) > 0 for all values of x, that means that the second derivative is always positive, which means that the first derivative (g'(x)) is always increasing as well -- but not at a constant rate. If g'(x) was increasing at a constant rate, g"(x) = 0. The rate of increase is increasing. We know that between g(4) and g(5), the g(x) increases by 6. This means that between g(5) and g(6), the increase must be greater than 6. g(5) + 6 = 18+6 = 24. Therefore, g(6) > 24, eliminating all of the options except for E.

MAKE SURE YOU HAVE HANDED IN ALL PARTS OF THE MIDTERM!!! 2nd Semester Seniors, BITCHEZ!

---------------------------------------------------------------------------------------------------------------------------- After we handed in all parts of the test, O'Brien brought our attention to three words written on the board:

Position

Displacement

Distance

He got out his masking tape and placed an X on his floor. He asked us to imagine that the floor was a Cartesian plane with the X marking the origin. We defined that any space to the right of the X was positive and any space to the left of the X was negative (note: directions are from the perspective of a student looking at the board). Then O'Brien called up two beautiful people: Alex (Wilder) and Becca and placed them as such:

 Observations about Alex:

-Alex's position is 5
-Alex's displacement from the origin is 5. 
-The displacement from Alex to Becca is -11.
-If Alex were to start at position 5, run outside to the parking lot, get into his car, drive to his house, eat a sandwich, shave, and then come back to position 5, his displacement would be 0.

Observations about Becca: 
-Becca's position is –6
-Becca's displacement from the origin is –6
-The displacement from Becca to Alex is 11.
-If Becca were to start at position 6, run and give Scotty a hug, then go back to position 6, her displacement would be 0.

Based on these observations, we created some definitions.

Definitions:

Position: Where you are at any time t; x(t)

Displacement: Difference between your end position and your start position

Distance: Absolute value of displacement

     After this discussion and wonderful simulation, we went to this link (it works best in Adobe Reader). The first question asked us what three closely related concepts we need to keep straight regarding motion. O'Brien told us the concepts were Position, Velocity, and Acceleration and provided us with the relationships: 

After that, we looked at and answered the the problems with our seatmates.

Question: If x(t) represents the position of a particle along the x-axis at any time t, then the following statements are true.
1) t=0
2) x(t)=0
3) v(t) = 0
4) right
5) negative
6) position
7) Instantaneous velocity is the velocity at a single moment (instant!) in time.
8) velocity
9) negative
10) velocity
11) First take the derivative of x(t) (this will give you an equation for velocity). Then find when the object is resting (i.e. when the velocity is 0).

The object is at rest at t = –1 and t = 3. Because our domain is 0 ≤ t ≤ 6, t = –1 is extraneous. To find the total distance traveled, we have to take the absolute value of the "differences in position" between all resting points. Translated, this means that we have to look at the displacement of the object between 0 ≤ t ≤ 3 and 3 ≤ t ≤ 6, take the absolute value of the displacement of both of these intervals, and then add the displacements together.

 
Please note that if we had taken the absolute value of the displacement of the object between t = 0 and t = 6, our total displacement would have been 18.

We ended the class with O'Brien passing out our IW: Unit 4 IW #4. 

THIS HANDOUT WILL BE DUE ON FRIDAY! 

TO EDDIE AND CROCKETT: This means you cannot leave it for the end of the quarter. 

The next scribe will be . . . Becca from Mecca, I got it from her! Becca Becca Becca Becca Becca from Mecca . . . . 

Scribe Post 9/14 (FTS)

REMEMBER: Changes will be made. Yay.

     On Friday, IWs #1-4 were due (though we only passed in three IWs as the first one was just to rememorize the unit circle). However, instead of passing them in immediately and starting the lesson, O'Brien asked if anyone had any questions on any of the IWs.

-----Review---------     
     We first addressed how to solve #31 on page 66 algebraically, which read:

O'Brien explained that this was a really good question to ask as we will probably have one similar to it during our quiz (which will cover IW #1-6) on Tuesday. He first confirmed that this is an indeterminate function for when we substitute in 0 for x, we have division by 0. He then reminded us of how we proved that  (see our proof from Anna's blog post here). When we factor out the
expression, we are left with: 

.

By the properties of limits (to review, click here and see property 3), we know that we can simply multiply the limits and of both of these factors to get the limit of our original function. We know that  and that when we substitute 0 in for x. , therefore . 

     We also looked at #33 on page 66, which read:. Here, we can separate the function into two functions --  and. We know that the first is 1 and the limit of the second is 0. 1 x 0=0=limit of the original problem. 

     Then we looked at #72, which read : . This problem was true -- to prepare for the quiz, just make sure you know that on questions like this, you are looking to break apart the fraction in order to get  so that you won't have an indeterminate function to work with. 

     

     The last problem that we looked at was #43, which was a picture of a graph and various true or false questions regarding right and left hand limits. To prepare for a question like this, make sure you understand that with limits, it does not matter what the function is at the value that x is approaching. What matters is what is around that area. Also, make sure that you know your left from your right. 

-------------------------------

     After this review, O'Brien started the lesson by talking about "Happy Limits". These are limits that are easily found because you can simply substitute the value that x is approaching into the equation, solve, and this number will be the limit of the function. Why is this, you ask?

BECAUSE THE FUNCTION IS CONTINUOUS AT THAT POINT.

This means that there are no bottomless holes, no sketchy gaps, no creepy asymptotes, no division-by-0. NOTHING. They are, in a word, beautiful.

Look:                                

I mean...who doesn't freak out when they see that? BUT because the function is continuous, we can just substitute in 3 and our final answer is:

Good stuff. 

     At this point, O'Brien explained that one of our objectives today was to look at continuity. From last year, we know that a function is continuous if we can draw it without picking up our writing instrument. He also provided us with the formal definition:

A function is continuous at a point c if 

as long as:

a) c must be in the domain of f(x) 

b)must exist

Along with this definition, O'Brien added:
To be considered continuous, a function must be continuous at every point on its domain. 

"Well what about asymptotes? And holes? And gaps?"

What about them? They aren't in the domain. It doesn't matter that there aren't points there.

::::UPDATE::::
Here is an example of another continuous function that we went over during the 40 minute class period.

Here f(x) is continuous. Yes, I realize that you cannot draw f(x) without picking up your pencil, which is how we defined continuous functions last year. The point is that there is an asymptote at x=3, meaning that 3 is not in the domain, meaning that it's okay that there are no points there.  On the other hand, if we were to define a point on the asymptote, such as (3,4), the function suddenly becomes discontinuous because it does not follow the rule stated in blue above.

As x approaches c (which, in this case, is 3), the limit is infinity. Yet f(3)=4. Because 4 is not equivalent to infinity (and because 4 never could be equivalent with infinity as infinity is a concept while 4 is a number), this function with a redefined point is not continuous.
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In this diagram, you can see that . As this fits the rule that we defined above, the function must be continuous. But what about at point b? Here there is a hole . . . is that an issue?
O'Brien explained that because the hole is at the endpoint of the function, it is sufficient to just have the one-sided limit (in this case, the left-sided limit) equal to f(c). However, this graph is not continuous at x=b. There is no point there. There is no graph. Along with this, point b is not actually in the domain of the function, so it doesn't matter that there is no point there. Of course, it is also not continuous at a plethora of other points such as any point to the right of b and any point to the left of a.


     Here's another example:

     This function is not continuous, as there is clearly a hole at point b. The limit of the function as x approaches b appears to be a few units below M. However, f(b) = M. Because this does not fit the rule, this function is not continuous. If you still don't really understand the rules of continuous functions, click here for a video by a very beautiful man and watch until 3:20.

     O'Brien said that all of the functions that we learned last year were continuous (quadratic, cubic, exponential, rational, logarithmic, etc) except for one function, which was the floor function (below):

BEEP BEEP BEEP BEEP BEEP BEEP *FIRE DRILL* BEEP BEEP BEEP BEEP BEEP BEEP



     When we came back from an uneventful fire drill (there wasn't even a firetruck), O'Brien drew this graph on the board:

We identified that the discontinuous points were at x=4, 3, 1, 2, and -1. Then O'Brien classified each discontinuity for us:

1) Removable Discontinuities: Removable discontinuities are where you can define the function at that point without breaking the vertical line test. X=-1, 2, and 4 are all removable because if we redefine those points, we can fill the holes. We are able to do this with all of these points because they have a limit i.e. the limit as x approaches 2 is 2.

2) Jump Discontinuities: Jump discontinuities are also known as non-removable discontinuities. We cannot redefine the function at these points because then it would not pass the vertical line test. Along with this, the right-hand and left-hand limits do not approach the same values at this point. X=1 is a jump discontinuity. i.e. At X=1, the left-hand limit is 2.5 and the right-hand limit is 1. If we were to redefine x=1 to equal 1, we would fail the vertical line test.

3) Infinite Discontinuities: An infinite discontinuity is also known as an asymptote, where the function is always discontinuous at one point. X=3 is an infinite discontinuity.

***If any of this is unclear, go to this page 80 in our textbook, which gives a pretty good overview of the different types of discontinuities. Also, if you watch this video (same as the previous one) from 3:20, you will get a pretty good idea as to how to recognize the different types of discontinuities.

     Before we moved on, O'Brien showed us an "oscillating discontinuity" by graphing

 in Geogebra. It looked like so:

As we zoomed in, we saw that there was an infinite number of oscillations as we got closer to 0.

Cool! I guess...if you're into that kind of stuff...

     It was at this time that we transitioned toward derivatives. Just for review, a derivative is the slope of a tangent line to a function. However, we'll revise this blog post in upcoming weeks when we learn more about derivatives. Instead, for this class, we directed our attention to our other objective which pertained to local linearity and limits. 

     We began by looking at this limit problem: .  We saw that this was an indeterminate function as when we substitute in 1 for x, we get division by 0. O'Brien asked us to type this equation into Geogebra. When asked how to do a limit in Geogebra, he instructed us to just type in the function because the limit command in Geogebra is weird. 

We typed in and got the following picture:

We created the slider k and then defined point A as (k,f(k)). We used a spreadsheet in Geogebra in order to see what happened as we got really close to 1. At x=1, Geogebra tells us that the limit of the graph is  . However, around 1, the limit is approximately -1.359. 
     O'Brien then showed us that we could find this without technology by looking at the numerator and the denominator separately. This is what the graphs (red)  and    (black) look like near x=1:

Notice both items look nearly linear (this explains local linearity -- as we zoom in close enough on a given point on any function, even if it is curvy, we will eventually see a straight line). Also notice that both functions go through (1,0). If we can find the slope of both of these lines, we can put the equations into point-slope form. We know that the slope of 2x-2 = 2. Therefore, the new equation of this is y=2(x-1). However, the red curve is harder to find the slope of. If we imagine that it's a bit more linear, we can see that we go over about 1 and go up 2 (almost 3 units). After asking Duncan, Cole, and Autumn for help, I finally realized how a slope of e (which is about 2.72) was a reasonable slope to start out with. When we plug this into the point-slope form, we get . To check that this is the right linear equation for  (red), we graphed both this and  (black) on the same graph.

These are obviously not the same line. Obviously. The line that we created has the wrong slope -- let's make it negative:

Much better! Now we know that our numerator is and that our denominator is . When we combine them, we'll get . We can cancel out the (x-1) from both the numerator and the denominator and we will be left with  which is the limit of our original function. In summation, we basically separated the numerator and the denominator into two fractions, zoomed in on the graph at x=1, found the slope of these straight lines, and used the slopes to get the limit of an indeterminate function.


Let's look at another example:
We are given:. We know that .  If we look at the graph of and compare it to the graph of , we can see that we have compressed  in by 2, thus making the slope steeper.

Note: sin(x)=blue

sin(2x)=black

If the slope of the numerator is 2 and the slope of the denominator is 1 (as y=x has a slope of 1), then the slope of the original function is 2, thus making the limit of the function (as x approaches 0) 2. 

     We can also look at the function:, we will see something very surprising. 

Althoughand  are not the same graph, they do have the same slope when we look at the graphs around 0. This is because we are only looking at the LOCAL LINEARITY of graphs, not the graph as a whole. In both cases, the limit is 2. 

     This method of finding the slope of an indeterminate function and using this as its limit is known as
L'Hôpital's Rule (also known as L'Hospítal's Rule). The official definition of this is as follows (from page 448 in our textbook):

Suppose that f(a)=g(a)=0, and that f '(a) and g '(a) exist, and that g '(a) does not equal 0, then 

.

For further explanations, go to this video.


     We ended the class by talking about why radians rock. (I'll admit, I was always skeptical for the fact that I learned the unit of degrees first and always found that to be rather simple. However, after this explanation, I was sold. Radians rock.)
We graphed (in radian mode) and went to the table. We wanted to see what happened to this graph as x approached 0. When x=0.000001 and when x= -0.000001, y=1. However, in degree mode, when x=0.000001 and when x=-0.000001, y=0.0174532... Luckily, Cal miraculously figured out that 0.0174532 was the same as . What would you prefer... or 1?

That's what I thought. 

During our next 80 minute class, we will also be talking about the Intermediate Value Theorem. For a brief intro, check out this beautiful man.

The homework for the forty minute class is to do IW #6 (see full assignment below) and to study for a QUIZ ON TUESDAY that will cover IW #1-6.
IW #6: p. 84/7, 11, 13, 14, 15, 23, 41, 43, 47, 54, 57, 59

Please note that Dunks, Cole, and Auts are all beautiful people. I probably would have died without them.

Update:::::::

The next scribe ended up being Coleson, as Crockett was "sick"...