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 . . .

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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 . . . .