Elazar and Connor | Final Exam Presentation

WORK AND FORCE

Force:
A push or pull which acts on an object. This doesn't necessarily mean the object will move, but the force is existent.

Imagine a water bottle on a table, if you push the bottle lightly it might not move. This is an example of a force acting on a force and being counteracted by friction. This leads to the ideas of force diagrams.

Think of all the forces acting on you now. What keeps you on Earth? What keeps you from falling through the ground? Are these things acting on you?

Before we take a look at work, recall what we learned about displacement. It is important to know that displacement is the difference between the final position and the initial position. Where the particle goes in between is not important. So displacement=X(final)-X(initial)


Work:

A force acting on a body which results in a displacement.

Work is related to force through calculus. Just like position is the antiderivative of velocity, work is the antiderivative of force. The only difference is the variable. With velocity and position the change is the change in time, while with work the change is with respect to the axis of displacement.

Constant force: something which doesn't change over time, gravity is a great example of this.
Equation for work when the force is constant:

Variable force: a force which varies over time, given as an equation.
Equation for work when the force is variable:

QUIZ

3/13/13

We started class watching ViHart’s video discussing the validity of tau rather than π to measure radians. From that we watched an incredibly confusing video about digamma or wau. This is really hard to comprehend until you find out that wau is really just one. This video of wau was just one huge pain in the ass. 

Next, it was time for the really difficult IW quiz.  

OB's naughty list: Elazar, Alex, Weston, Eliot, Connor, Lexi, Noah, Cooper, Megan, Francie

OB's explanation of the calendar:

Submit IWs by thursday because OB will be gone so the zero of the missing assignments will sit there for a long time. Over the weekend the assignment is to finish up IW 5. The answers to IW 5 will be posted sometime soon. IW 6 and 7 will be free response questions. OB predicts that we'll have about 3 hours to work on 6-7 free response problems. All of the answers will be available online. The next quiz will be Friday 3/22 and will include questions directly from IWs 6 and 7. There won't be any notes or blog post, but if you do these problems the quiz will be like free points. These questions will also be on the test in two weeks. OB wants us to take them seriously. According to OB we’ll have about 3 hours to do 6-7 free response questions. All IWs and supercorrections will be due by the end of the quarter which is three weeks away. 4th quarter starts with a follow up test. There will be two quizzes before spring break. These quizzes will cover multiple choice and free response questions from the IW. There will be extra credit available over spring break. In April 2 quizzes right before spring break. they will be MC and free response AP questions from old IW. Extra credit will be available over spring break. After spring break we will take a full AP exam. OB recommends we do the  IWs to simulate the test. After May 8th we have a project which will last nearly a full month. No IW after may 8th! Less than two months until the end of AP Calculus! We are about 57 days from the AP test.

Onto IW 5, we did problem 31 from page 411. This set of problems is question 4 on IW 5. Problem 31 part a starts off by finding the volume of the shape enclosed between the y-axis, y=x^2, and y=1. If you have been doing your IW this week 31 shouldn't be too difficult, as long as you remember it's top - bottom. Part b has you find the area of the object when you rotate around an axis which the area isn't touching. This is difficult to imagine conceptually, remember that you need to subtract to volume between the area and the axis of rotation.

31. Find the volume of the solid generated by revolving the region bounded by the parabola y=x^2 and the line y=1 about 

a) the line y = 1.

b) the line y = 2.

c) the line y = –1. 

Here is a graph of the problem situation:

a)

b)

c)

Here is an interesting video covering a similar topic, but with a little more rewriting of the integrand.

IW 5

Scribe Post 12/3

After going over the second quiz we went into today's lesson. We learned two words which we will start to see more frequently in the book and on the AP Exam.

Linearization: Another way of saying the tangent line.

An interesting applet to get an even better understanding of the tangent line:
http://calculusapplets.com/linearapprox.html

Differential:

find dy if x=1 and dx=0.01

We then worked on Exploration 8-3: Maximal Cylinder in a Cone Problem. For some reason we all struggled with this even though O'Brien said that it's really easy. Here are the answers:
1.
V(0) = 0
V(1) = 9π
V(2) = 24π
V(3) = 27π
V(4) = 0
2.

3.

4.

5.

6.
radius: 8/3
height: 4
volume: (128π/9)

7.
-find f'(x), solve for 0
-sub that value into f(x) and the function for height

8. You know what you learned better than I do.

9/28/12 The Overview of Unit 2

We started off class with OB telling us about derivatives. We looked at four different perspectives of the derivative: verbal, numeric, graphic and algebraic.

Verbal:


The derivative is the slope of the tangent line. Use local linearity to find the tangent line. The slope is a rate of change. The derivative is also known as instantaneous rate of change, it is the speed at that instant. Any quantity per quantity is a rate of change: cm^3/min, $ spent/year. Anything that is changing is calculus. Calculus: the ultimate applicable mathematics.

Numeric:


See question 4 from the test. The derivative is the slope of a set of points that are really really close to each other. To do this take a set of data such as:

Now you use

and sub in each value for x-final and x-initial where x-initial is 1 and x-final is 1.0001. This gives



m=6 so the derivative is 6.

This process is called the difference quotient, aka the slope.

The symmetric difference quotient takes just a little bit above and just a little bit below x when you wish to find f'(x) of function f(x). Use

From here use the difference quotient to find f'(x).

Graphing calculators can do this for you in a much simpler way by using nDeriv. It can be found by hitting math --> 8. Make sure you are not using MathPrint otherwise it is confusing. There are 4 things which need to be defined: the function, variable, the value for the variable, and some small number h. If nothing is entered for h then the default for h is 0.001

nDeriv(f(x)* ,x, 2, 0.0001)

*on your calculator use y1 from the YVARS menu.

Graphic:

take the graph of the function f(x) in this case:

Add the point (2, f(2)) which is point A then use the add point tool and make a random point on the graph so that it can be dragged around this is point B. Next put a line between these two points. This line is called a secant line.

Drag B so that it is on top of A which is supposed to give the tangent line at point A. The problem is that this is undefined 0/0, fortunately we have been learning about limits for a month! In this case to find the tangent line of A* we write our own limit:

*remember that A is really (2,4)

This gives us the slope of the tangent line which is the derivative at point A. Now we get to the more interesting part: we realize that there is a derivative for every point of f(x). We need to figure out how to find this function which gives the derivative for every point of f(x). This is incredibly easy on Geogebra, in the bottom text box you simply write f'(x) after you have graphed f(x). Unfortunately, I'm doubtful that we will be allowed to use this on opportunity days, there is an algebraic way. Before we get to that look at the graph of the derivative function (the blue line):

It is possible to have f'(x) traced by a point by entering (x(B), b) into geogebra. Where B is the point which can be dragged and b is the slope of the tangent line to B. b can be found by typing slo and selecting Slope[ <Line> ].


Algebraic:


We've done the hard work while learning about limits, the theoretical underpinning of integrals and derivatives. Find the derivative of

start x at x=c


It is not possible to simply sub in because of the removable discontinuity. Multiply by the conjugate FUFOO of the numerator. Also, this is (above) is the definition of a derivative. PatrickJMT has a helpful video

Which gives:

The (x-c)s cancel

So,

A general thing to know for the function f(x) at x is:

Another way to look at question 16 from the test is

So,

In part of problem 16 we were looking for when x=2. If it is plugged into the derivative the answer is

IW 1


p. 105/7, 9, 13-17, 25, 26, 27, 37

Scribe is Cooper

UPDATE:
There are much easier ways to derive functions. These ways include the power rule, quotient rule, and product rule.

The power rule: 

This is extremely useful when deriving polynomial functions.

The quotient rule:
Supposed there are two functions u and v so that

Once you have put in the correct values simplifying is not even necessary! Although it is recommended that we become used to simplifying this so that we improve our algebra and do better on the AP test.

The product rule:
Similar to the quotient rule because there are two functions joined together. Still using functions u and v.

Again, it is not necessary to simplify beyond this but we should become used to it because the multiple choice part of the AP test will not leave the possible answers as "messy" this, according to OB.

These ways of taking the derivative of the function are much easier than solving a limit to figure out the derivative.