Scribe Post 3/11/13

We walked into class today and, being the devoted math students we are, were thrilled to see today's goal on the board accompanied by three juicy problems to help us acquire this new and stimulating knowledge. You can imagine my jubilance when I was told I had the honor of scribing for today's class and recording each and every happening on this glorious day of mathematics. The whole class was enthusiastic to dig right into these three succulent example problems, but we were forced to put our excitement on hold, as Sarah had a question about last night's IW (which we should all probably get cracking on).

Sarah's question was this: does it matter whether we look at an area problem with vertical rectangles or horizontal rectangles?

OB decided to answer this question by taking a look at question 3 on page 399, which told us to find the area of this ugly thing below. OB sketched it up on the board and plotted some points by plugging numbers into the equation.

We decided to try solving this problem by using both horizontal rectangles and vertical rectangles. OB pointed out that if we use horizontal rectangles, we are summing rectangles from y=0 to y=1 on the y-axis and if we use vertical rectangles we are summing rectangles from x=0 to x=1 on the x-axis. The working for both methods is shown below:

Horizontal:                                                                                           Vertical:

Same-Same!

The lesson to learn from this problem is that areas can be found both with horizontal and vertical rectangles, but sometimes one is easier than the other.

From here we moved onto volume and looked at the first problem in our set of three: 

When we sketch this area, we get something that looks like:

OB pointed out that here we are "summing discs" that look like the one above that have a radius of x^2-4x+5 and a thickness of dx. To find the volume of the resulting shape, we can simply sum up the volumes of all these discs, shown below (the volume of a disc is its area multiplied by its thickness)

Fully satisfied with this answer, we moved onto the next problem:

We're summing discs again, but this time the discs have a hole in the middle! They're washers (or pineapple slices, or bagel chips, etc)! How do we find the area of a washer? The area of the big circle minus the area of the little circle. The calculus working for this problem is shown below:

After finding this solution, we went on a bit of a tangent about the volume of a cone. We all know (questionable) the volume of a cone is (πhr^2)/3, but it has never really been proven to us until now because those silly geometry teachers don't know calculus! Work is shown below:

From here we moved on to our last and final problem of the day:

Our representative rectangle looks like this, with a length of 2√y. This means that each square will have side lengths of 2√y, and therefore an area of (2√y)^2.

To find the volume we simply sum up the areas of all the squares:

This is all well and nice, but what if the cross-sections were something other than squares, like equilateral triangles? We figured this one out as well.

OB proclaimed that we can use any cross-section as long as it can be described. In other words, as long as we can find the cross-sectional area of each slice, we can find the volume of the entire shape.

After working through these three tough problems, OB gave us ten minutes to watch the videos he posted on "Volume Links" to help us visualize some of these shapes. Some watched the videos. Others discussed the AP Lit assignment.

Once the ooh-ing and ahh-ing from the videos had died down, OB gave us the rest of class to work on IW #4.

Here is my copy of IW #4 after those 10 minutes of hard work:

3/12/13 40 min period:

During the 40 minute period, we went over questions from the IWs. Solutions are below:

p410 / 9

p410 / 19

p410 / 23

NEXT SCRIBE IS EBEN

Scribe Post 10/12

We began class with a quiz on IW 1-4, which took up a good 40 minutes. Note that IW 1-4 were due this class and need to be handed in ASAP to Mr. OB if not already.

After the quiz, we tested our skills with derivatives in the Derivative Matching Game, an action-packed, thrilling adventure in which you match the graph of a function to its derivative and when you win a bunch of colorful shapes follow your mouse around the screen.

A topic of much confusion in AP Calculus has been the notation of derivatives, primarily due to the laziness of physicists all around the world. We thoroughly covered the notation of derivatives used both in physics and in mathematics, and decided that it was much easier to write the mathematical notation. Below is a list of derivative notation that goes from the original function (y) to its fourth derivative. Examples of how physicists derived their notation are shown for the first and second derivative, but can be applied to all.

AP Physics scholars already knew this, but for those who did not torture themselves with the vast amount of work that is AP Physics we went over the relationships between position, velocity, and acceleration by applying the definition of a derivative as a rate of change. Velocity is the rate of change in position, so therefore it is the derivative of position. In other words, the instantaneous rate of change (x/t) of an x-t graph at a specific moment is equal to the velocity at that moment. In turn, acceleration is the rate of change in velocity, and so therefore is the derivative of velocity. Just like before, the instantaneous rate of change (v/t) of a v-t graph at a specific point in time is equal to the acceleration at that point in time. These derivatives can usually be found by utilizing the oh-so-powerful power rule. We also learned that to go backwards (from acceleration to velocity, and from velocity to position), we use something called the antiderivative, which can usually be found by using the power rule in reverse. For a more complex look at antiderivatives, go here. The relationships between position (x), velocity (v), and acceleration (a) are shown below by Mr. OB (top) and myself (bottom):

For practice with this "kinematic stack" as it is called in physics, we turned our textbooks open to page 136 and looked at problems 10 and 11. 

Problem 10:

From this graph, we can tell that particle P is moving right when the slope is positive, and left when the slope is negative. Therefore, it is moving right from 0–1, standing still from 1–2, moving left from 2–3, standing still from 3–5, and moving left again from 5–6. The velocity graph is simply the graph of the derivative, which looks like this:

Note that the ends of the line segments are not defined because their are no derivatives at the corners of the function.

Problem 11:

Problem 11 had us analyze the above graph, and asked us:

a) When does the body reverse direction?

b) When (approximately) is the body moving at a constant speed?

c) Graph the body's speed for 0 ≤ t ≤ 10.

d) Graph the acceleration, where defined.

Because the body is moving forward when the velocity is positive and backward when it is negative, the body changes direction whenever the graph crosses the t-axis, so in this case at t = 2 and t = 7. The body is moving at a constant speed when the velocity is constant, or in other words a horizontal line on the v-t graph. This occurs between t = 3 and t = 6. 

Because speed is the absolute value of velocity, the graph of the body's speed just looks like this:

The acceleration graph is simply the derivative of the velocity graph, and so looks like this:

After we finished analyzing these two problems, we turned to page 138 in our textbooks and began problem 28, an economics problem:

Mr. OB pointed out that economists like to make their work sound a lot more complicated than it is, so they came up with a term called "marginal revenue," which is in fact just the derivative of the revenue function. If you really want to get into the economics side of things, you can go to this website. We didn't get a chance to finish this problem, as the announcements came on, but we did get to get a graph up on Geogebra:

Here, Mr. OB cleverly noted that as the number of office desks made (x) increases, the revenue increases, but at a decreasing rate. In other words, you get less and less revenue for each office desk you made. A few conjectures were mades as to why this is, including the "new factor," supply and demand, and the cost of materials to make the desks.

Mr. OB said that we will continue this problem next class, as we were rudely interrupted by the announcements.

Note: Homework for next class is IW #5 (not #4 like it says on iCal): p. 135 / 9, 13, 15, 19, 23, 27, 40, 42, 43

40 Min:

Mr. OB clarified the fact that we have a 35 min class on Tuesday, no class on Wednesday,  and then a quiz on Thursday, so it is important to ask any questions on the IW soon.

We then proceeded to finish up the marginal revenue problem. We found the derivative r'(x) by using the quotient rule as shown below:

When we graph this function (green) alongside the original revenue function (blue), we get a graph that looks like this:

To find the increase in revenue from making 5 desks to making 6 desks, we can just do r(6) - r(5) = $47.62. Mr. OB took a jab at economists when he said "It's not rocket science, just economics." To find the limit of r'(x) as x approaches infinity, we can simply look at the graph and see that it is 0. This number represents the fact that as a company produces more and more desks, the marginal revenue will decrease with each desk, but will never actually reach zero.

The last math we did in class was to look at question 13 from IW #5:

We found (a) relatively quickly by using the power rule and the relationships between position, velocity, and acceleration:

Mr. OB then pointed out that there are two ways to find (b): the pre-calc method and the calc method. We, being the suave-sophisticated math students we are (Ms. Seibert flashback), of course used both. The pre-calc method was to find the vertex at b/2a, because it was a parabola, and the calc method was to find the time when the velocity was equal to zero. We got 15 seconds for both methods, which are shown below:

We found the rock's highest point by plugging 15 seconds (the vertex of the parabola) back into the original position equation, which gave us 24(15) - 0.8(15)^2 = 180 meters.

To find the time at half this distance (90 meters), we can simply set the position equation equal to 90 and solve for t by using "quadratic formula magic" as Mr. OB called it:

For part (d), because the acceleration is constant the trip down should take the same amount of time as the trip up, so we can simply do 15 • 2 = 30 seconds.

There is a cool tool on the graphing calculator to actually see the motion of the rock, which is described on page 133 of our textbook:

To see our rock in question 13, we simply need to set our calculator to parametric mode, and then use the test tool (replacing values above with values from question 13). We can even adjust the graph to have a little rock at the end to get the full, high-definition picture.

At the very end of class, someone mentioned the Austrian man, Felix Baumgartner, who dropped out of the stratosphere in a space suit. To watch him accomplish this amazing feat, watch this video.

Scribe for next class: Caleb

Update (in addition to the white day update):

Tips for the derivative matching game (which has been on several quizzes):

- Look for maxes and mins (for any horizontal value on the function that has a tangent line with a slope     of zero, the graph of the derivative will cross the horizontal axis at that value)

- Look for +/– slope (when the slope of the function is positive, the vertical values of the derivative will be positive, and the same goes for negative)

–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

Remember that the dy/dx notation means take the derivative of y with respect to x. These variables can be replaced with any other variable (oftentimes in physics we take the derivative with respect to time, so for example dx/dt)