Scribe Post - October 5, 2010

We started off class today with the Supercorrection follow up test. This took 25 minutes. If you studied, you should have scored 100 percent, since the questions were almost identical to those on the unit test.

Next, Mr. O'Brien asked us to take a look at our 4th quarter projects from last year (you can find them here). Although we didn't know it when we first did them, these projects have some direct relations to calculus. We first looked at our calculation of acceleration of the BMW in the project:




Since this number is the change in velocity over time, it is actually the derivative of velocity. Therefore, we can write it like this:





Then we realized that velocity is actually a derivative of displacement. Thinking about it that way, you could write acceleration as the following: a derivative of a derivative, or a "higher order derivative."





Because acceleration is a second derivative, we can use the power rule in reverse to find the original function - raise the power of x by one, and divide it out of the coefficient. This is called anti-differentiating. The following steps show the process of taking a higher order derivative and anti-differentiating back to the equation:













(In the above steps, a=acceleration, v=velocity, and x=displacement)

Here's a graph of a, v, and x to get a another perspective on the matter:






a is red, v blue, and x "williams purple."











Another point we made from looking back through our projects is that Velocity is a vector: it is not only a speed but also a direction. The problem we run into here is that a vector can be negative, and in a real life application, you can't be moving in a negative direction. This is because direction is relative: moving negatively doesn't make sense, but you can be moving north, south, east or west. To simplify, in real life, we say that speed is the absolute value of velocity.

Within our projects, there was a third and final connection to calculus: limits. We used the idea of upper and lower estimates to come up with a set of values in between. Each time we filled out a table of values, we used smaller intervals, until finally we came up with a fairly accurate estimation of the distance traveled by the BMW.



Next, we looked at rates of change in a real-world context using economics. Economists use the term "marginal cost" to refer to small increments of change in x, the independent variable (or number of units produced), and how they affect c, the overall cost of production. Here's an example:




The above equation is a cost function, where c is the cost of producing x items. The derivative of this cost function is what we call the marginal cost:




This equation is used to find the additional cost (in any given amount of units) of producing one more item. So, if we were working in 100s of dollars, and we wanted to make 7 more items but were worried about costs, we could use this marginal cost equation to come up with an answer:




Since we were working in hundreds of dollars, our answer is $1900. This is the cost of each of 7 additional items.


Now that we'd seen derivatives from a real world perspective, Mr. O'B decided to test how well we could recognize them graphically. We split up the class into several groups and each group was given a mysterious brown envelope. Unfortunately, it turned out that they just contained a bunch of slips of paper: six yellow ones with graphs of functions on them and six blue ones with the derivatives of these functions. O'B also added three extra random derivative graphs per package, just to be tricky.
The objective: match each function with its derivative.
The reward: a round of IOUs for the winning group.

After ten minutes or so of hard work, Kayla, Nick and I came up with the winning combination. Once we were finished gloating over our prize, we were asked to explain how we came up with our answers. As a class, we went over a few different ways to distinguish which derivative matches each function.

Based on our first function/derivative set, we thought that the derivative just went down a power. This method worked in our simple case of a cubic function with a square derivative, but would be very tricky to use for more complicated graphs.


The second suggestion was that a derivative had the same number of zeros as its function, or perhaps one less zero. This was quickly disproved due to the fact that a function has the same derivative no matter how it is shifted. Therefore, a function which never crosses the x axis could have a derivative with numerous zeros. For example, the graph of (x^2+2) has no zeros, while its derivative, 2x, crosses at (0,0). The graph shows the function f(x) in green and its derivative f '(x) in orange.










The final (and best) idea that we came up with involves the slope of a function. It's important here to know where the derivative graph comes from. The easiest way to explain it is that the Y values of the derivative graph come from the numeric slope of the original function at the same values of x. Therefore, a turning point in a function where the slope is zero corresponds to a Y value of zero on its derivative graph. This is illustrated in the picture above: at x=0, the green line is changing from a negative slope to positive... in other words its slope is zero. This corresponds with the x-intercept at x=0 of the orange derivative line.

It took me a while to come up with that last paragraph, and I'm still not sure how clear I've made the concept. If its still fuzzy in your mind, here's another explanation.

If you want some more practice drawing derivatives based on their functions, try this applet.

One last method of comparison is the symmetry of a function: an odd function (one with origin symmetry) will always have an even derivative. This rule is obviously only useful in certain cases, but should you come upon a function with origin symmetry you'll have the tools you need.


Homework assignment #3 is on the iCal. The next scribe will be T-Mac.

UPDATE:
Reading back through this blog post, I find it interesting how far I've come in understanding this material in such a short time. I'm actually surprised at how much sense my explanation of the derivative graph makes, because I remember not knowing what I was talking about. With so many different derivatives, forms, and things to memorize this year, its been a bit overwhelming. I find it comforting, though, to look back and realize that I get it.