This week we talked about cross sections. Basically, we were defining a volume based on an area between functions and the width or length of the distances between them. We could define them using infinitely many 2-D shapes and calculating the volume based on the area of the shapes. As far as I could tell, this seems a lot less complicated than the rotation volumes, simply because we didn't have to worry about manipulating the functions to be defined with different variables when rotating around different axes. That really takes the complexity out of the equation, as most of it is plugging and chugging anyways. The fact that we came into this week with more experience in working with integrals made the whole thing a lot easier to understand as well. I hope that the rest of calc is this easy.
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Lately, we've been working on rotated solids. The basic concept is that you rotate the area under a graph around a certain axis. Generally, depending on the variable by which the function is defined, and the axis that it is rotated around, you would choose the disk method, or the shell method. The disk method is generally easier if the function is defined by x and rotated on a horizontal axis, or if the function is defined by y and rotated on a vertical axis. The volume is calculated by adding the volume of infinitely small disks perpendicular to the axis of rotation. The shell method is generally easier if the function is defined by y and rotated on a horizontal axis, or if the function is defined by x and rotated on a vertical axis. The volume is calculated by adding the volume of infinitely small cylinders parallel to the axis of rotation. Initially, I had some trouble visualizing the situation, but after a little practice, it became pretty easy.
So this week, the primary topic of discussion was slope fields. A slope field is a graphical representation of the slopes when given the derivative of a function. They are pretty versatile because they can represent the slopes of not only a function with a derivative with only one variable, they can represent a function with a derivative defined by both y and x. It's a concept that, truth be told, I don't really know the purpose of, but I'm going to trust that it's something that, like the Fundamental Theorem, will reveal it's purpose and usefulness once I become more knowledgeable in calc. I also tried to be more active in the CCC's this week. It is a bit hard for me because I'm not very good at explaining things, and I don't really need too much help, but I try my best to explain the concepts people need help with, and I just ask Eric for assistance if I can't find a good example or way to explain something.
I think that this section was primarily review, as we had, albeit briefly, covered it before. It was a cool concept, making derivatives a lot easier by factoring out a common u value. At first, it was kind of confusing, because I was confused about when we had to take the derivative of u, and the actual function, and when we were supposed to plug u in the function or take it out. Basically I was a bit confused on taking the derivative of f(u) with respect to x or with respect to u. But as with almost everything else, once I did a couple of problems, and watched the examples done during the lesson, it was pretty easy. On a semi related note, I was really satisfied with how I did on the short answer portion of the quiz, and I'm glad I caught the double negative in the last problem, as that would have made my grade lower than it already is.
I think that in learning about the Fundamental Theorem, I mostly used deductive reasoning. In general, as a student, I like to find out the "why", instead of finding out the "how". That being said though, this section was really disappointing. Maybe my expectation of a mind blowing new concept or a revolutionary formula was too high, but as far as I can tell, it's basically saying: The derivative of the anti-derivative of a function is the function. Whoopdie-freaking-do. Even the book says that "if any equation deserves to be called the Fundamental Theorem of Calculus, this equation is surely the one." Perhaps I don't understand Calculus well enough to recognize the gravity of the equation, but it just seems redundant. So, to answer your question, I think it fits within calculus because it's true. The derivative of the anti-derivative of a function is the function, but I just don't see the significance.
I think that these topics are getting more and more interesting. There have been a lot of "Aha!" moments this week, because I have been able to see the connection between what we learned now, and what we learned in the past month or so. The biggest "Aha" moment for me was when we were finding the area under a curve using an integral. I was really confused when you were going over the proofs and stuff, but when I realized that the area underneath was just the position of the point when the curve itself was the velocity, I realized that we only needed to take the anti-derivative of a function to get the area. I also realized that the limit of the sum was just a fancy way of saying "add up all of the infinitely small rectangles under the curve. After figuring that stuff out, the rest of the lessons came together, and it was smooth sailing from that point onwards.
This week the big topic of conversation in the classroom was optimization. Basically, we would be given a certain limit on the perimeter, surface area or something similar to that, and we would have to figure out the maximum area of a certain shape that fits the criteria. I don't mean to sound arrogant or anything, but this week was pretty easy for me. This mainly comes from the fact that the math competitions I do have quite a bit of optimization problems, so I'm pretty experienced when it comes to this sort of thing. The only thing that was really new for me was using shapes on a plane as the limiters of area. It initially threw me for a loop, but after doing a couple of problems, I can figure them out quite easily now. Also, doing the problems using derivatives instead of using a graphing calculator was also pretty new. That's really about it this week.
This week we learned about implicit differentiation. Implicit differentiation is just deriving a function with both x and y as variables. Deriving those aren't necessarily more difficult, they just take a bit more work to apply the concepts and simplify. Other than that, it's more or less the same as deriving a normal function. We also learned about the derivative of the Inverse trig functions. Again, this wasn't a hugely difficult concept, the only thing that really was learned were the new formulas. Lastly, we learned about the derivative of Exponential and Logarithmic functions. This was also just memorizing the formulas (or writing them down on cardstock). While it may seem lazy for me to say this, this week wasn't really that mind-blowing in terms of new concepts. All we really did was learn new derivatives of different types of functions. After that, we applied them to the problems using the rules we already knew, that's pretty much it, and I don't think that we need to discuss it too heavily. So....Toodles!
This week we covered the way to find the anti derivative of a function. This is done by substituting the variable "u" in to replace a smaller function within the big function, and finding the derivative of u with respect to x. Then, you divide the original function by the derivative of u. Finally, you use the reverse of the power rule to find the derivative of the function with respect to u. Lastly, you substitute the function that u represents in for u, and simplify. Because of the fact the the derivative of the function only symbolizes the slope of the line, you cannot be sure of where the actual function is in terms of how much it is above or below the answer, so you put a "+c" at the end of the original function to represent this unknown y intercept. Most of the errors from me this week have resulted from performing my math really sloppily, and just making really stupid mistakes. I also have a really bad habit of making the mistake of not adding the "+c" to the end of the function. Overall, this was the most I have struggled with a concept in a long time, although I don't think it's because it's particularly difficult, but because everything else has been relatively easy for the last 4 years.
Maybe it's called going off on a tangent because it's a derivative of the original conversation. What's the derivative of optimus? Optimus Prime!
This week we covered the derivatives of trigonometric functions, as well as the chain rule for derivatives. We learned about how the derivatives of a trig function changed when the inside of the function changed. After that we learned how that related to the chain rule. One concept I didn't understand very well was why the derivative of sin(u(x)) where u(x) is an arbitrary function wasn't cos(u(x)) all the time. After talking about it with my parents, I realized that cos(u(x)) represented the derivative of sin(x) at the point (u(x), sin(x)). On a side note, I really like how all of the trigonometric derivatives could be derived from the sine and the cosine derivatives, that makes memorization a bit easier, and it's nice to know that if I ever forget or lose my cardstock, I can always fall back on that. Overall, this derivative stuff is getting pretty interesting. |
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April 2016
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