As you can tell, Gavin is enjoying how to lean to use cross sections. So cross sections are similar to the shell method, but instead of rotating around an axis or line, you just put an infinite amount of semi-circles or squares on the graph. When getting the volume using cross sections of a graph with semi-circles, you have 1/2 pi and multiply that by the integral of the top equation minus the bottom, all divided by 2. Then take that and square it. When using squares, you just take the top minus the bottom and square it.
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The shell method is when you take an enclosed area in a graph and rotate it along the x-axis, y-axis, or and other vertical or horizontal line. The picture below provides an example and hoe to solve it. To solve for the area, you have to have the integral from a to b of x times f(x)dx. Then multiply the whole answer by 2pi. If the problem rotates along the x-axis, the you have to change the equation into terms of y.
Slope fields was a new big concept that we learned this week and I feel that they are not that hard to figure out. Slope fields are when you graph the slopes of a derivative at certain points. After you graph a bunch of points then you can tell what the original graph looks like. So, in the slope field above, you can see the graphed slopes at each point. When x=-3 the slope is 2 and when x=-1 the slope is 0. As you can see, all you do is graph the slopes of every point.
This week I learned on how to solve complicated integrals by using "U" substitution. Here is an example: First, I found the U which was (x-4) and then take the derivative of the U. When using "U" substitution, you have to find new bounds for the U. How you find the bounds is you plug in the original bounds into the U function. Now at the second step, you take the anti-derivative of the original function that has U substituted in. With the new anti-derivative, you plug in your new U upper bound into the equation and then minus the lower bound plugged into the equation. The answer ends up being 64/3.
I am more of an inductive learner when it comes to things. I am a person who finds patterns in math and from those patterns, I use to solve other ones. Inductive learning helped me with the Fundamental Theorem because I saw the pattern that helped me and my group solve all of the example questions. This theorem is fundamental because it is one of the basics for calculus. I feel that this theorem ties everything together that we learned throughout the year. We use derivatives, anti-derivatives, and integrals.
At first I was confused on how integrals work. After spending most of the week on them, I am starting to understand them more. So first we learned that the area underneath a curve is equal to the distance traveled. I understood that fairly well from using the LRAM MRAM and RRAM. Example down below: This concept was easy for me to wrap my mind around. LRAM stood for the rectangles wound have their left corner touching the graph, RRAM stood for the rectangles wound have their right corner touching the graph, and MRAM stood for the rectangles wound have their center touching the graph. But, as soon as we started to get into using smaller and smaller rectangles to find the area, I got somewhat lost. Now I know about the calculator function that makes those problems that were hard a breeze. Thank you " fnInt( ".
Overall, all I learned this week was optimization. We worked on different problems with different scenarios. One example was to design a 1 liter oil can that takes up the least amount of material to make. And it is given that 1000cm^3=1 liter. When always doing optimization problems, I first draw a picture of what I am solving for because it helps me picture what is happening. To solve these optimization problems, you need to have two functions because you need to substitute one into the other in order to solve for a variable. I know the volume of the oil can and I am trying to solve for the surface area. Volume for a cylinder equals Πr^2 times h. Where r=radius and h=height. Surface area = (2Πr^2) + (2Πrh). Solve the volume equation for h and then plug that new equation into the Surface Area equation. Take the derivative of the new equation made and then solve for r. once you have r, plug the r values into the first equation to solve for h.
We did not do much this week other than reviewing and taking a quiz. But on Friday, we learned how to maximize and area/volume by algebra. An example is down below: So in class we had a 10 by 12 piece of paper and what we needed to do was to cut the corners out to maximize the volume of the box created. So, "X" became the variable that stood for how deep to make the cut. Because volume is l x w x h, you would just plug your variables into the equation. So, the height is equal to the distance that you cut out of the corners which is "X" and the length is equal to 12 minus the two corners that you cut out which is "2X". The width is equal to 10 minus the two corners that your cut out which is "2X" also. The final equation that you get is: X(12-2X)(10-2X). Plug that equation into your graphing calculator and determine what "X" is at the max. "X" would have to be 1.81 in order to maximize the volume.
So, I learned this week that you can determine the concavity of a function by taking its derivative and finding inflection points. Here is an example below. First, I simplified the equation at set it equal to 0. Then you have to determine what x value you put in will give you a 0 as an output. So, when x=0 or 3, then the slope will be 0. Those will be your critical points that will help you determine when the graph is increasing and decreasing. So you put a point in to the left of 0(like -1) and plug that into the equation. if the answer is negative then the graph is decreasing, visa versa. Then find a point between 0 and 3 and then find a point after 3. So the graph is decreasing before 0 and between 0 and 3, but then it starts to increase after 3. Then take the derivative of the function to determine where the inflection points are.
This week, I learned how to take the derivative of inverse trig functions. It is not that hard because all you have to do is follow a pattern that all inverse trig functions have. For example, the derivative of the inverse sin(u)= 1/(square root(1-u^2)). Here's an example problem down below. This is what makes these problems a breeze to do. All you do it plug in the X^2 into the derivative of the inverse sin(u). You get 1/(square root(1-(X^2)^2)). The chain rule still applies, so you multiply by the derivate of X^2 which is 2X. This shows how easy these problems are.
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April 2016
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