Hello all AP Calculus blog viewers! Long time no blog. Throughout the last couple weeks our class finished chapter 6, tested, and are in the works of chapter 7. The test on chapter 6 went very well and it was mostly over a continuation of definite and indefinite integrals. Section 7.1 is about the functions of position, velocity, acceleration, and jerk. 7.2 includes finding the area between two curves. To do this, one must first find the place where the two curves intersect. Then, take the integral of the first curve (making the intersection points the bounds) and subtract it from the integral of the second curve. The first curve is always the curve that is above the other when you are in the first two quadrants. There are different circumstances when the line is a or b. Almost every day of this week we talked about section 7.3. In this section we learned that when you have a curve and you want to rotate it about an axis or solid line, you must figure out the radius of your line at any point (Usually just the function), square that number or function, and then take the integral with whichever bounds need to be used for the specific example. It sounds very complex when reading it in text, but if you need further help with visualizing the subject, use this link: http://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/812d1354-ca6b-424e-ae7b-248360575cbd.gif. Thank you for reading my blog and have a great rest of your day!
Hello all Advanced Placement Calculus blog viewers. It's been a relatively short week in math class. Monday we received the day off from school, and I had to miss Wednesday for a doctors appointment. We started off on Tuesday by taking a quiz about slope fields and integrals with U substitution. The quiz went well, but it took a lot of hard meticulous work to complete it. Personally, I thought it was one of the hardest quizzes of the year. The next day, we started talking about the separation of variables within derivatives. By separating these variables, it allows you to find the original equation from which you took the derivative. For example, If you have the equation dy/dx=(xy)^2, if you separate the variables within the equation and get y^-2(dy)=(dx)x^2. You then can take the antiderivative of both sides and find the C of the equation. Then, with this information you can find the original equation of F(x). Hello all blog viewers! This week was a very successful week in Advanced Placement Calculus. During Monday and Tuesday, we reviewed old lessons on U substitution. It makes a lot more sense why we learned anti derivatives the way we did in the first trimester in class. It was so we could later use that previous knowledge to understand integrals more and have new ways to solve for integrals. Using U substitution allows for one to solve harder integrals that you wouldn't be able to solve by simple anti-derivative rules. During Monday and Tuesday we also were able to reflect upon the previous chapter when Mr. Cresswell handed back chapter 5 tests. I did very well on it and feel very confident with most of the material. Later in the week we started to discuss slope fields. These fields allow for one to view what a graph of a unique function looks like by using the different slopes of the derivative. For example, when using dot paper and graphing the function from the derivative of dy/dx=x+y, you plug in different coordinates to find different slopes. The slope at (1,1)=2 and so on. This field is very useful when trying to envision things one would not be able to do with normal analysis tools. We have a quiz coming up on Monday and I plan to do a few homework assignments over the weekend to prepare for it. Go Math! Throughout most of calculus, I have used the method of deductive reasoning to understand new concepts . This is because it is much easier for me to understand any specific type of proof or theorem by looking at strict definitions and formulas rather than doing explorations and learning through simpler geometry. however, after being introduced to material by formulas it did help solidify it by using the simple explorations. Specifically when doing the fundamental theorem of calculus exploration I was able to use deductive reasoning throughout the worksheet to complete it and then use inductive reasoning to understand the whole theorem. The best of both worlds right? The word fundamental is defined as forming a necessary base or core. Most of calculus application and fundamental analysis is based of the use of derivatives right? Well what a coincidence! There is a theorem that specifically explains that there is a derivative and anti-derivative for any function. This seems pretty "fundamental" to any application of a derivative. Do you know what the best part of the theorem is? It ties directly into what we are learning about in class this chapter. Integrals. Through analytically analyzing of an integral, one is able to find that the integrand is actually f'(x) of an original function f(x). This is what the whole theorem implies. That there is a derivative and anti-derivative to any function! Go math.
Hello everyone, Long time no talk. This week in AP calculus, the class just continued to learn and practice with integrals (the area underneath the curve). On Tuesday we took an AP style quiz. Although the quiz wasn't worth any actual points, it was a good test to see where I personally stood in the class. I think that I am in good shape going into the end of the school year to take the test, but I have a ton of things that I will need to improve my understanding on and practice more. On Wednesday the class took a quiz over integrals. The quiz was over the material from the past two weeks: Raymond sums, integrals, rules of integrals, you name it. Even though it wasn't the most difficult quiz that we've taken in the class, I took 1/2 of the time taking the quiz trying to figure out the extra credit question. It asked you to prove that on an interval a to b of a linear function, that f(c)=(f(a)+f(b)0/2. After long consideration of how to solve the question, i had to give up and ask Mr. Cresswell how to complete it. It turns out that I was a long way away. To solve it, you must use an equation such as y=cx and then integrate this equation, do some careful algebraic analysis, and then end up with the original proof.
hey all blog Viewers! How has your week been? This was our first week back from winter break and oh it was quite a shock. We started to learn about Integrals and the area underneath the curve. We first did this by using Raymond sums (rectangular approximation method). I wasn't a huge fan of this method since it wasn't exact. During these few days we spent on Raymond sums, I figured out how to use integrals. It is actually very simple. You can think of it this way. If you have a graph of velocity with relation to time and you can find distance by multiplying velocity in time. Which is the area underneath the curve! So basically, you can find position by taking the antiderivative of velocity. You can do this with all functions. It's the same as taking the integral. If you were looking at the integral of the function 5x on the interval (a,b) , you would take the anti derivative of it and get 5/2(x)^2. Them you plug in B and subtract that area from a. This gives you the integral. Very easy with some practice. We also took a look at the mean value theorem. This helps you find an average value of any function without much help from the calculus. We are getting ready to quiz and I feel very confident. Go math!
Merry Christmas and Happy Holidays to all blog viewers. In the previous week we took the test over optimization, related rates, and the non calculator curve sketching. I did very well on the test and aced it! None of the questions were too hard. The related rates that I struggled on the week before the test was cleared up by Mr. Cresswell and videos on YouTube. Also, the questions on the test were much easier than questions on the quizzes. One of the questions I vividly remember from the test involved related rates and the volume of a cylinder. You also had to relate this go a rectangle where x was the circumference of the cone and y was the height. You wanted to know the rate at which the volume changed at a certain length of the circumference x. You can find this setting the radius equal to x/(2pi). Next you plug that back into the volume equation after you take the derivative of the equation. You then are able to find your rate. It's much simpler when you have a picture to follow along with. Thank you for viewing my blog and have a happy holiday. Go math!
Hello all blog viewers! This week in AP calculus I accomplished many things! On Monday I finished the optimization quiz and did very well on it. I got an 18/19 on the quiz and feel very confident for the test that we have coming up on Monday. Since I missed so much school the previous week, I had to go right into the related rates quiz on Tuesday after finishing up optimization. I definitely struggled with the quiz, but I ended up being able to figure out all of the problems after some thinking. One of the problems I struggled with most was a problem involving water dripping out of cone cup. I ended up figuring out that you have to get rid of one of the variables by using similar triangles and plugging all of the numbers back in. Out of all of the tests we have taken in AP calculus, I feel the least confident going into this one. This is because there are so many different types of questions that could be tested on related rates and optimization that it is hard to prepare for it. I look forward to seeing my score on the test so I can see how much I've learned in the second trimester. Go math!
Hello math world! I apologize for my absence throughout Thanksgiving break. Many new things are starting to go on in AP calculus. We learned about optimization this last week and had a quiz on it. We also started to learn about related rates near the end of the week. Optimization is when you want to maximize or minimize a certain value. These sections fall under the applications of Derivatives section because they force you to use Derivatives in order to optimize or relate your rates. Unfortunately I couldn't be at school 3/5 days this week, so I was only able to start taking the quiz today. I'm finding it much harder to understand than things we learned in the first trimester of class. This weekend I'm going to try and watch a few videos in order to grasp the concept more. An example of an optimization problem is if you have 30 square inches of text on a paper, and you have margins that are 2 & 3 inch long, how can you minute the amount of paper used (what are the dimensions of the paper). You want to set up two equations: x×y=30 and (x+4)(y+6)= Area of full paper. Now, you want to be able to isolate one variable, so you can find by using the first equation that 30/y=x. Now after this step you plug 30/y back into the second equation for x, and you then multiply out your equation, take the derivative, and find the zeroes of that derivative. These zeroes will give you the minimization of y, and you go and plug y back in to the original equations to find the dimensions of the paper. These problems are much more time consuming than problems strictly containing derivative usage. I hope that throughout next week I can get caught back up in the work and have a very successful trimester. Go math!
|
Archives
February 2015
Categories |