1) The lecture was very informative but also moved a bit quickly. A lot of information was presented in a short amount of time and students had (it seemed) to be constantly aware of what was going on. Not paying attention for even a few moments can be dangerous. The class size was also very large. I expect a lecture in college to be very similar to this.
2) Courses I would Take:
- Biological Engineering
- Civil and Environmental Engineering
- Economics
- Electrical Engineering and Computer Science
- Global Studies and Languages
- Mathematics
- Mechanical Engineering
- Physics
There is some value to a university even if lecture videos are online because attending a lecture in person truly gives you the "university experience." You can physically interact with others, ask questions, and make study groups if you need assistance. Watching the lecture video should really only be used after attending the lecture itself in order to reinforce what you just learned.
Thursday, December 11, 2014
Thursday, December 4, 2014
Assignment #10
1) Logistic equations represent growth over time which is perfect for describing populations because populations can grow over time as well. Environments cannot continue to grow forever however and logistic equations solve this problem by implementing a carrying capacity.
2) The point is when the limited is divided in half (L/2). This point represents the point of maximum growth of the logistic curve.
3) The first step is to always separate the different variables (usually x and y) so that we can integrate the equation. If we do not separate the variables then we do not know what we are solving for and the whole understanding of the problem becomes null.
2) The point is when the limited is divided in half (L/2). This point represents the point of maximum growth of the logistic curve.
3) The first step is to always separate the different variables (usually x and y) so that we can integrate the equation. If we do not separate the variables then we do not know what we are solving for and the whole understanding of the problem becomes null.
Thursday, November 20, 2014
Assignment #9
The video discusses the splitting and averaging of surfaces
in order to create the animations we see in Pixar movies. Pascal’s Triangle is
used in order to create smooth curves and shapes (for the limits) for various
objects. For surfaces however, Pascal’s Triangle does not work and a different
set of mathematical tools must be used in order to figure out weights that will
generate smooth objects. Tony (the non-British man that is talking) discusses approaching
infinite in order to create these smooth shapes. By splitting and averaging
shapes an infinite number of times, Tony is saying that the two points will get
closer until they reach a specific limit and come together at the shape’s
original midpoint. Weights are carefully chosen in order to produce the
smoothest surfaces necessary.
Wednesday, November 5, 2014
Assignment #8
1)
A) ʃsin u du = -cos u + C
B) ʃcos u du = sin u +C
C) ʃtan u du = -ln |cos u| + C
D) ʃcot u du = ln |sin u| + C
E) ʃsec u du = ln |sec u + tan u| +C
F) ʃcsc u du = -ln |csc u + cot u| + C
2)
You would set "u" equal to "2x" because it is inside the function "u^1/2". du would equal 2dx but it cannot be achieved because of the (4x+1)dx.
A) ʃsin u du = -cos u + C
B) ʃcos u du = sin u +C
C) ʃtan u du = -ln |cos u| + C
D) ʃcot u du = ln |sin u| + C
E) ʃsec u du = ln |sec u + tan u| +C
F) ʃcsc u du = -ln |csc u + cot u| + C
2)
You would set "u" equal to "2x" because it is inside the function "u^1/2". du would equal 2dx but it cannot be achieved because of the (4x+1)dx.
Tuesday, October 21, 2014
Assignment #7
1) The general solution to (x^n dx) is [ (x^(n+1)) / (n+1) ] + C. The C is very important because it represents a constant that will differentiate one integral from another.
2)
sin(x) dx = -cos(x) - It is going in reverse (up instead of down) of S,C,-S,-C
cos(x) dx = sin(x) - It is going in reverse (up instead of down) of S,C,-S,-C
sec^2(x) dx = tan(x) - Memorization, since secant is squared I think of tan
sec(x)tan(x) dx = sec(x) - Memorization, since there is a sec and a tan I think of sec
csc^2(x) dx = -cot(x) - Memorization, since co secant is squared I think of a negative co tangent
Integral csc(x)cot(x) dx = -csc(x) - Memorization, since there is a co secant and a co tangent i think of a negative co secant
2)
sin(x) dx = -cos(x) - It is going in reverse (up instead of down) of S,C,-S,-C
cos(x) dx = sin(x) - It is going in reverse (up instead of down) of S,C,-S,-C
sec^2(x) dx = tan(x) - Memorization, since secant is squared I think of tan
sec(x)tan(x) dx = sec(x) - Memorization, since there is a sec and a tan I think of sec
csc^2(x) dx = -cot(x) - Memorization, since co secant is squared I think of a negative co tangent
Integral csc(x)cot(x) dx = -csc(x) - Memorization, since there is a co secant and a co tangent i think of a negative co secant
Thursday, October 2, 2014
Assignment #6
The Second Derivative test is used to determine at what x-values a differentiable function can have relative extrema by seeing if the critical value plugged into it makes the equation positive or negative. If the critical number makes the function a negative the function would be concave down with a relative maximum. If the critical number makes the function a positive the function is concave up with a relative minimum. The First Derivative test is used to find these critical values that can be used in the Second Derivative test to determine the function's concavity, relative minimum, and relative maximum. 
Tuesday, September 23, 2014
Assignment #5
Since the pumping rate can be modeled using a differentiable function, the functions of "f(3)=30" and "f(5)=30" are the same. This means that f(3)=f(5) and there's a different value for an x value between 3 and 5, in this case, "f(4)=25." There must be at least one point in between 3 and 5 where the rate is zero according to Rolle's Theorem.
Monday, September 22, 2014
Assignment #4
http://apcentral.collegeboard.com/apc/public/repository/ap07_calculus_ab_frq.pdf
Question five deals with related rates because a quantity is changing (the volume of the balloon) in relation to the other quantities in the problem (the balloon's radius). This is also all in respect to time.
Question five deals with related rates because a quantity is changing (the volume of the balloon) in relation to the other quantities in the problem (the balloon's radius). This is also all in respect to time.
Sunday, September 14, 2014
Assignment #3
A derivative example that can be solved
using the chain rule is "f(x)=(3x+1)^2." The derivative of this is
"f '(x)= 6(3x+1)." To use the chain rule on “f(x)=(3x+1)^2,” you
must multiply the whole function by two and subtract “1” from the exponent. You
will start off by getting “2(3x+1).” You must then find the derivative of what
is in the parenthesis, 3x+1, and multiply that to “2(3x+1).” The derivative of
"3x+1" is "3." You may then simplify the function by
multiplying the “2” and the “3.” If you wish to, you can even simplify the
function further by distributing the “6” to the “3x+1.”
Monday, September 8, 2014
Assignment #2
The Intermediate Value Theorem says that a function must be continuous over a given interval. If f(x) (a function) is continuous over a given interval, then there is a value of that function, such as x, that lies within that interval as well. It is quite important to establish that the function is continuous because, otherwise, it will have holes, jumps, gaps, etc. This theorem is considered an existence theorem because it implies that a number exists but it does not give us the exact number.
Wednesday, September 3, 2014
Assignment 1
I am studying AP Calculus BC because I enjoy Math and wish to challenge myself in Calculus. I hope to get a better understanding of Calculus in general and obtain a great score on the AP Exam. Mathematics is in many professions, especially Business. I wish to use what I learned in AP Calculus BC to help me in the Business field.
Andreas Tsoumpariotis
Andreas Tsoumpariotis
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