Assignment #11

1) Lectures are very large and move quite rapidly. I think that this is how lectures in college will occur. You will have to constantly be paying attention, making sure you take down each bit of information the professor states.

2) Some classes I would take during college would be Electrical Engineering and Computer Science, Mathematics, Finance, and other business classes. I think that when the professor posts their lecture online is extremely helpful because it allows you to go back in-case you missed any information, and can give you the ability to examine and go over your notes you took during the lecture thoroughly.

Assignment #10

1: Logistic equations are used to represent growth over time. These equations are used primarily to describe population growth within a specific area. Environments can only hold a specific amount of animals or people or other things because their materials are limited and do not continuously grow. Logistic equations solve this problem by creating a carrying capacity for these locations

2: The point is limited when it is divided in half. This point then becomes known as the point of maximum growth on the logistic curve.

3: First you separate the different variables allowing the equation to then be able to get integrated. If the variables are not separated then you cannot solve for those variables making the problem incomplete.

Assignment #9

The video explains how animators at Pixar use the splitting and averaging of surfaces to make 3d animations used within their movies. They use Pascals triangle, which is used to create smooth curves and shapes for the animations. Pascals triangle does not work for all surfaces used in animation and other equations are required. They discuss how by splitting and averaging the shapes an  infinite amount of time, the two points used will infinitely continue to get closer until they reach a specific limit and come together at the shapes original midpoint.

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 have to set u equal to 2x. This is because it is within the function of U^1/2. du then equals 2dx, but it cant because of (4x+1)dx

Assignment #7

1) The general solution to (x^n dx) is {x^(n+1)]+C. The C is important because it states whether or not there was a constant present before the integral was taken.

2)
A.  sin(x)dx=-cos(x) S C -S -C  :going right to left      south carolina - south -carolina
B.  cos(x)dx=sin(x) S C -S -C  :going right to left        south carolina - south -carolina
C.  sec^2(x)dx=tan(x)     - it sounds easy to remember if you say it to yourself a few times quickly
D.  csc^2(x)dx=-cot(x)     - opposite of sec^2    all of the C's get negated
E.  sec(x)tan(x)dx=sec(x)   -  its a pattern secxtanxsecx
F.  csc(x)cot(x)dx=-cot(x)  - its a pattern except it gets negated cscxcotx -cscx

Assignment #6

The second derivative test is used to determine whether or not at specific x values in a differentiable function can have relative extrema. This is determined by plugging in the critical numbers of the equation, making the equation positive or negative. If it is negative there it is concave down and has a relative max, if it is positive then it is concave up creating a relative min. To do all this you must use the first derivative, which finds the critical numbers allowing you to determine the concavity, relative max and relative min.

Assignment #5

Due to the use of a differentiable function, f(3)=30 and f(5)=30 therefore f(3)=f(5). This also means that there is a different value for x between 3 and 5, which is f(4)=25. According to Rolle's Theorem, there must be at least one point between 3 and 5 where the rate is equal to 0