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.

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.

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

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.

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.

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.

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.”