2005 Free Response

1. The amount of sand being removed from the sand is represented by the integral or anitderivative of R(t).

2. From the given we know that the beach contains 2500 cubic yards of sand when t=0.To find the amount of sand in a given time we need to know the net change of sand meaning the amount of sand leftover when sand is being removed. When finding the change we then just add it to the initial amount of 2500. The integral of S(t) is the amount of sand being added and the integral R(t) is the amount of sand being removed.
3. The rate at which the amount of sand on the beach is changing can be calculated by the formula S(t)-R(t). S(4)-R(4)= 1.909 cubic yards/hrs.

4. To find were where the sand on the beach is at its minimum we graph the equation S(t)-R(t) and find the x-interceps. The minimum is where t=5.118 since t=o,6 are both maximums. To find the minimum value we input t=5.118 into the equation Y(5.118)=2500+fnint (S(t)-R(t),t,0,5.118) which comes up to be 2492.369 cubic yards.

Mean Value Theorem II

f ' (c)= f(b)-f(a)/b-a
1. In the graph the slope of point c is parallel to the slope of chord AB which shows that both have the same slope. When solving for f ' (c) you are basically looking for a point in the graph that has the same slope between two points in the graph.

2. In order to meet the conditions of the Mean Value theorem a function needs to be continous and differentiable. In this example the function in the graph is discontinous. The y- axis is an asymptote.The mean value theorem does not apply here because there is no point in the function that has the same slope of points A and B.

3. The function 7sin(x^2) does not satisfy the conditions of the mean value theorem because the graph is continuous but not differentiable. The slope of points A and B is zero but there is no point in the function with a slope of zero. As seen in the graph the peaks are undifferentiable with an undefined slope which means the mean value theorem does not work in this case.

Mean Value Theorem

f ' (c)= f(b)-f(a)/b-a

1. In the graph the slope of point c is parallel to the slope of chord AB which shows that both have the same slope. When solving for f ' (c) you are basically looking for a point in the graph that has the same slope between two points in the graph.

2. In order to meet the conditions of the Mean Value theorem a function needs to be continous and differentiable. In this example the function in the graph is discontinous. The mean value theorem does not apply here because there is no point in the function that has the same slope of points A and B.

In this example the function is continous but not differentiable at x = 3. The function does not satisfy the conditions of the mean value theorem because the slope of points A and B is zero but the slope of point x=3 is undefined. There is no point in the function with a slope of zero.

The Function f(x) from the graph f '(x)

1. The function f(x) is increasing in the interval (-2,0) U (0,2), since the graph is showing the derivative of f(x) when f(x) has a positive slope f(x) is increasing.

The function f(x) is decreasing in the interval (-infinity,-2) U (2, infinity), because a function is decreasing when the slope is negative.

2. There is local minimum at x=-2. Extremas can be found in critical points when f '(x)=0 or undefined. x=-2 is a local minimum because the slope of f(x) changes from being negative to positive.

There is local maximum at x=2 because at this point the slope of f(x) changes from being positive to negative.

3. The function is concave up in the interval (-infinity, -5/4) U (0,5/4). A function is concave up where f ''(x)>0.
The function is concave down in the interval (-5/4,0) U (5/4, infinity). A function is concave down where f ''(x)<0.
I can tell from the graph f ' (x) where f ''(x) is positive and negative by looking where the slope is positive and where the slope is negative since f ''(x) is derivative of f ' (x).

4. f(x) is a power function of 5 because the graph of f '(x) is to the power of 4 so its anti-derivative has to have a power of five.

Growth vs. Fixed

1. Which mindset do you think you are a part of when it comes to "intelligence"? According to the reading, what tells you that you are of this mindset?
I consider myself to have a growth mindset when it comes to intelligence typically because I always tell myself that the more you know the smarter you get but also have some characteristics of a fixed mindset. In the reading a person who has a growth mindset is described as someone who does not back down from challenges. I take challenges to see how strong I am in a certain area but am also cautious about the challenges I choose. I always try to learn from people who have had success to see if there is anything I can relate to.

2. How has this mindset helped or hurt you in math?
A growth mindset has helped me put an effort into my work such as like in math problems. I always make things as simple as I can make them because the simpler it is the easier it gets.

3. What is your reaction to finding out that the brain is just a big muscle that can be trained?
I was not very surprised because the brain is just like any other muscle the more you put it to the challenge the stronger it gets. Some people train their muscles (brains) more than others which are why some are more successful than others.

4. How do you see this new piece of information affecting your future?
The more a person works at something the better the outcome is. This information will help me in being able to understand others by considering the way they think.