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.

Algebra vs. Calculus

1.The difference between finding the limit of a function as x=c and actually plugging in the number x=c, is that the limit of a function as x approaches c could not be a value of the function if the function is not continuous at c. In a limit you are finding the value as x is approaching the constant c which can be different to what the actual value is. When plugging in x=c to the function you are finding the exact value of the function. The best way to see the difference is through a function that is not continuous for example a piecewise function which can have several different functions with different limits when x=c or a different value when x=c.
The two cases are the same when the function is continuous were the limit when x approaches c is the same as the value of the function when x=c.

2. The similarities when finding the derivative of a line and the slope is that the value is the same in the case of a line because in both functions you are finding the change in y over the change in x. In a derivative when finding the value we use limit as h approaches 0 f(a+h)-f(a)/h meaning that we are assuming that the difference as h approaches a is infinitely small.
Derivatives are mainly used to find the slope of a point in a curve by finding the slope of its tangent line.

Limits

The whole concept of how limits worked took me a while to figure out mainly the part of discontinuity and continuity. I have trouble in trying to figure out whether a piecewise function is continuous or discontinuous.

One of the biggest problems I have is trying to figure out End behavior models for trigonometric functions without using a calculator for example problem number 38 on pg 95 which is f(x)=ln[abs(x)]+sinx. I was able to figure out the right end behavior model knowing that sinx was not really significant because the smallest value is -1 and that the biggest value is 1 so I knew that ln[abs(x)] was the right end behavior model.

I still have trouble in trying to figure out how to solve for horizontal asymptotes for a function. I am not sure whether there is a way to solve for it without having to graph it.

I was really confident about the limits test after seeing the practice test we got primarily because I did not see anything that gave me too much problem. After taking the test I was blindsided by a couple of problems on the test which is still bugging me that I was not able to solve which was one about explaining why a trigonometric function was continuous and to graph a sketch. The other one was the free response which had a piecewise function with a function having a variable on it.

Life after high school

There are many majors that I found a lot of interest in but the ones I found the most interesting were in the region of Biological and Biomedical Sciences and also in the region of engineering. I have not really decided what I want to be but know it is between a surgeon and an engineer.

The first major I am interested in is Biomedical Sciences which is basically biology but more in depth. It also deals with how it relates to the body and health, medicine, and disease. People who study this major research for new ways of treating people with various diseases and also analyze technical procedures to improve human human health.

The second major I am interested in is Biological engineering which deals with knowing how to incorporate engineering to helping solve medical problems. People in this field look for new ways to improve machinery to aid in diagnosing and treating certain diseases. I am very interested in majoring in this area particularly because it deals with engineering and medical issues.

The third major I am interested in is Applied Mathematics because it is a more in depth area of mathematics and requires a lot of thinking and reasoning. People in this field use math and statistics to help solve problems in engineering. I am pretty strong in Math so I think I would be very successful in this field.

The first college I am interested in UC Los Angeles because they offer various majors in the area of biology and health. It is also a top university and has a very good medical school. The university is also highly ranked because it has a lot of resources.

The second college that I am interested in is UC Berkely which offers various majors in the area of engineering. It is a large university in a small setting and is very well known around the nation as a top university. It has a lot of facilities in engineering and is popular in giving degrees in the engineering area.

The third college I am interested in is California Institute of Technology. It is a top university and is extremely competitive but offers many degrees in the area of math. Admission is very low because they only accept a small number of students but the university offers many resources and aid.

Tips and Hints

1. Share how you remember transformations. Do you have any tips or hints that help you remember?

I picked up transformations fairly quickly by just seeing how functions shifted and changed. The hardest part for me in learning transformations was to memorize how the parent function looked like. I remember parent functions by picturing a sketch of it on my head for example how it looks like without the axis whether it is curvy or straight. I then just memorize three points one would be a positive value for x a negative value and zero. I know that most functions have some type of particular pattern and would just use that to find other points in the graph. For example I remember the function f(x)=cosx by knowing that it is wavy and the function intersects points

(-pie/2,0), (0,1), (pie/2), and (pie,-1). I then just finish the graph of by seeing the wavy pattern and finding the rest of the points. In terms of knowing how a graph shifts whether it is up, down , side to side I just see in the equation what is being changed whether it is the input or output. I know that if the input is being changed by adding or subtracting the function shifts side to side for the x values and if it is multiplied by negative 1 it flips from side to side,etc. In terms of remembering what happens to the graph when the function is being change I just remember that if you change the input the x values are being affected and the same for the output.

2. Trigonometry is a lot more simple if you remember the basis of it which is the unit circle. To me knowing the unit circle makes trigonometry a lot more simple. I memorized the unit circle by memorizing the values in the first quadrant and knowing how it changes from quadrant to quadrant. For example the tip Ms.Hwang gave us helped me a lot in finding the radian values for the 2nd, 3rd, and 4th quadrants for example in the second quadrant the value is one less the denominator, for the second one more the denominator, and for the fourth two times the denominator subtracted by one.

3. I Still struggle in trying to remember how the graphs for the inverse functions look like and how their domains change. I also sometimes still get confused on whether what changes should i do to a graph when the function is changed.

Logs and Inverses

A major concept that I understood was that a logarithm is the inverse of an exponential function, for example the function 2^x =8 To find the value of x you would need to do the inverse of an exponential function to find the value of the exponent(x). It is basically the same thing as when you are trying to find the value x in the equation x+2=5, To get x by itself you would do the inverse of adding 2 which is subtracting. Now, back to logs in trying to solve the equation 2^x=8 you would take the inverse which is the log base 2 (2^x)=log base 2(8). You would then get x =log base 2 (8),you can simply this more by knowing that 8 is the same thing as 2^3 which would make it log base 2 (2^3) which would make it 3.



Another major concept I understood was the rules of exponents and logarithms for example in logs- log (XY)=log (X) + log (y) , log (X^5)=5log(x), log (X/Y)=log (X)-log(Y). these rules allow you to write and manipulate logs in many different forms. For example you can manipulate log (10) into log (5)+log (2)=log (5x2)=log (10). These rules are very helpful when trying to find a value in an exponential equation or in a logarithm.



A very important concept that I understood about inverses and functions is that f(f^-1(X))=x and f^-1(f(x))=x. This is very helpful in checking whether or not an inverse function is the inverse of a function. I also understood that a function has to pass the vertical line test which states that if a vertical line was passed through a graphed function it should only intersect the function at one point as seen in the image.
The domain of a function would be what the possible x-values for x can be when inserting into a function or graphically. The domain will give you a boundary about what the inputs(x) can be. The range of a function is the possible y-values of a function meaning a boundary of what the outputs(y) will be.

A function that is one-to-one is any function that passes the horizontal and vertical line test which means it is any function that has an inverse. Passing a horizontal line test on a function is the same thing as passing a vertical line test on its inverse function. Graphically when a function is graphed its inverse will be symmetrical to line y=x.

I had some trouble in trying to figure out an answer to a math problem we got in our homework pg 44 number 35 & 36. I am not really sure whether there are different rules when solving an exponential function that has addition on it.

EVEN AND ODD FUNCTIONS

EVEN FUNCTION

The formula for an even function is defined as f(-x)=f(x) for a function of y=f(x). Graphically an even function would be symmetric about the y- axis as seen in the image.

Since y=f(x) and according to the formula for an even function f(-x)=f(x) then for an even function y=f(-x). For an even function it does not matter whether the input for x is negative or positive because the outcome (y) be the same, which is why there is symmetry about the y-axis. As seen in the example for an even function f(x)=x^2 the input can be negative or positive but the outcome will be the same. f(2)=2^2=4 , f(-2)=(-2)^2=4.

ODD FUNCTION

The formula for an odd function is f(-x)=-f(x) for a function of y=f(x). An Odd function if graphed will have symmetry about the origin as seen in the image. The formula basically says that if a negative value is imputed into a function the outcome(y) will be the negative value of the function for a positive value f(-x)=-y. For example in the odd function f(x)=x^3 if -2 is imputed the result would be f(-2)=(-2)^3, which is f(-2)=-8.

If a positive value is imputed the outcome will be positive. From the formula we can see this

f(-x)=-f(x)in seeing what the value for a positive number would be -f(-x)=f(x),-(-y)=f(x), y=f(x).

For example in the odd function f(x)=x^3 if two is imputed the value would be f(2)=2^3,

f(2)=8

About Me

Hello my name is Ivan 

I enjoy eating food and like to watch TV with my dog 

I decided to take this class to see how strong I am in math and also wanted a challenge.