2005 FR 5

R(t)= 2+5sin(4piT/25)------removes sand
S(t)= 15t/(1+3t)-----------adds sand

1. The amount of sand that will be removed from Sandy beach can be expressed by the integral of the function R(t) on the interval 0 to 6.
Calculator: fnInt(2+5sin(4piT/25),T,0,6) = 31.81593137 = approx. 31.816
Considering that the units is cubic yards, the answer is 31.816 cubic yards.

2. To find the TOTAL amount of sand at any given time, we must subtract the integral of R(t) from the integral of S(t), since fnInt(R(t)) is the amount of sand being hauled OUT of Sandy Beach and fnInt(S(t)) is the amount of sand being hauled IN to Sandy Beach. However, the integrals of these functions must have a certain limitation, an interval. Since we are looking for the integral from t=0 up to any period of time, we can name this variable x. Therefore, we are finding the integral from 0 to x of (S(t)- R(t)) [calculator:(S(t)- R(t),T,0,x]. Also, since there are 2500 cubic yards of sand at t=0, 2500 must be added to the equation.
Calculator: Y(t)= fnInt((15t/(1+3t)-(2+5sin(4piT/25)),T,0,x)+ 2500
or in simpler terms: Y(t)= fnInt(S(t)-R(t),T,0,x) +2500

3. To find the rate at which the total amount of sand is changing at t=4, we use the derivative of Y(t), Y'(t). Since Y(t)= fnInt(S(t)-R(t),T,0,x) +2500, we use nDeriv( to find its derivative., which is S(t)-R(t). Since we are looking for the instantaneous rate of change at t=4, we plug in 4 into S(t)-R(t).
R(4)-S(4)=4.615384615- 6.524135262=-1.908750647=approx -1.909
Since we are finding a rate, the units is cubic yards per hour. The answer is -1.909 cubic yards per hour.

4. So, now we must find what point in time, in between 0hrs and 6hrs, the amount of sand reaches an all-time low. Since the amount of sand on a beach cannot be negative (that just wouldn't be reasonable), we are looking for the smallest positive output of Y(t) on the interval 0-6. First, we must find the critical points, so we use the equation of the derivative of Y(t), Y'(t), and set it equal to 0. We are looking for when the outputs of Y'(t) switch from negative to positive.
0=S(t)-R(t)=Y'(t)
Using my calculator to graph S(t)-R(t), i use 2nd calc, Zero to find when the graph has a y value equal to 0.
Y'(t)=0 at t=5.1178653=approx. 5.118
When testing critical points, the end points must also be checked. This means the critical points are t=0, t=6, and t=5.118. When looking at the graph of Y'(t), it becomes clear that t=0 and t=6 are both maximums. Before t=5.118, the outputs of y'(t) are negative, while after t=5.118, the outputs of y'(t) are positive. This means t=5.118 is definitely a minimum.
The amount of sand on Sandy Beach reaches a minimum at t=approx.5.118 hours
To find the actual minimum value, we must plug in t=5.118 into Y(t)= fnInt(S(t)-R(t),T,0,x) +2500.
Calculator: fnInt(S(t)-R(t),T,0,5.118) +2500= 2492.369483=approx. 2492.369 cubic yards

Revision :]

1. The Mean Value Theorem

The Mean Value Theorem states the following:

• Given that f(x) is a continuous and differentiable function, there is a guaranteed point where the slope of the tangent line at point c, is equal to the slope of the secant line passing through the two end points of an interval. If the interval is [a,b], the secant line passes through the points a & b.
• The secant line and tangent line are parallel, for lines with the same slope are always parallel. They will never intersect!

For example...

The graph of f(x)= sin(x) + 2 on the interval [0,3], proves the Mean Value Theorem. The function is continuous and differentiable on the given interval, a requirement of the Mean Value Theorem. Also, the slope of the function's tangent line (y=3) is parallel to its secant line (y=2). Both of these lines have a slope of 0. The tangent line touches the function sin(x) at point c (3). The secant line passes through points x=0 (point a) and x=3 (point b), the two endpoints on the interval (0,3) [a,b0].







2. Why does the Mean Value Theorem ONLY work for continuous and differentiable functions ?

Okay, let's get down to the basics.

A continuous function is defined as a function for which small changes in the input result in small changes in the output.
Continuous functions have the following properties:

• For all the points on a continuous graph, the limit from the right equals the limit from the left. Therefore, the limit at each point exists.
• [The limit as x approaches a number is the actual output (the y value) at that x].
  

• Also, on a closed interval, every continuous has an absolute maximum and absolute minimum.
• A function may be continuous, but have a point where it is not differentiable [a point where the slope is undefined due to a corner, cusp, jump discontinuity, oscillating discontinuity, or infinite discontinuity (asymptote)].

A function f(x) is differentiable at a point if its derivative exists at that point.
Differentiable functions have the following qualities:

• The slope must be defined at a particular point in the function

• If f is differentiable at a point c, then f is continuous at that point c. (but NOT vice versa).

Continuous but not DIFFERENTIABLE:
Consider the function f(x) = x^(7/9) + 1 on the interval [-1,1]. It is continuous at every point in this interval and also differentiable at every point except x=0, where there is a cusp. At x=0, the slope is undefined. This function fails the Mean Value Theorem first and foremost because there is no tangent line parallel to the secant line y=2. Technically, the line y=1 does not exist, since as i said before, x=0 is undifferentiable. However, if the slope at x=0 existed, it would be 0 and be parallel to the secant line x=2.
This is why a graph must be differentiable and continuous to have a guaranteed point c.









Differentiable but not CONTINUOUS:

Consider the graph of f(x)=(sin(x)/x on the interval [-2,2]. This graph is differentiable at every point except x=0, where it is also discontinuous. This graph fails the Mean Value Theorem because it does not have a guaranteed point c. Point c in this case would the tangent line at y=0, but since the function is neither differentiable nor continuous there, the "tangent line" does not exist. Only the secant line that goes through both -2 and 2 exists.

A very mean value theorem !

The Mean Value Theorem !

The Mean Value Theorem relates the slope of a secant line to the slope of a tangent line. According to the Mean Value Theorem, if a function, f, is continuous on and differentiable on , there is a guaranteed point c in in which

In other words, there is a guaranteed point on the interval [a,b] where the slope of the tangent line through point c (f'(c)) is equal to the average slope throughout the interval [a,b] [f(b) - f(a)]/(b - a). This also means that the secant line and tangent line are parallel, for lines with the same slope are always parallel.

This can be shown graphically by utilizing the function f(x)= x^2.

This function is both continuous and differentiable in the interval [0,2], as shown to the left.

So, to prove that the Mean Value Theorem applies to this function, we must find the slope of the secant line by using the part of the theorem that reads [f(b) - f(a)]/(b - a). Let's use the interval [0,2], where a=0 and b=2.

[f(2) - f(0)]/[2 - 0]

= (4-0)/2

= 2

This is shown to the left, where the green line represents the secant line of the function f(x)=x^2.

The slope of the secant line is 2. According to the Mean Value Theorem, we must be able to find a point between x=0 and x=2 where the slope is 2. To do so, we must find the slope of the function f(x).

Take the derivative of f(x)=x^2.

f'(x)=2x.

Now we have found f'(x) and [f(b) - f(a)]/(b - a), the slope of the tangent line and slope of the secant line, respectively. Now, in accordance to the Mean Value Theorem, we must set them equal to one another to find point c, where the slope of the tangent line is equal to that of the secant line.

2x=2

x=1.

This means that at x=1, the slope of the tangent line is equal to the slope of the secant line. At x=1, both the tangent line and secant line are parallel to one another. Since x=1 lies within the interval [0,2], this proves the Mean Value Theorem.

Now, we need to find the equation of the tangent line. We know the slope is 2 and x=1.

We need to find a point, so we can plug in 1 into the original function f(x)= x^2.

f(1)=1

so the point is (1,1)

use y-y1=m(x-x1)

y-1=2(x-1)

y=2x-2+1

y=2x-1.

The equation of the tangent line is y=2x-1.

The Mean Value Theorem only applies to continuous, differentiable functions!

The Mean Value Theorem only works for differentiable and continuous functions. This is so because a discontinuous function or one that is not differentiable at one or more points does not guarantee a point c where the slope of the tangent line is equal to the slope of the tangent line. In fact, the place where that point c would have been may not be continuous due to an asymptote, cusp, jump, or corner. The point where that c would have been may also not be differentiable, meaning the slope of the tangent line does not exist. Sometimes, the point where the c would have been is neither continuous nor differentiable. For these reasons, a function must be continuous AND differentiable in order for the Mean Value Theorem to apply.

An example of a function in which there is no GUARANTEED point c is f(x)= [sin(x)]/x.

The table in the lower right hand corner suggests the function is not differentiable at x=0.

[sin(x)]/x FAILS the Mean Value Theorem because it is discontinuous at the point x=0.

Another good counterexample to the Mean Value Theorem would be f(x)=| x |.

The Mean Value Theorem FAILS because f(x)=| x | is not differentiable at x=0, where there is a corner.

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

1. Since the function increases when the slope is positive, or when f'(x)>0, this function is increasing in the interval (-2,0)U(0,2). Here, the slope is positive, since the outputs of the graph f'(x) are positive. Thus, the function is constantly increasing. Since a function is decreasing when the slope is negative, or when f'(x)<0, this function is decreasing in the interval (-infinity, -2)U(2, infinity). Here, the slope is negative, since the outputs of the graph f'(x) are negative. Therefore, the graph is constantly decreasing.

2. There is a relative extrema at the points (-2,0) and (2,0). Here, the slopes changes from positive to negative or negative to positive.
At the point (-2,0), slope changes from negative to positive. This is a local/relative minimum.
At the point (2,0), slope changes from positive to negative. This is a local/relative maximum.

3. The function is concave up when f"(x)>0 and concave down when f"(x)<0. In order to find where the graph of f(x) is concave up or concave down, one must take the derivative of the function f'(x), or, in other words, find its slope.
f(x) is concave up in the interval (-infinity, -1.2)U(1.2, infinity). Here, f'(x)>0 (above the x-axis).
f(x) is concave down in the interval (-1.2,0)U(0,1.2). Here, f'(x)<0 (below the x-axis).

4. Since the graph of f'(x) has 4 "bumps" (i don't know how else to explain it), this means that f(x), from which f'(x) is derived from, must be one power higher. Therefore, the graph of f(x) must be an x^5 function. Also, from using my data from f'(x) and f"(x), i was able to draw the graph of f(x). f'(x) provided me with the slope at certain intervals, and f"(x) provided me with the concavity at certain intervals. This helped me finally conclude that the graph of f(x) is indeed an x^5 function.

Mindsets !

1. When it comes to "intelligence," i believe i am a mongrelization of both the fixed and growth mindsets. I am a little bit of both because i tend to avoid challenges when i am simply not in the mood, or lazy. For instance, i dropped AP U.S. History, something i am not necessarily proud of, yet something i did. I believe the reason i dropped the class was because i was too lazy to read text about a bunch of dead men. Quite frankly, i saw no point in doing so. I would much rather spend my time reading interesting books that will entertain and enlighten me. May i say, The Lost Symbol by Dan Brown is an excellent example! I sometimes see other people's success as threatening and daunting. However, i use this feeling of intimidation to achieve greater heights than them and prove myself capable of achieving such an honor as well. For instance, Michael Phelps always makes me want to go to a swimming pool and learn how to swim, even if it does happen to be 5 o' clock in the morning! I might be intimidated by his amazing swimming skills, but i find inspiration in them. Sometimes i think about how satisfying it would be to put him to shame! On the other hand, i never give up! If i am confronted by an obstacle, i simply build a bridge and get over it! I never give up on my goals and do not stop trying until i achieve them. I know that if i try, i will get somewhere. Being somewhere is better than being at square one and going nowhere. Although i must admit some criticism makes me want to cry, i find most criticism is incredibly helpful. By embracing my mistakes and accepting i did something wrong, i can pave the road to being more open minded and even more oftenly correct. I learn from others' mistakes as well as my own, and always take into consideration the circumstances under which these mistakes were made. I try my absolute hardest toavoid these circumstances.
   
2. My conflicting mindsets have helped and hurt me in math. My persistence and constant effort has worked miracles. There have been times when i have worked incessantly on one homework assignment for an entire night. My guinea pig has gone days without any attention from me, all because i was busy deciphering the secret language of calculus. My mongrelized mindset has also, quite sadly, hurt me in math. There have been times when i was too fatigued, both mentally and physically, to try my best on a particular day. I must also admit, i have ignored useful feedback, simply on the account that i was skeptical!
   
3. When i read that the brain was like any other muscle in the human body that can be trained and exercised, i said to myself, "I knew it!" I really do believe that the more you use your brain, the stronger associations you will be able to make, and the more critical thinking processes you will be able to undergo. A human brain needs at least 8 hours of exercise. Anything less will threaten one's intelligence (according to my philosophies).
4. I can definitely see this information impacting my future. I can see myself changing my mindset. I can see myself becoming more aware of my own mental processes and fine tuning them to reach higher standards. I can see myself reading the newspaper for fun, solving mathematical equations, (such as Quantam gravity and the effects of it on the "Big Rip"), and seeing the many connections present in the world, how everything came from something.