HELLO!!!

So right now i'm in my statistics class and Krystal is sleeping right next to me hhaaha. This is a wild class. Random students are in here.
And it's a shortened day so the bell is about to ring.
Well, GOODBYE and I hope I pass the ap statistics test, but i highly doubt it.

Calculus :O

After so much struggling, we've finally SURVIVED AP Calculus.!!!!

Thanks to all the GREAT friends who helped us overcome obstacles,

and ESPECIALLY to a GREAT teacher, Ms. Hwang!! (:

Let's Hope we PASS this Test!!!!8]

2005 FR 5 D:

The tide removes sand from Sandy Point Beach at a rate modeled by the function R, given by

A pumping station adds sand to the beach at a rate modeled by the function S, given by

Both R(t) and S(t) have units of cubic yards per hour and t is measured in hours for . At time t=0, the beach contains 2500 cubic yards of sand.
(a) How much sand will the tide remove from the beach during this 6-hour period? Indicate units of measure.

fnInt(R(t),t,0,6)

fnInt(2+5sin(4pi t/25),t,0,6)=approximately 31.815 or 31.816 cubic yards of sand.

(b) Write an expression for Y(t), the total number of cubic yards of sand on the beach at time t.

Y(t)= fnInt(S(t)-R(t)dt)+2500

Y(t)=fnInt((15t/1+3t)-(2+5sin(4pi t/25)),t,0,x)+2500

(c) Find the rate at which the total amount of sand on the beach is changing at time t=4.

So you need the rate of change of the equation in part b, at t=4, since the equation is the total amount of sand on the beach, and you need to know the rate at which this is changing.

If you dont want to find the derivative, just RECALL that...

When told the given equations, they are already in the units cubic yards per hour, which is the rate at which it is changing., therefore you just need to subtract the rate at which the tide removes sand from the rate at which the tide adds sand. (S(t)-R(t))

But if you want to take the derivative of Y(t)...

Since you know that Y(t) is the antiderivative, that means that the derivative would just equal the original equation "f(x)", which in this case is S(t)-R(t). <--- It is ALREADY the derivative (Rate of Change)

same thing if you do....

Y'(4)=nDeriv(fnInt(S(t)-R(t))+2500... The integral (fnInt cancels out with the nDeriv, because if you take the antiderivative and then the derivative, then you are left with S(t)-R(t), because of the Fundamental Theorem of Calculus Part 1. Taking the derivative of 2500 is just 0, since it is a constant.

so S(4)-R(4)= 4.6154-6.5241= approximately -1.908 or -1.909 cubic yards/hour

(d) For , at what time t is the amount of sand on the beach a minimum? What is the minimum value? Justify your answers.

Y'(t)=S(t)-R(t)

0=S(t)-R(t)

(The GRAPH of Y(t))

Using the Calculator, you find that the critical point is t=5.118 (the 2 curves intersect at this point twice if looking at both equations, but if you're looking at the graph of S(t)-R(t), the zero is t=5.118) and the endpoints are t=0 and 6.

By looking at the graph of Y'(t)=S(t)-R(t), its is clear that x=0 is a maximum because Y'(t) is negative after that point and it is also obvious that x=6 is a maximum because Y'(t) is positive before that point. By looking at x=5.118, you see that Y'(t) changes from negative to positive at that point, and therefore it is a minimum.

The Minimum Value of sand can be found by plugging in the minimum value to the equation that outputs "the total number of cubic yards of sand on the beach at time t.", also known as Y(t), the equation we found in part b. (Wow this equation really comes in handy! (:).

So Y(5.118)= fnInt(S(5.118)-R(5.118),t,0,5.118) +2500. (You want the area of the graph from 0 to 5.118 in order to get the amount of sand at the time t.)

Y(5.118)= -7.630517431+2500 = approximately 2492.369 cubic yards of sand.

Well what I did for part D to find Y(5.118), I first plotted the 2 equations of S(t) and R(t) in the "y=" in the calculator. I put R(t) into the "Y1" and the S(t) into "Y2". Next, I just pressed "math+9" to add "fnInt(", then i clicked "VARS"-->"Y-VARS" --->Function---> "Y2". Then I added the subtraction sign and added "Y1" using the same method. Now I have Y2-Y1, which is equal to S(t)-R(t). Then you just add the conditions ",x,0,5.118)", NOT up to 6. Then just hit enter and add "2500" to get your answer.! TADA!

Im not so sure if this is correct after seeing that some have the same answers, but many others do not. If I am mistaken, feel free to correct me, since it is best to learn from failed attempts (:

Mean Value Theorem Edits :&

Well I only put the corrections here, and the rest of the post is still on the original! (:

1. ----------------------
f'(c)= [f(b)-f(a)]/(b-a)

• This means that the slope of the secant line at 2 points A and B, on a continuous and differentiable function between the intervals [a,b], is guaranteed to be equal to the slope of the tangent line at one point, c. On the graph, c may be the midpoint of [a,b].
  
• Because the slopes of the tangent and secant lines are equal, the lines will be PARALLEL to each other.


• Lets try another function that ISN'T f(x)=x^2, because Ms. Hwang doesn't like it's simplicity (:


• So this time, how about... f(x)=e^x

As you can see, the slope of the tangent line at c=0 is equal to the slope of the secant line between the interval (-1,1), so the lines are PARALLEL.

**I don't like how that one turned out, since the secant line looks weird...,so I shall use another example.. f(x)= -sin(x+pi)-1 between the interval (pi, 2pi).

The graph shows that between the interval (pi, 2pi), there is a tangent line at point c=3pi/2 with the equation y=-2 and a secant line, y=-1. The function, being both differentiable and continuous, allows the slope of the tangent line to be equal to the slope of the secant line, therefore allowing the Mean Value Theorem to apply. Both slopes are equal to 0.

2. -----------------------

It only works for continuous and differentiable functions because if between the interval [a,b] the function is discontinuous(because of a jump or asymptote,) or not differentiable (because of a jump, asymptote, cusp,corner...) then there may not be a tangent line at point c (or there may not even be a point c on the function if it is discontinuous where point c should be) that is parallel to the secant line going through points A and B.

• Since I had an equation involving abs of x, (f(x)=abs (x +1)+3), and im not sure whether I have to re-do this one, I will anyway.


• So... ANOTHER example of a CONTINUOUS but NOT DIFFERENTIABLE function is f(x)=x^(2/3)+1 on (-1,1).

There should be a tangent line (y=1) parallel to the secant line (y=2) at point c=0, but it is not possible for such a tangent line to exist due to the fact that the function is not differentiable at point c.(the slope of the line before c=-1 and the slope of the line after c=-1 are not equal) despite the continuity. Whenever there is a cusp at point c, the mean value theorem does not apply because f ' (c) does not exist.

Ohhhh AND to answer your question ms Hwang, related to my LAST post, the secant line of the second equation is y=1

A Very Mean Value Theorem! xD

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

1. Explain what this means GRAPHICALLY by showing a good example!

• This means that the slope of the secant line at 2 points A and B, on a continuous and differentiable function between the intervals [a,b], is guaranteed to be equal to the slope of the tangent line at one point, c. On the graph, c may be the midpoint of [a,b].
  
• Because the slopes of the tangent and secant lines are equal, the lines will be PARALLEL to each other.
  
• This is proven through the graph of the function f(x)=x^2

The secant line (green line) goes through the points (0,0) and (2,4), where a=0 and b=2. It's slope should be equal to at least one slope of a tangent line(blue line) at a point c. In our case c=1, which is the midpoint between a=0 and b=2.

slope of secant line=[f(b)-f(a)]/(b-a)
=[f(2)-f(0)]/(2-0) ----use(f(x)=x^2)
=(4-0)/2
=4/2
=2

You know that c=1 because

f'(c)= 2 (2 is the slope of the secant line)
2c=2 (slope of tangent is 2x[or 2c],found by taking derivative of f(x)= x^2)
c= 1

f'(c)=2c
f'(1)=2(1)
f'(1)=2 (Slope of tangent line)

2=2
Slope of tangent line=slope of secant line (PARALLEL)


2. Explain why this only works for continuous and differentiable functions.

• It only works for continuous and differentiable functions because if between the interval [a,b] the function is discontinuous(because of a jump or asymptote,) or not differentiable (because of a jump, asymptote, cusp,corner...) then there may not be a tangent line at point c (or there may not even be a point c on the function if it is discontinuous where point c should be) that is parallel to the secant line going through points A and B.
  
• An example of a discontinuous function that fails the Mean Value Theorem is 1/abs(x).
  
  
  
  A point c does not exist between the interval [-2,2] where f'(c)=[f(b)-f(a)]/b-a.
  
  (NOTE: A function may be discontinuous between an interval [a,b] and STILL have a tangent line at a point c that is equal to [f(b)-f(a)]/b-a]!!! Because of the discontinuity, there is just not a GUARANTEED point c, since the Mean Value Theorem fails).
  
• An example of a continuous but not differentiable function that fails the Mean Value Theorem is abs(x+1)+3 on (-2,0).
  
  

There should be a tangent line parallel to the secant line (green line y=4) at point c=-1, as shown in the picture (blue line), but it is not possible for the tangent line to actually exist because the function is not differentiable at the point c (the slope of the line before c=-1 and the slope of the line after c=-1 are not equal)despite the continuity.

The Function f(x) from the Graph f'(x) (:

1. Where is the function, f(x), increasing? Where is it decreasing? How can you tell from this graph? Explain.

• f(x) is increasing between (-2,0)U(0,2). This is due to the fact that when f '(x)>0, the graph of f(x) is increasing.
• f(x) is decreasing from (- infinity, -2)U(2, infinity). I can tell from this graph because whenever f '(x)< 0, it means that the graph of f(x) is decreasing.
• (On the graph f '(x), anything above the x axis is where the graph f(x) is increasing and anything below the x axis is where f(x) is decreasing. **Remember that f '(x) graphs the slope of f(x).)

2. Where is there an extrema? Explain. (There are no endpoints.)

If you're talking about the extrema on the graph f(x)....

• Local Minimum: x=-2 (-2,0) A Local minimum occurs at the point where f ' (x)< 0 and then changes to f ' (x)>0, also known as a critical point ( f ' (x)=0 or undefined ) [the function has to change from negative to positive slope in order for it to be a local minimum).
• Local Maximum: x=2 (2,0) A Local maximum occurs at the point where f ' (x)>0 and then changes to f ' (x)<0, which is also a critical point. (the function has to change from positive slope to negative slope in order for it to be a local maximum).

(Way clearer explanation than on my test, >:\)

3. Where is the function, f(x). concave up? Where is it concave down? How can you tell from this graph?

• Concave up: (-infinity, -1.25)U(1.25, infinity). The function is concave up whenever f "(x)>0
• Concave down: (-1.25, 0)U(0, 1.25). The function is concave down whenever f " (x)<0

4. Sketch the graph f(x) on a sheet of paper. Which power function could it be? Explain your reasoning.

The graph f(x) appears to be an x^5. Since the graph of f ' (x) looks like an x^4, my prediction is reasonable since an x^4 is the derivative of an x^5 graph.

Mindsets :O

1. Well I believe that I have components of both mindsets in me, depending on certain subjects. For example, sometimes I don't like taking challenges, which shows the Fixed Mindset. Like at this moment, I hate the challenge of taking AP English, and I want a way out of it, yet at the same time, I know that it will help me improve my analytical writing and I want to stay. The problem is that I don't like writing essays, and since the class is based on essay writing, I find it very stressing.
At the same time, I am enthusiastic about taking the AP Calculus challenge, probably because I love math and I want to be able to know more, which shows the Growth Mindset. If I have trouble, I know that I can always ask for help, and even if I fail, i'll get right back up and keep attempting.
On the other hand, I do believe that intelligence can be developed. Also, i've recently learned to not care about how people see me. I only care about how I see myself, and If I fail, its my problem and no one elses because at least i'll know that I tried. Yes, it may discourage me and may take me a while to recover, but that doesn't mean that life should end there. There is still plenty of time to get back up and keep adding effort in order to succeed. That's another thing: I don't see effort as useless. It is always important to work hard even if it doesn't get you anywhere at that moment. Later on you'll realize that it made you into a stronger person and gave you more confidence to keep trying. Also, I like receiving criticism. It tells me what i'm doing wrong so that I can change and improve on what i'm doing wrong. There always has to be criticism because not everyone is perfect and if they were, then what is the point of going to school? You go to school to learn and fail, and then get back up, fail, and keep trying until you succeed.
2. This mindset has helped me in math because I'm always up for the mathematical challenge. So far, i've reached calculus, and im enjoying every bit of the challenge. I receive criticism, and it just makes me put in more effort to fix my errors and understand the lesson even further. Obstacles help me grow, even though they may make me gloomy at first. Once I master it, i get ecstatic and realize that I am one step closer to overcoming the challenge. I know that if I need help, I can always count on my other classmates to help me. That is the perk of having classmates that may have succeeded more in that certain area of math.
3. Well, by finding out that you can train your brain, I was given more hope to keep going and learn more.This has reasurred me that school does have a purpose and it is okay to fail, because it just makes me stronger as a person. Everyone struggles, and it just takes time to adjust to it.
4. Now, I feel more confident in accepting challenges and overcoming obstacles. It's not going to be easy, but slowly i'll be able to let go and allow myself to take the risk, even if it leads to failure. No one is perfect, and everyone has had an experience that has allowed them to see this. This doesn't mean that I have to give up if I fail. I have to keep trying, no matter how difficult it may be. I know that I feel crazy writing this right now, but I know that it is the right thing to do. Yes, I may not follow my advice all the time, but it takes some time to get used to it and let it soak in. I've done it before, like in AP Biology. If I didn't get something, I would keep trying to understand until I finally did. It was a struggle that I overcame. This will allow me to succeed in college. I've heard about how difficult it is, and if I fail, I have to keep trying. This article will truly affect my future in a positive way.