2005 FrEe rESPOnSe #5

The tide removes sand from Sandy Point Beach at a rate modeled by the function R, given byA 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 0 greaterthanorequalsto x lessthanorequalsto 6. 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(2+5sin(4piet/25),x,0,6)= 31.816 cubic yards


---We use function R(t) because it just asking about the sand that is being removed.



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



Y(t)= 2500+fnInt( S(t)-R(t), x,0,6)
-----It's the integral of function S(t) minus R(t) from 0 to 6 because we want to know how many cubic yards are going to be plus the ones that are already at time t=0

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

Plug in t=4 into
S(t) - R(t)=
4.615 - 6.524 = -1.909 cubic yards/hour

----- We plug t=4 into the rate functions so we know the rate at which it's changing at that time.

(d) For 0 greaterthanorequalsto x lessthanorequalsto 6, at what time t is the amount of sand on the beach a minimum? What is the minimum value? Justify your answers.

------We set S(t) - R(t) = 0 , so we can find the critical points. These rate functions are the derivative of the volume of sand on the beach.
The critical points are
X=0
X= 6
X= 5.12, here the functions change sign from negative to positive so there is a minimum.
So at time t= 5.12 there is a minimum amount of sand on the beach.
To know exactly how much it's there at that time, we have to plug in t= 5.12 into Y(t) likethis

Y(t)= 2500+fnInt( S(5.12)-R(5.12), x,0,6)
* I couldn't find it cuz i have to know the exact antiderivative for the integral and it's hard (ithink,,__ help here pleasee) -_-

Super MEAN value theorem !

1.- Analytically the Mean Value Theorem it's where
f'(c) = [f(b) - f(a)]/(b - a).
By breaking down the formula that defines the Mean Value Theorem

(f'(c) = [f(b) - f(a)]/(b - a)) we have that:
-Tangent line's slope at point c defined by f'(c) =
-Secant line slope defined by [f(b) - f(a)]/(b -a).

This means that in the interval [a,b] there will be a point c whose tangent line's slope will be the same as the slope from the secant line between points a and b, which will make both lines parallel to each other.

Graphically ( to understand it better):
I'm going to use f(x)= √x +2 as an example.
The red line it's f(x)= √x +2

The green line is the secant line ([f(b) - f(a)]/(b -a)) between the interval [1,25] where a=1 and b=25.
Secant line equation= f(x)= 1/6x+2.84
Secant line slope= 1/6

Now according to the Mean Value Theorem there is going to be a point c in between this interval whose tangent line's slope will be the same as the slope of this green secant line.

In order to find that point c, we look for the derivative of f(x)= √x +2 which is the slope for f(x)= √x +2. The derivative is f'(x)= 1/2√x and we set it equals to the slope from the secant line, like this

1/6=1/2√x
and at the end we get x= 9
9 it's our point c where its tanget line will have the same slope as the secant line!

In this way, after plugging in x=9 in f(x)= √x +2 we get the coordinates (9, 5), which is the point for the tanget line.

The pink line it's the tangent line at point c which is (9,5) that has the same slope as the secant line in the interval [1,25] thus making them parallel to each other.

Tangent line equation= f(x)= 1/6x+3.5
Tangent slope= 1/6

2.- However, the Mean Value Theorem fails if in the interval [a,b] the function is discontinuous or not differentiable.

DISCONTINUOUS:

Here the Mean Value Theorem Fails because in the interval [a,b] the function is discontinuous and there is no guaranteed point c between [a,b] because the function has a gap and it doesn't satisfy the continuity requirement. (NOTE: L in this graph is just to prove the Intermediate Value Theorem, which is different from the Mean Value Theorem. I used the graph due to the discontinuity it shows.)

However, I fixed the graph and this one to the right its clearer!!!








NOT DIFFERENTIABLE:
.
The Mean Value Theorem fails in the interval [-4,4] because it has a corner on x=0 which makes the function not differentiable at that point. This means that a tangent line at x=0 doesn't exist.

A very MEAN Value Theorem!!!!!!!!

1.- Analytically the Mean Value Theorem it's where

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

Graphically:

This means that in the interval [a,b] there will be a point c whose tangent line's slope will be the same as the slope from the secant line between points a and b, which will make both lines parallel to each other.

By breaking down the formula that defines the Mean Value Theorem (f'(c) = [f(b) - f(a)]/(b - a)) we have that:

Tangent line's slope at point c defined by f'(c) =

Secant line slope defined by [(b) - f(a)]/(b -a).

2.- However, the Mean Value Theorem fails if in the interval [a,b] the function is discontinuous or not differentiable.

DISCONTINUOUS:

Here the Mean Value Theorem Fails because in the interval [a,b] the function is discontinuous and there won't be any point c between [a,b] because it just doesn't exist.

NOT DIFFERENTIABLE:

The Mean Value Theorem fails in the interval [-4,4] because it has a corner on x=0 which makes the function not differentiable at that point. This means that a tangent line at x=0 doesn't exist.

From f(x) to f'(x)

1.) Where is the function, f(x), increasing? Where is it decreasing? How can you tell from this graph? Explain
The function f(x) is increasing in the intervals (-2,0) U (0,2) because there all the outputs for f '(x) are positive, which means that f(x) had a positive slope in all those x values at that interval and therefore it was increasing. The graph of f(x) is decreasing in the intervals: (-infinity, -2) U (2, infinity), because at those intervals the outputs for f '(x) are negative and therefore f(x) had a negative slope while it decreased.

2.) Where is there an extrema? Explain. (There are no endpoints.)
There is an extrema at x= -2 , 0, 2
There is an extrema at those values of x because there f '(x) is zero and since wherever f(x) has a slope of its tangen line (f '(x)) equals to zero this means that the graph has reached a plateau, which is like a high stable elevation. And because f '(x) is the derivative/slope from f(x) , at that plateau of f(x) the slope it's going to be zero. Also we can know the extrema by using the critical points, which happen when f ' (x) = to zero or it's undefined or at endpoints. However, we don' t have endpoints but we can see that f '(x) =0 at x= -2, 0, 2 as said before.

3.) Where is the function, f(x), concave up? Where is it concave down? How can you tell from this graph?
The function f(x) is concave up at the intervals (-infinity, -1.20) U (0, 1.20) because there is where f ' (x) has a positive slope and it's decreasing and we know by our concavity test that f(x) it's going to be concave up when f ''(x), which is the derivative of f ' (x), it's greater than zero (f ''(x)>0). This means that it's going to be concave up when f ' (x) it's increasing. The graph it's going to be concave down in the intervals (-1.20, 0) U (1.20, infinity) which is where f ' it's decreasing. And when f ' (x) is decreasing, f ''(x) it's going to be negative ( 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 should look like an f(x)= x^5 since f ' (x) it's x^4 , and according to the pattern if you go from f(x) to f ' (x) the latter will drop down a degree but if we go from f ' (x) to f(x) then I would add one degree to find f(x).

My MinDsEt

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?

When it comes to intelligence, I believe I mostly have a growth mindset and a little of a fixed mindset. I say this because I always like to learn new things and that implies that I have to be willing to take challenges so I can improve myself. I truly agree that the mind is such a powerful muscle that can be trained. As I've heard we don't really get to use all the "free memory" in our brains so we can continue to learn more and more to become wiser. That means that our intelligence depends on how much we are willing to learn in our lives. I also put lots of effort in my work because I know at the end it's somehow going to pay off and the feeling I get after doing things right and with patience is satisfactory. I would say I have a little of a fixed mindset because sometimes I take criticisms as something very bad and that lets me down for a while. As they explain, I get to feel bad about myself but later on my spirit comes back and I move on. However, I 've let go many important things out of my hands when I feel down do to criticism and negative feedback.

2.- How has this mindset helped or hurt you in math?

Well the growth mindset has helped me a lot throughout my calculos class because everyday I'm ready to get that other piece of information that will expand my mind and I'll add another gold coin of knowledge to my treasure. Even though sometimes I don't get to understand things, I'm very perseverant and try to understand until I find the meaning behind it. However, sometimes when I do a math problem in public and I get it wrong I feel that everyone it's all over me saying that I'm wrong and many other things that make me feel bad. And maybe that's why I don't really like to do problems in public and now more than ever I dont't have to let that control me because of the "question card" thing that Miss Hwang has assigned to us.

3.- What is your reaction to finding out that the brain is just a big muscle that can be trained?
My reaction has always been very impressive and exciting because I know a have a powerful machine in which I can input all the information I want.

4.-How do you see this new piece of information affecting your future?
I see that this new information will help me for the better in my future because now I know some parts in which I'm having problems with. I want to reach my full potential and for that I have to fix some of the areas in which I'm having problems by having more confidence in myself and getting all criticisms and negative feedback in a positive and constructive way.