2005 FR 5

a)How much sand will the tide remove from the beach during this 6-hour period? Indicate units of measure.
S(t)= yrs^3/ hrs - adds
R(t)= yrs^3/ hrs - removes
t=0 there is 2500 yrs^3 of sand
0∫6 R(t)= 31.816 yrs^3 of sand is removed

b)Write an expression for Y(t), the total number of cubic yards of sand on the beach at time t.
Y(t)= yrs^3 (the amount of sand not the rate or acceleration)
Y(t)= [∫S(t)-R(t)dt]+2500
*S(t)is being add and R(t)is being removed therefore you subtract
*you add 2500 once you find the integral because 2500 yrs^3 is the starting amount

c) Find the rate at which the total amount of sand on the beach is changing at time t=4.
Y(t)= yrs^3 of sand (outputs amount not the rate)
Y'(t)= yrs^3/ hr (outputs rate of which the sand on the beach is changing over time)
Y'(t)= S(t)- R(t)
Y'(4)=S(4)- R(4)
Y'(4)=4.6154 - 6.5241
Y'(4)= -1.9087 yrs^3/ hrs

d)For [0, 6] at what time t is the amount of sand on the beach a minimum? What is the minimum value? Justify your answers.
* mins at critical points and/or ending points
Critical Points
Y(t)= yrs^3
Y'(t)= yrs^3/ hrs
Y'(t)= 0
Y'(5.118) =0 t=5.118 hrs
Before Y'(5.118) Y'(t)<0 and After Y'(5.118) Y'(t)>0
Y(5.118)= 2492.37 yrs^3 of sand
Y(0)= 2500 yrs^3 of sand
Y(6)= 2493.28 yrs^3 of sand
The minimum is at t=5.118 with a total of 2492.37 yrs^3 of sand

The Mean Value Theorem

If a function is continues and differentiable within the closed intervals [a, b] then:
f'(c)= [f(b)- f(a)]/ b-a
IN other words this mean that there is a point between the closed intervals of which the instantaneous slope is equal to the average slope of the closed intervals.
For Example:


f(x)=X^2, [1, 4] (its both differentiable and continues between the closed intervals)
the average slope OR in other word the secant line is 4x
To find the secant line you have to use the formula [f(b)- f(a)]/ b-a it give you the slope which is 4. Then you use the point-slope form y-y1=m(x-x1)using either one of the close interval points (1,1) or (4, 16).

According to the mean value theorem there is a point between the close intervals [1,4] of which the instantaneous slope or in other word the tangent line is parallel to the secant line.
Now you use the slope you got and set it equal to f'(c)
f'(x)= 2x
4=2x
x=2
Now you got to find the tangent line at x=2
you got the slope m=4 (parallel lines have the same slope) and the point (2,4)
use the point-slope form to find the tangent line
y=4x-4

As you can see the secant line of the closed intervals and the tangent line of which is parallel is within the closed intervals making the Mean Value Theorem TRUE

The Theorem FAILS when the function between the interval is not differentiable or continuous
Example (not differentiable):
f(x)= abs (x-6)+3 , [2, 10]
the secant line is y=7
the tangent line that is parallel is when x=3 but at that point the function is not differentiable therefore contradicting the theorum (f'(c)= [f(b)- f(a)]/ b-a)






Example (when discontinuous):
f(X)= tan (x) [0, 3.14]
The secant line is y=0
the tangent line that is parallel to the closed intervals does not exist because sec^2(X)can't equal 0

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

1. f(x) is increasing at -2When the output is of the graph f,(x) is positive the function f(x) is increasing when the output for the graph of f'(x) is negative he function f(x) is decreasing when the output for the graph is zero it means f(x) is not decreasing or increasing it is just at rest therefore the intervals in which f(x) is increasing those not include -2, 0, and 2
2. The extremas are at x=-2 and 2
at x=-2 there is a local min. and at x=2 there is a local max
at x=-2 in the graph f'(x) the output is zero from the left of x=-2 it is negative and from the right it is positive therefore making it minimum. At x=2 the output is zero, from the left of x=2 it is positive and from the right it is negative making it a maximum
3. its concaved up at (-infinity , -3/2) U (0, 1)
its concaved down at (-3/2, 0)U (1, positive infinity)
Since the graph shows f'(x) taking the derivative (in otherwords the slope) of f'(x) will give you f''(x). When f''(x) is positive it is concaved up when f''(x) is negative it is concaved down
4.f(x)could equal to x^5. f'(x) is x^4 when using the antiderivative of f'(x)=x^4 and f(x) should be x^5.

Set your mind to success (the true success)

1. Which mindset so you think you are a part of when it comes to "intelligence"?
I am a mix of both a fixed mindset and a growth mindset. When it comes to challenges I could be up for them but, it sadly depends the challenge and the mood I'm in (if I'm too stressed or frustrated chances are I'll give up). When it comes to obstacles i try finding other ways around it and if not I'll just confront it. I'll use what i know to get through it and if i don't have all that is needed I'll try asking for help although i don't really like to. I always try to do my best but, when i don't see results quickly i start losing hope. I really hate asking questions and giving my opinion because i don't like people saying I'm wrong because then i feel pretty stupid. Criticism I don't see as an insult i see it a downer. When it comes to the success of others i can be pretty envious but, then again i let them have their moment. The success of others also encourages me to do better sort of like a friendly competition
2.How has this mindset helped or hurt you in math?
Well because i like math i don't really think its hurt me. Yeah there are times when one problem takes like a half page to do and that can really frustrate but, in the end i get over it. Or sometimes when you do a silly mistakes like confuse a + for a - then you have to see where it is that you did your mistake and then do the problem all over again.
3.What is your reaction to finding out that the brain is just like a big muscle that can be trained?
It useful to know because now you won't go saying your stupid and can't do anything about it. You need to just keep trying, don't give up and see the positive outcomes of your hard work and effort.
4.How do you see this new piece of information affecting your future?
Well now i know that the more challenges i take, the more obstacles i try to confront the stronger my brain will become. All challenges will harvest to an improvement of my minds conditions, all i got to do is try my best and not give up. I also got to learn how ask questions and take opinions to succeed success.