2005 FR 5

Rate sand is removed: R(t) = 2+5sin(4πt/25)
Rate sand is added: S(t) = 15t/(1+3t)
Initial: 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.

₀∫⁶ 2+5sin(4πt/25)dt = (2t + (125/4π)cos(4πt/25))]⁶₀ ≈ 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) = 2500 + 0 ∫ 6 (2+5sin(4πt/25)) - (15t/(1+3t))dt

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

Rate sand is added - Rate sand is removed, when t=4
15t/(1+3t) - 2+5sin(4πt/25)
15(4)/(1+3(4)) - 2+5sin(4π(4)/25)
4.615 - 6.524
- 1.909 cubic yards of sand per 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) = 0 and when t = 0 or 6 are critical points where as a minimum or a maximum may occur.
Y'(t) = S(t) - R(t) = 0
Y'(5.118) ≈ 0
Y(5.118) ≈ 2492.369 cubic yards of sand

Mean Value THeorem

1.) The purple line has a slope that is defined as [f(b) - f(a)]/(b - a) which is parallel to the yellow line which has a slope defined as f '(c).

A,B and C are outputs or points of y = f(x). The line that A and B produces is the secant line in which, according to the Mean Value Theorem, is parallel to a tangent line that is somewhere between point A and B on the graph of f(x).

2.) The figure below displays a removable discontinuity and an corner.
There is a jump between x = 0 and x = 3. The blue line illustrates the slope between points A and B which clearly shows that the slope of A and B cannot have a parallel tangent on the function. There is a corner between x = 3 and x = 7. The purple line illustrates the slope between points C and D which is 0. x = 5 is the center between these two points and has a slope that is undefined. THEREFORE, the Mean Value Theorem only applies to continuous and differentiable functions.

f(x) from f'(x)

1 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Increasing - (-2, 0 )U( 0, 2 )
Since f '(x) displays the slope of f (x), whenever f '(x) is positive, is when f (x) is increasing.

Decreasing - ( -infinity, -2)U(2, infinity)
Whenever f '(x) is negative, f (x) is decreasing.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
A extrema occurs whenever f '(x) = 0 or undefined therefore in this derivative,when x = 0 and around -5/4 and 5/4 occurs a possibility of an extrema. Most likely when x = -5/4 and 5/4 are definite extremas because it is where f (x) changes its slope from negative to positive or positive to negative. When x = 0, there is probably no extrema because the slope does change after but rather says positive.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

3 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Concaves up - ( - infinity, -5/4)U( 0, 5/4)
Concaves down - ( -5/4, 0)U( 5/4, infinity)
f(x) concaves up whenever f "(x) is positive and concaves down whenever f "(x) is negative. Whenever the slope of f '(x) is positive is where f (x) concaves up and whenever the slope of f '(x) is negative is where f (x) concaves down.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

4~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The power function is x^5. Since f '(x) is a x^4 function because the function changes its slope 4 times, the antiderivative of x^4 is x^5 thus the function has to at least reach a 5th power.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

My mindset

I think I am sort of in between of a fixed and growth mindset. I usually take on any challenges that come my way but I don't really try to find any. For obstacles, I tend to try it out for a while but then give up after because it's too time consuming. Sometimes I come back to it and try it again but would rather do something more "entertaining". I don't see effort as useless but rather see it as a opportunity to show my potential. Kind of hard to do since I'm a lazy guy. Criticism... Typically, I dislike negative criticism because it discourages me. After a while, I get over it and use that criticism to motivate myself to try better and sometimes don't take the criticism into consideration. For the success of others, I guess I feel sort of threatened. For a guy that lives on competition, the success of others tends to make me want to surpass them more.

Well, I can come up with one thing that had hurt me in calculus. In the homework, if there is a problem that I can't seem to figure out (meaning that it taking me about 30 minutes and still couldn't find a way to solve it), I just leave the work that I've done trying to figure it out and go on to the next problem.

The brain as a muscle? I guess it sort of works. A headache from doing word problems is kind of like soreness after running 4 - 6 miles except that the soreness lasts longer than a headache.

I honestly don't see this new info affecting my future. I just see it as extra information.