(a) How much sand will the tide remove from the beach during this 6-hour period? Indicate units of measure.
R(t) represents the rate of sand being removed. The integral of R(t) will represent the amount of sand being removed.
6
∫R(t)= 2 + 5sin(4πt/25) dx
0
fnInt(2+5sin(4πt/25),t,0,6) ≈ 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.
Since R(t) represents the rate of sand being removed, subtract R(t) from S(t), which represents the rate of sand being pumped.
*Find the integral of both to find just the amount of sand.
**Add 2500 because that was the amount of sand originaly there at t=0.
Y(t)= 2500 + fnInt( S(t)-R(t) ),x,0,6
(c) Find the rate at which the total amount of sand on the beach is changing at time t=4.
Y(t) represents the total number of cubic yards of sand on the beach (cubic yards), so the derivative of Y(t) will represent the rate at which the total number of sand is changing (cubic yards/hour)
Y'(t) is just S(t)-R(t).
Y'(4) = 15(4)/1+3(4) - 2 + 5sin(4(4)π/25)
Y'(4) = 60/13 - 2 + sin(16π/25)
Y'(4) = 4.615 - 6.524
Y'(4) ≈ -1.909 cubic yards/hour
(d) For 0≤t≤6, at what time t is the amount of sand on the beach a minimum? What is the minimum value? Justify your answers.
From what I remember from class, you:
Graph both S(t) and R(t). Use your graphing calculator for this :p
Then, find the point where they intersect.
*The point of intersection shows where the amount of sand being removed is the same as the sand being pumped at 0≤t≤6.
The point of intersection is at (5.1178, 4.6943)
*4.6943 is the amount of sand at t=5.1178
**Add 2500 because that was the amount of sand originaly there at t=0.
So, the grand total cubic yeards of sand is 2504.6943
Sunday, April 4, 2010
Saturday, March 6, 2010
Mean Value Theorem o.O (remix)
1. The graph below shows f(x)=x-sin(x)
There are two lines in this graph, the red and black line.
The red line is the slope at point c, f'(c)
The black line is connected by the points a and b.
It has a slope of f(b)-f(a) / b-a
These two lines are parallel to each other.
2. The graph below shows f(x)=1 / (x-5)^2
This function has an infinite discontinuity as x approaches 5.
This fails the Mean Value Theorem because there is not a tangent line at x=5 that would be parallel to the line joining points A and B
3. The graph below shows f(x)=sin(1/x)
This function is continuous at at all points but it is not differentiable at x=0
This is an oscillating discontinuity; it occurs at a value of x, (in this case, x→0), near to which a function refuses to settle down.
This fails the Mean Value Theorem because there is not a definite slope for point C.
There are two lines in this graph, the red and black line.
The red line is the slope at point c, f'(c)
The black line is connected by the points a and b.
It has a slope of f(b)-f(a) / b-a
These two lines are parallel to each other.
2. The graph below shows f(x)=1 / (x-5)^2
This function has an infinite discontinuity as x approaches 5.
This fails the Mean Value Theorem because there is not a tangent line at x=5 that would be parallel to the line joining points A and B
3. The graph below shows f(x)=sin(1/x)
This function is continuous at at all points but it is not differentiable at x=0This is an oscillating discontinuity; it occurs at a value of x, (in this case, x→0), near to which a function refuses to settle down.
This fails the Mean Value Theorem because there is not a definite slope for point C.
Saturday, February 13, 2010
f(x) to f'(x)
1. increasing on (-2, 0)U(0, 2)
decreasing on (-∞, -2)U(2, ∞)
when you look at the graph of f'(x), the positive outputs are where f(x) is increasing.
this applies the same when looking at the negative outputs, where is f(x) is decreasing
2. if i were to round it,
local max at x=±1
local min at x=0
the local points are where the graph switches from a negative slope to a positive slope, or vice-versa
in this case, starting from the left, or negative side...
-the graph is going in a positive slope until x=-1, where is reaches its max and starts going down, or in a negative direction
-once it reaches x=0, the graph then starts going up, or in a positive direction
3. if i round, then..
concave up at (-∞, -1)U(0, 1)
concave down (-1,0)U(1, ∞)
where f'(x) outputs a negative slope, then that is where f"(x) is going to go down, or in a negative direction
-same thing when looking at the positive slope, but it will go in a positive direction
4. it has to be x to the fifth power because f'(x) shows that it's slope changes slope 4 times,
this means f'(x)=x^4, so
when finding the anti derivative, you add a power, so
f(x)= x^5
decreasing on (-∞, -2)U(2, ∞)
when you look at the graph of f'(x), the positive outputs are where f(x) is increasing.
this applies the same when looking at the negative outputs, where is f(x) is decreasing
2. if i were to round it,
local max at x=±1
local min at x=0
the local points are where the graph switches from a negative slope to a positive slope, or vice-versa
in this case, starting from the left, or negative side...
-the graph is going in a positive slope until x=-1, where is reaches its max and starts going down, or in a negative direction
-once it reaches x=0, the graph then starts going up, or in a positive direction
3. if i round, then..
concave up at (-∞, -1)U(0, 1)
concave down (-1,0)U(1, ∞)
where f'(x) outputs a negative slope, then that is where f"(x) is going to go down, or in a negative direction
-same thing when looking at the positive slope, but it will go in a positive direction
4. it has to be x to the fifth power because f'(x) shows that it's slope changes slope 4 times,
this means f'(x)=x^4, so
when finding the anti derivative, you add a power, so
f(x)= x^5
Friday, January 15, 2010
Mindset =)
1. I can be put under both Growth and Fixed mindset but if I had to choose one that fits me more, then it will be Fixed mindset.
I see challenges and obstacle as a way to improve my knowledge. Even if I don't get the right answer or do something the right way, I at least learned something: a way not to do it.
2. These mindsets have both helped and hurt me in math.
It has helped me by learning from my mistakes and going through with challenges.
It has hurt me when I don't want to use a lot of effort for just one problem.
3. Knowing that the brain is a muscle that can be trained is interesting. People should work out their brain as much as possible.
They should also not get discouraged when they think that they are not getting smarter.
Its like real exercise, you don't get buff overnight. It takes time.
4. I see this affecting my future as I can go either down the good path or the bad path.
The good path is achieving my full potential and not giving up when things looks impossible.
The bad path is not trying to solve my problems, not taking things seriously, or afraid to fail.
I do believe I can go down the good path and do the best I can do. Life is not meant to thrown away but to live and learn.
I see challenges and obstacle as a way to improve my knowledge. Even if I don't get the right answer or do something the right way, I at least learned something: a way not to do it.
2. These mindsets have both helped and hurt me in math.
It has helped me by learning from my mistakes and going through with challenges.
It has hurt me when I don't want to use a lot of effort for just one problem.
3. Knowing that the brain is a muscle that can be trained is interesting. People should work out their brain as much as possible.
They should also not get discouraged when they think that they are not getting smarter.
Its like real exercise, you don't get buff overnight. It takes time.
4. I see this affecting my future as I can go either down the good path or the bad path.
The good path is achieving my full potential and not giving up when things looks impossible.
The bad path is not trying to solve my problems, not taking things seriously, or afraid to fail.
I do believe I can go down the good path and do the best I can do. Life is not meant to thrown away but to live and learn.
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