Hey everyone, this is my DEV project for Calculus. I made two versions, both using spresent and a blog for those of you who can't run the spresent. I've picked 4 different types of problems including related rates, optimizations, finding the definite intergral, and volume revolutions. Enjoy.
Tuesday, June 5, 2007
Question 1
Little Tom has a talent to roll perfect spherical shape snowballs. He enters a snowball building contest at DMCI. Calculus students notice that the rate of which the radius of the sphere shaped snowballs Little Tom is rolling changes 3 cm per second.
a) At what rate is the volume changing when the radius is equal to 16 centimeters?
b) If the snowman Little Tom built had a volume totaling 729 cm3. Given r(0)= 0, approximately how long did it take him to build it?
Solution
a) Answer dV/dt = 3072π cm3 per second.
From the question we can tell that this is going to be a related rates question. We can solve this by:
• Separating the "general" and "particular" information.
"General information" is within the problem and true at all times.
"Particular information" is true only at a particular instant.
e.g. in this problem, the general information is that the radius is growing at a rate of 2 centimeters per second. So, if we let the radius of the sphere be represented by r, we can say that r'(t)= 2 cm/s.
note:
It is important to separate out these two types of information because information about the particular case is used at the very last step of solving the problem so you don’t want to confuse yourself.
• Identify the known rate and the desired rate
In this case we know dr/dt, and we want to find the rate of change of volume, dV/dt.
• Find a relationship using geometry or trigonometry that relates both the known and unknown rate of change
In this case we’re working with spheres, and we’re trying to find its volume so we’re going to be using the equation
V= (4/3)πr3
note:
From using this equation we get both Volume and radius of the sphere represented in the same equation and they are both changing with time so it is best to write it
V(t) = (4/3)π[r(t)]3
• Now differentiate both sides of the equation above with respect to time.
V(t) = (4/3)π[r(t)]3
dV/dt = 4π[r(t)]2 dr/dt
• Then just plug in the values as we know r(t) and dr/dt.
dV/dt = 4π[16]2(3)
b) Answer t = ([(3/4π)729](1/3)) / 3 seconds or approx 1.8611 seconds.
With the given information we know that V= 729 cm3, and r(0) = 0, r'(t)= 3.
• Antidifferientiate r'(t) so we can an equation for r(t).
∫ r'(t) dt = r(t) = 3t + C
• Since we are given a point on r(t), we can pinpoint our graph to eliminate C.
r(0) = 3(o) + C
C = 0
• Now that we have an equation for r(t), we can use it in the sphere-volume equation.
729 cm3 = (4/3)π[r(t)]3
r(t) = [(3/4π) 729](1/3)
3t = [(3/4π) 729](1/3)
t = ([(3/4π) 729](1/3)) / 3
a) At what rate is the volume changing when the radius is equal to 16 centimeters?
b) If the snowman Little Tom built had a volume totaling 729 cm3. Given r(0)= 0, approximately how long did it take him to build it?
Solution
a) Answer dV/dt = 3072π cm3 per second.
From the question we can tell that this is going to be a related rates question. We can solve this by:
• Separating the "general" and "particular" information.
"General information" is within the problem and true at all times.
"Particular information" is true only at a particular instant.
e.g. in this problem, the general information is that the radius is growing at a rate of 2 centimeters per second. So, if we let the radius of the sphere be represented by r, we can say that r'(t)= 2 cm/s.
note:
It is important to separate out these two types of information because information about the particular case is used at the very last step of solving the problem so you don’t want to confuse yourself.
• Identify the known rate and the desired rate
In this case we know dr/dt, and we want to find the rate of change of volume, dV/dt.
• Find a relationship using geometry or trigonometry that relates both the known and unknown rate of change
In this case we’re working with spheres, and we’re trying to find its volume so we’re going to be using the equation
V= (4/3)πr3
note:
From using this equation we get both Volume and radius of the sphere represented in the same equation and they are both changing with time so it is best to write it
V(t) = (4/3)π[r(t)]3
• Now differentiate both sides of the equation above with respect to time.
V(t) = (4/3)π[r(t)]3
dV/dt = 4π[r(t)]2 dr/dt
• Then just plug in the values as we know r(t) and dr/dt.
dV/dt = 4π[16]2(3)
b) Answer t = ([(3/4π)729](1/3)) / 3 seconds or approx 1.8611 seconds.
With the given information we know that V= 729 cm3, and r(0) = 0, r'(t)= 3.
• Antidifferientiate r'(t) so we can an equation for r(t).
∫ r'(t) dt = r(t) = 3t + C
• Since we are given a point on r(t), we can pinpoint our graph to eliminate C.
r(0) = 3(o) + C
C = 0
• Now that we have an equation for r(t), we can use it in the sphere-volume equation.
729 cm3 = (4/3)π[r(t)]3
r(t) = [(3/4π) 729](1/3)
3t = [(3/4π) 729](1/3)
t = ([(3/4π) 729](1/3)) / 3
Sunday, June 3, 2007
Question 2
Find the point on the graph of y = x2 that is the smallest distance from the point (0, 4).
Solution
Answer at point x= +√(7/2), with a value of 2.8343.
The question is an optimization question however we are looking for a minimum value so it is really called a minimization question.
∙ Like other optimization problems you need to find what is always true using geometry or an equation, and identify your constraint.
In this case you will be using the distance formula D = √(y2 + x2). And your constraint is y = x2. However, square roots may be difficult to work with so you'd want to square both sides of the distance formula to get
D2 = (y2-4)2 + x2
∙ And substituting y for x2
D2 = (x2-4)2 + x2
∙ Find the derivative of the equation in order to find its roots.
D2'(x) = 2(x2-4)(2x) + 2x
D2'(x) = 4x3-16x + 2x
D2'(x) = 2x(2x2-7)
D2'(x) = 0 at x = ±√(7/2) and x=0
∙ Using a number line we can pinpoint where a minimum is located.
D2'(x) = ___-______+_____-_____+___
-√(7/2) 0 +√(7/2)
∙ A minimum occurs when the derivative changes from negative to positive.
In this case it happens at x= -√(7/2) and x= +√(7/2). So we would have to investigates these two values.
∙ So going back to using our original distance formula equation and plugging in the values
D(x) = √((x-4)2) + x2)
D(-√(7/2)) = √((-√(7/2))-4)2) - √(7/2)2) = 6.1617
D(+√(7/2)) = √((+√(7/2))-4)2) + √(7/2)2) = 2.8343
∙ Therefore the minimum value is at x = +√(7/2).
Solution
Answer at point x= +√(7/2), with a value of 2.8343.
The question is an optimization question however we are looking for a minimum value so it is really called a minimization question.
∙ Like other optimization problems you need to find what is always true using geometry or an equation, and identify your constraint.
In this case you will be using the distance formula D = √(y2 + x2). And your constraint is y = x2. However, square roots may be difficult to work with so you'd want to square both sides of the distance formula to get
D2 = (y2-4)2 + x2
∙ And substituting y for x2
D2 = (x2-4)2 + x2
∙ Find the derivative of the equation in order to find its roots.
D2'(x) = 2(x2-4)(2x) + 2x
D2'(x) = 4x3-16x + 2x
D2'(x) = 2x(2x2-7)
D2'(x) = 0 at x = ±√(7/2) and x=0
∙ Using a number line we can pinpoint where a minimum is located.
D2'(x) = ___-______+_____-_____+___
-√(7/2) 0 +√(7/2)
∙ A minimum occurs when the derivative changes from negative to positive.
In this case it happens at x= -√(7/2) and x= +√(7/2). So we would have to investigates these two values.
∙ So going back to using our original distance formula equation and plugging in the values
D(x) = √((x-4)2) + x2)
D(-√(7/2)) = √((-√(7/2))-4)2) - √(7/2)2) = 6.1617
D(+√(7/2)) = √((+√(7/2))-4)2) + √(7/2)2) = 2.8343
∙ Therefore the minimum value is at x = +√(7/2).
Question 3
An indoor swimming pool holds 120 gallons of water. Water is pumped into a swimming pool at 5 gallons per minute and is drained out of the tank at the rate of √(t+1) gallon per minute for when the drainer is on. At time t = 0, the tank contains 2 gallons of water.
a) Suppose that the drain was left on while the water is being pumped. How many gallons of water drained out of the pool from time t=0 to t=3 minutes? And how much water is in the pool at t=3 minutes?
b) The drainer is turned off at t=8 minutes. Find the exact time t, when the pool is at its maximum.
c) Find the average rate of change at which the water fills up the pool to its maximum, under conditions in part b).
Solution
a) Answers are
D(t) = [0 ∫ 3] √(t+1) dt = (14/3) or approx 4.6667
and P(t) = [0 ∫ 3] 5 dt - D(t) + 2 = (37/3) or approx 12.3333
These types of problems involve finding the definite integral.
∙ By integrating a rate gives you the total change in the parent function.
Integrating the rate at which water is drained over an interval will give us the total water drained.
∙ Let D(t) = total water drained in gallons.
By integrating the rate of which water is pumped in gives the total water added to the pool. Therefore by taking the total water pumped in subtracted from total water drained and adding the water you already have in the pool at t=0, gives you the total water you have in the pool in the given time interval. Let P(t)= total water in pool in gallons.
b) Answer t=(1/5)[120 - (1/3)] or approx 23.9333 minutes.
It is given that the maximum of what the pool holds is 120 gallons. Therefore P(t)=120 at its max.
∙ So just by using the equation of P(t) given the time interval of water drains [0,8] we get
P(t) = [0 ∫ t] 5 dt - [0 ∫ 8] √(t+1) dt + 2
120 = [0 ∫ t] 5 dt - [0 ∫ 8] √(t+1) dt + 2
∙ The first Fundamental Theorem of Calculus states that [a ∫ b] f(x) dx = F(b) - F(a).
Which means the integral of a rate change over any given interval is the total change in the parent function over that interval.
∙ And applying the Fundamental Theorem of Calculus we get
120 = [5(t) - 5(0)] - [(1/2)((8)+1)(-1/2) - (1/2)((0)+1)(1/2)] + 2
120 = 5t - (1/2)(1/3) - (1/2)
5t = 120 + (1/6) - (1/2)
t = (1/5)[120 - (2/6)]
c) Answer 4.3593 gallons per minute.
When you want to find the average rate of change, you'd multiply the time frame in the interval with the integral in question, using this formula A(x)=[1/(b-a)]([a ∫ b] f(x) dx).
∙ Let A(x)= the average rate of change of total water filling the pool
A(x) = 1/(t-0) ([0 ∫ t] 5 dt) - ([0 ∫ 8] √(t+1) dt) + 2)
note:
Remember you are finding a rate, so don't forget to have the correct units. If it is a rate then there is a value in respect to another, e.g. miles per hour.
∙ Recall that t = (1/5)[120 - (2/6)], so just plugging in the value we get the average rate of change.
a) Suppose that the drain was left on while the water is being pumped. How many gallons of water drained out of the pool from time t=0 to t=3 minutes? And how much water is in the pool at t=3 minutes?
b) The drainer is turned off at t=8 minutes. Find the exact time t, when the pool is at its maximum.
c) Find the average rate of change at which the water fills up the pool to its maximum, under conditions in part b).
Solution
a) Answers are
D(t) = [0 ∫ 3] √(t+1) dt = (14/3) or approx 4.6667
and P(t) = [0 ∫ 3] 5 dt - D(t) + 2 = (37/3) or approx 12.3333
These types of problems involve finding the definite integral.
∙ By integrating a rate gives you the total change in the parent function.
Integrating the rate at which water is drained over an interval will give us the total water drained.
∙ Let D(t) = total water drained in gallons.
By integrating the rate of which water is pumped in gives the total water added to the pool. Therefore by taking the total water pumped in subtracted from total water drained and adding the water you already have in the pool at t=0, gives you the total water you have in the pool in the given time interval. Let P(t)= total water in pool in gallons.
b) Answer t=(1/5)[120 - (1/3)] or approx 23.9333 minutes.
It is given that the maximum of what the pool holds is 120 gallons. Therefore P(t)=120 at its max.
∙ So just by using the equation of P(t) given the time interval of water drains [0,8] we get
P(t) = [0 ∫ t] 5 dt - [0 ∫ 8] √(t+1) dt + 2
120 = [0 ∫ t] 5 dt - [0 ∫ 8] √(t+1) dt + 2
∙ The first Fundamental Theorem of Calculus states that [a ∫ b] f(x) dx = F(b) - F(a).
Which means the integral of a rate change over any given interval is the total change in the parent function over that interval.
∙ And applying the Fundamental Theorem of Calculus we get
120 = [5(t) - 5(0)] - [(1/2)((8)+1)(-1/2) - (1/2)((0)+1)(1/2)] + 2
120 = 5t - (1/2)(1/3) - (1/2)
5t = 120 + (1/6) - (1/2)
t = (1/5)[120 - (2/6)]
c) Answer 4.3593 gallons per minute.
When you want to find the average rate of change, you'd multiply the time frame in the interval with the integral in question, using this formula A(x)=[1/(b-a)]([a ∫ b] f(x) dx).
∙ Let A(x)= the average rate of change of total water filling the pool
A(x) = 1/(t-0) ([0 ∫ t] 5 dt) - ([0 ∫ 8] √(t+1) dt) + 2)
note:
Remember you are finding a rate, so don't forget to have the correct units. If it is a rate then there is a value in respect to another, e.g. miles per hour.
∙ Recall that t = (1/5)[120 - (2/6)], so just plugging in the value we get the average rate of change.
Question 4
Let Q be the shaded region in the first quadrant enclosed by the graph y= 2cos(x), and y=ex2 and the y-axis as shown in the figure.
a) Find the area of region Q.
b) Find the volume of the solid generated when Q is revolved about the x-axis.
c) The region Q is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a square. Find the volume of this solid.
Solution
a) Answer, [0 ∫ a] (2cosx - ex2 ) dx = 0.4566
For these questions you find which graph is furthest on top visually, and subtract it with the bottom graph.
• You would also need to find the points of intersections to be able to find the area in between.
x-intersects at x=0 and x=0.67031574 when 2cosx=e^(x2). Let a=0.67031574.
In this case y=2cosx is on top and y=ex2 is on the bottom.
• By taking the integral of (2cosx)-(ex2) we get the area of Q.
b) Answer, Volume = π [ 0 ∫ a ] ( (2cosx)2 - (ex2)2 dx = 1.3751
Visually picturing the area being revolved around the x-axis.
• We see that it makes washers (Circular figure with a hole in the center).
• The area of the washer at one a particular point can be expressed as πR2 - πr2 where R is the bigger radius and r is the smaller one.
Factoring out pi, we get π(R2 - r2). In this case the bigger radius is (2cosx - 1), where as the smaller radius is (ex2 - 1)2.
• By taking the integral of π(R2 - r2), will give you the total area over the designated time interval.
Keep in mind that the interval starts from 0 because we are only worried about the shaded region highlighted in the first quadrant. If it were to be the whole area in between the graphs then that would be from -a to a.
c) Answer, Volume = [ 0 ∫ a ] [2cosx - ex2]2 dx = 0.3687
From the picture we can clearly see what the question is asking.
• By taking the integral of the area of the base of the solid squared, will give you the volume of the solid.
The solid generated is a cube.
a) Find the area of region Q.
b) Find the volume of the solid generated when Q is revolved about the x-axis.
c) The region Q is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a square. Find the volume of this solid.
Solution
a) Answer, [0 ∫ a] (2cosx - ex2 ) dx = 0.4566
For these questions you find which graph is furthest on top visually, and subtract it with the bottom graph.
• You would also need to find the points of intersections to be able to find the area in between.
x-intersects at x=0 and x=0.67031574 when 2cosx=e^(x2). Let a=0.67031574.
In this case y=2cosx is on top and y=ex2 is on the bottom.
• By taking the integral of (2cosx)-(ex2) we get the area of Q.
b) Answer, Volume = π [ 0 ∫ a ] ( (2cosx)2 - (ex2)2 dx = 1.3751
Visually picturing the area being revolved around the x-axis.
• We see that it makes washers (Circular figure with a hole in the center).
• The area of the washer at one a particular point can be expressed as πR2 - πr2 where R is the bigger radius and r is the smaller one.
Factoring out pi, we get π(R2 - r2). In this case the bigger radius is (2cosx - 1), where as the smaller radius is (ex2 - 1)2.
• By taking the integral of π(R2 - r2), will give you the total area over the designated time interval.
Keep in mind that the interval starts from 0 because we are only worried about the shaded region highlighted in the first quadrant. If it were to be the whole area in between the graphs then that would be from -a to a.
c) Answer, Volume = [ 0 ∫ a ] [2cosx - ex2]2 dx = 0.3687
From the picture we can clearly see what the question is asking.
• By taking the integral of the area of the base of the solid squared, will give you the volume of the solid.
The solid generated is a cube.
Reflection
Why did you choose the concepts you did to create your problem set?
Since I started the DEV project after my exam it was a bit easier to create and choose the concepts of my problem set. I chose the questions that I had the most trouble with on the AP exam. By tackling the questions I thought to be the hardest will really make it stick in my brain for when I take the course at the university. I wanted to challenge myself, and I did.
How do these problems provide an overview of your best mathematical understanding of what you have learned so far?
The questions I chose provides the best overview of my mathematical understanding as it shows that I am able to learn and annotate the most troubling problems that I think were. The problems I chose also span through all units of the course which also tells you about what I have learned so far.
Did you learn anything from this assignment?
Yes, I learned how to use spresent, and neat features of the blog I didn't know about. More valuably, I developed a greater understanding for the problems I chose to do through detailed annotations.
Was it educationally valuable to you? (Be honest with this. If you got nothing out of this assignment then say that, but be specific about what you didn't like and offer a suggestion to improve it in the future)
Educationally valuable, yes. I like the fact that being forced to focus on 4 set a problems will make you learn it inside and out. And for university it will be hard to forget because there was quite a bit a time invested. However, the time spent talking about the DEV project or actually doing it could have been used as time in preparation for the exam. For example, doing wiki assignments instead (because you don't have to spend time publishing to make it look nice, but still carries same thorough examination). Another possible suggestion is to make the DEV project partnered work. Not only is it less weight on your shoulders but by and by talking it out with one another will increases chances that the question is done right, and it will stick more in your head. Learning is a conversation after all ;).
Since I started the DEV project after my exam it was a bit easier to create and choose the concepts of my problem set. I chose the questions that I had the most trouble with on the AP exam. By tackling the questions I thought to be the hardest will really make it stick in my brain for when I take the course at the university. I wanted to challenge myself, and I did.
How do these problems provide an overview of your best mathematical understanding of what you have learned so far?
The questions I chose provides the best overview of my mathematical understanding as it shows that I am able to learn and annotate the most troubling problems that I think were. The problems I chose also span through all units of the course which also tells you about what I have learned so far.
Did you learn anything from this assignment?
Yes, I learned how to use spresent, and neat features of the blog I didn't know about. More valuably, I developed a greater understanding for the problems I chose to do through detailed annotations.
Was it educationally valuable to you? (Be honest with this. If you got nothing out of this assignment then say that, but be specific about what you didn't like and offer a suggestion to improve it in the future)
Educationally valuable, yes. I like the fact that being forced to focus on 4 set a problems will make you learn it inside and out. And for university it will be hard to forget because there was quite a bit a time invested. However, the time spent talking about the DEV project or actually doing it could have been used as time in preparation for the exam. For example, doing wiki assignments instead (because you don't have to spend time publishing to make it look nice, but still carries same thorough examination). Another possible suggestion is to make the DEV project partnered work. Not only is it less weight on your shoulders but by and by talking it out with one another will increases chances that the question is done right, and it will stick more in your head. Learning is a conversation after all ;).
Subscribe to:
Posts (Atom)