Friday, June 10, 2011
Wednesday, June 8, 2011
integration by parts
1. introductory to integration by parts (wikipedia definition)
Integration by parts is a rule that transforms the integral of products of functions into other simpler integrals. The rule can be derived in one line by simply integrating the product rule of differentiation.
If u = f(x), v = g(x), and the differentials du = f '(x) dx and dv = g'(x) dx, then integration by parts states that
When to use it:
- When the integrand is the product of two functions and substitution does not work
- If f(x) is continuous and differentiable and if g(x) is continuous
Integration by parts derived:
We know that the equation used in integration by parts is derived from the product rule as all integrands that can be solved using this are composed of two separate functions...
(f(x)g(x))'=f(x)g'(x)+f'(x)g(x)
integrating both sides:
f(x)g(x)=f(x)g'(x) dx+f'(x)g(x) dx
suppose there is a given interval from a to b:
abf(x)g'(x) dx=f(x)g(x)ab-abf'(x)g(x) dx
with common notation:
f(x)g(x)ab=f(b)g(b)-f(a)g(a)
using the product rule and the fundamental theorem of calculus the rule is shown to be true:
f(b)g(b)-f(a)g(a)=abddx(f(x)g(x)) dx
=abf'(x)g(x) dx+abf(x)g'(x) dx
rule stated in indefinite integral form:
f(x)g'(x) dx=f(x)g(x)-f'(x)g(x) dx
2. how integration by parts helps/quickens integrating
Integration by parts is a rule that transforms the integral of products of functions into other simpler integrals. The rule can be derived in one line by simply integrating the product rule of differentiation.
If u = f(x), v = g(x), and the differentials du = f '(x) dx and dv = g'(x) dx, then integration by parts states that
When to use it:
- When the integrand is the product of two functions and substitution does not work
- If f(x) is continuous and differentiable and if g(x) is continuous
Integration by parts derived:
We know that the equation used in integration by parts is derived from the product rule as all integrands that can be solved using this are composed of two separate functions...
(f(x)g(x))'=f(x)g'(x)+f'(x)g(x)
integrating both sides:
f(x)g(x)=f(x)g'(x) dx+f'(x)g(x) dx
suppose there is a given interval from a to b:
abf(x)g'(x) dx=f(x)g(x)ab-abf'(x)g(x) dx
with common notation:
f(x)g(x)ab=f(b)g(b)-f(a)g(a)
using the product rule and the fundamental theorem of calculus the rule is shown to be true:
f(b)g(b)-f(a)g(a)=abddx(f(x)g(x)) dx
=abf'(x)g(x) dx+abf(x)g'(x) dx
rule stated in indefinite integral form:
f(x)g'(x) dx=f(x)g(x)-f'(x)g(x) dx
2. how integration by parts helps/quickens integrating
This strategy separates a very complicated integral into two individual integrals which will become easier to evaluate in relation to the initial function.
3. steps/guidelines to integrating by parts
a. begin with the part of the function which has the easiest and most simplified derivative (preferably 1) and label it “u”
b. label the derivative of u “du”
c. let the remainder of the original function be “dv”
d. integrate dv to find “v”
e. sub values into
f. integrate u and v
g. use algebra to solve
4. give easy example using guidelines
xe2x dx=?
step a: u=x
step b: du=1dx
step c: dv=e2x
step d: e2x dx=12e2x
step e: xe2x dx=x(12e2x)-12e2x dx
step f:
xe2x dx=x(12e2x)-12e2x dx
=x2e2x-12e2x dx
=x2e2x-1212e2x+C
=x2e2x-14e2x+C
5. give tedious example using guidelines
excos(x) dx
step a: u=cos(x)
step b: du=-sin(x) dx
step c: dv=ex dx
step d: ex dx =ex
step e: excos(x) dx=excos(x)+exsin(x) dx
before step f we must integrate by parts again...
step a: u=sin(x)
step b: du=cos(x) dx
step c: dv=ex dx
step d: exdx=ex
step e: exsin(x) dx=exsin(x)-excos(x) dx
step f: exsin(x) dx=exsin(x)-excos(x) dx
excos(x) dx=excos(x)+exsin(x)-excos(x) dx
2excos(x) dx=ex(sin(x)+cos(x))+C
excos(x) dx=ex(sin(x)+cos(x))2+C
to further help your understanding visit:
8. turn on v1210 and call it good!
Force and Work
We started out by defining force to the class. We did this by talking about Newton’s 3 laws.
I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.
II. The relationship between an object's mass m, its acceleration a, and the applied force F is F = ma. Acceleration and force are vectors (as indicated by their symbols being displayed in slant bold font); in this law the direction of the force vector is the same as the direction of the acceleration vector.
III. For every action there is an equal and opposite reaction.
The first law was what we used at first by using the equation of the sum of the forces all equaling zero. We used this to show how to set up the equations as well as solving for any variable such as a specific type of force.
We went on to explain the different types of forces.
Force of gravity – This is the product of the mass of an object multiplied by the gravitational constant of 9.81 m/s^2. This is true for any object close to the Earth’s surface. This force is always directed directly down.
Normal Force – This is the force put on an object from the surface that it is on. It is always perpendicular to the surface. It can be calculated by using Newton’s 2nd law, the equation of the sum of the forces equals’ mass times acceleration.
Force of friction – This is the force that is caused by the restriction of an object from moving across a surface. It can calculated by the equation of normal force times mew. Mew is the coefficient of friction between the object and surface. It ranges from 0 to 1, higher being more restrictive.
The force of tension – This is a force given by a rope pulling on an object or an object hanging up a string or something that would have tension. This can be calculated again my Newton’s laws.
When solving for these forces using Newton’s laws we need to set it up correctly than it is simple algebra processes; at least in the beginning of learning physics. What we need to know is that the forces in the x direction and y direction are completely independent of each other. Therefore, we need to use the equation twice, once for each direction if it is needed to solve a problem. Also, we need to know that when we set on the sum of the forces on the left side of the equation we need to separate the forces directed in the upward and downward direction; or facing the right or to the left direction. This is why making a force diagram help; which means to draw all of the forces about a dot to show what directions each of them are facing.
All of these forces are measured in Newton’s
Next we went on to work. We explained that most problems can be answered by doing either forces or energy. We first described the types of energies and their equations.
Kinetic Energy – 1/2mv^2 ß this is used for an object that is in motion; m is mass and v is velocity
Gravitational Potential Energy – mgh ß This is for an object that is hanging by something, or just has the potential to fall due to the force of gravity, m is mass, g is 9.81, and h is the height of the object.
Thermal energy – Fx ß This is used to calculate something such as the work done by the force of friction. F is the force and x is delta x, so how far it travels.
Elastic Energy – 1/2kx^2 ß This is used for an object working against a spring. K is the spring constant for an ideal spring, and x again is how far the object travels while against the spring.
In solving a problem with energy it is a simple process at first. You need to select two different places in the movement of an object. Then you decide energies are present at those 2 spots. You set them equal to each other and then fill in the variables that you know and use algebra to solve for whichever variable you’re looking for.
We next went over practice problems and took the quiz
Here are links that help
http://www.physicsclassroom.com/class/newtlaws/u2l2a.cfm
http://theory.uwinnipeg.ca/physics/force/index.html
http://phun.physics.virginia.edu/topics/energy.html
I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.
II. The relationship between an object's mass m, its acceleration a, and the applied force F is F = ma. Acceleration and force are vectors (as indicated by their symbols being displayed in slant bold font); in this law the direction of the force vector is the same as the direction of the acceleration vector.
III. For every action there is an equal and opposite reaction.
The first law was what we used at first by using the equation of the sum of the forces all equaling zero. We used this to show how to set up the equations as well as solving for any variable such as a specific type of force.
We went on to explain the different types of forces.
Force of gravity – This is the product of the mass of an object multiplied by the gravitational constant of 9.81 m/s^2. This is true for any object close to the Earth’s surface. This force is always directed directly down.
Normal Force – This is the force put on an object from the surface that it is on. It is always perpendicular to the surface. It can be calculated by using Newton’s 2nd law, the equation of the sum of the forces equals’ mass times acceleration.
Force of friction – This is the force that is caused by the restriction of an object from moving across a surface. It can calculated by the equation of normal force times mew. Mew is the coefficient of friction between the object and surface. It ranges from 0 to 1, higher being more restrictive.
The force of tension – This is a force given by a rope pulling on an object or an object hanging up a string or something that would have tension. This can be calculated again my Newton’s laws.
When solving for these forces using Newton’s laws we need to set it up correctly than it is simple algebra processes; at least in the beginning of learning physics. What we need to know is that the forces in the x direction and y direction are completely independent of each other. Therefore, we need to use the equation twice, once for each direction if it is needed to solve a problem. Also, we need to know that when we set on the sum of the forces on the left side of the equation we need to separate the forces directed in the upward and downward direction; or facing the right or to the left direction. This is why making a force diagram help; which means to draw all of the forces about a dot to show what directions each of them are facing.
All of these forces are measured in Newton’s
Next we went on to work. We explained that most problems can be answered by doing either forces or energy. We first described the types of energies and their equations.
Kinetic Energy – 1/2mv^2 ß this is used for an object that is in motion; m is mass and v is velocity
Gravitational Potential Energy – mgh ß This is for an object that is hanging by something, or just has the potential to fall due to the force of gravity, m is mass, g is 9.81, and h is the height of the object.
Thermal energy – Fx ß This is used to calculate something such as the work done by the force of friction. F is the force and x is delta x, so how far it travels.
Elastic Energy – 1/2kx^2 ß This is used for an object working against a spring. K is the spring constant for an ideal spring, and x again is how far the object travels while against the spring.
In solving a problem with energy it is a simple process at first. You need to select two different places in the movement of an object. Then you decide energies are present at those 2 spots. You set them equal to each other and then fill in the variables that you know and use algebra to solve for whichever variable you’re looking for.
We next went over practice problems and took the quiz
Here are links that help
http://www.physicsclassroom.com/class/newtlaws/u2l2a.cfm
http://theory.uwinnipeg.ca/physics/force/index.html
http://phun.physics.virginia.edu/topics/energy.html
Tuesday, May 31, 2011
Friday, May 27, 2011
Thursday, May 26, 2011
Wednesday, May 11, 2011
Student Feedback
Your honest feedback is appreciated:
http://www.surveygizmo.com/s3/536408/Student-Feedback-for-Mr-O-Brien-2011
http://www.surveygizmo.com/s3/536408/Student-Feedback-for-Mr-O-Brien-2011
Thursday, May 5, 2011
Tuesday, April 12, 2011
Friday, April 8, 2011
Wednesday, April 6, 2011
Tuesday, April 5, 2011
Monday, April 4, 2011
Thursday, March 31, 2011
AP Exam Revision Day 1
Our shared revision guide
Answer form for 1998 Section I Part A (55 minutes, 28 questions, non-calculator, beginning on p. 125)
Answer form for 1998 Section I Part A (55 minutes, 28 questions, non-calculator, beginning on p. 125)
Wednesday, March 30, 2011
Wednesday, March 23, 2011
Monday, March 21, 2011
Scribe Post 3/21/11
Scribe Post
Update:
Occasionally you can actually solve cross-sectional area problems without heavy integral work (as we saw on the Unit 5 test, where there was a problem that actually involved constructing half of a sphere). So, just pay attention to what shapes are being created in each problem, and before you decide to take a more complex approach, make sure there isn't a sleeker way involving a simple generic shape equation.
Friday, March 18, 2011
Assignment #7 Even Answers
46.
a. π[3 – ln4]
b. 17π/3
c. π[ln(4)-1]
52.
V = 36π/5, so W = 8.5 * 36π/5 = 192 g
a. π[3 – ln4]
b. 17π/3
c. π[ln(4)-1]
52.
V = 36π/5, so W = 8.5 * 36π/5 = 192 g
Wednesday, March 16, 2011
Extra Credit Opportunity!
Watch this video:
http://vihart.com/blog/pi-is-still-wrong/
Check out the Tau Manifesto:
http://tauday.com/
Share your thoughts in the comments...
http://vihart.com/blog/pi-is-still-wrong/
Check out the Tau Manifesto:
http://tauday.com/
Share your thoughts in the comments...
Tuesday, March 15, 2011
Monday, March 14, 2011
Thursday, March 10, 2011
Trig, Khan Academy, and class
1. Take this 2 minute practice trig quiz (15 questions...).
2. Take the class quiz.
3. Please register for the Khan Academy. Use my email address under the Add a Coach link.
4. Let's do some practice...
5. Please watch this video before next class.
Tuesday, March 8, 2011
Sunday, March 6, 2011
Wednesday, March 2, 2011
Monday, February 28, 2011
Unit 5 Assignment #1
Assignment #1
When you are finished, check and correct your work by using the Solution Guides:
2007 Solution Guide
2008 Solution Guide
Indicate your original score out of 9 for each question.
When you are finished, check and correct your work by using the Solution Guides:
2007 Solution Guide
2008 Solution Guide
Indicate your original score out of 9 for each question.
Friday, February 18, 2011
February Vacation Extra Credit Problems
Six February FRQ Extra Credit Problems
When you are finished, check and correct your work by using the Solution Guides:
2007B Solution Guide
2008B Solution Guide
Indicate your original score out of 9 for each question.
When you are finished, check and correct your work by using the Solution Guides:
2007B Solution Guide
2008B Solution Guide
Indicate your original score out of 9 for each question.
Monday, February 14, 2011
Monday, February 7, 2011
Thursday, February 3, 2011
Wednesday, February 2, 2011
Tuesday, February 1, 2011
Assignment #5 & #6 solutions
Assignment #5 Riemann Sums for Definite Integrals Solutions
Assignment #6:
Assignment #6 Exploration: Some Very Special Riemann Sums
Trapezoidal Rule Exploration Solutions
A Motion Antiderivative Problem Solutions
Some Very Special Riemann Sums Solutions
p. 339/30: 53.36 KW/Weekend (using odd-numbered hours)
Assignment #6:
Assignment #6 Exploration: Some Very Special Riemann Sums
Trapezoidal Rule Exploration Solutions
A Motion Antiderivative Problem Solutions
Some Very Special Riemann Sums Solutions
p. 339/30: 53.36 KW/Weekend (using odd-numbered hours)
Friday, January 28, 2011
Thursday, January 27, 2011
Monday, January 24, 2011
Wednesday, January 19, 2011
Tuesday, January 11, 2011
Monday, January 10, 2011
Scribe Post 1/10/11
So, Firefox just crashed and deleted my whole scribe post when I had one word left. Fun night. We started off class with a brief discussion of the midterm and how it is going to work. Basically, it will include non-calc and calc free-response questions so be prepared for both. Also, homework is due Friday, except for today's assignment which is due Wednesday, which is also the day we have a quiz on Diffy Qs.
In class today, we began by talking about #6 on the slope fields packet. We also talked about the hot tub word problem. Here are a few helpful hints to keep in mind when working on Diffy Q problems:
-If you have a hard time with slope fields, make a table of values!
-You only need to add "C" to one side of your anti-derivative equation, as if you add c to both sides then subtract from one side, you still have a constant.
-Just because you are taking the anti-derivative of a fraction doesn't mean the anti-derivative is necessarily the natural log function
We then talked about Julia moving around the room. This taught us about position, displacement, and distance. We learned that position was Julia's location at a particular moment (e.g. position was -8 when Julia was at -8). Displacement was how far Julia traveled from her start location to her ending location, or final position minus initial position (e.g. Julia started at -8 and ended at -1 so her displacement was 7). We then decided that distance is how far Julia traveled total, and it must always be positive, as you can't un-travel. Julia's distance was 13, as it was the total of the absolute value of each leg of her trip (she started at -8, went to 2, then went back to -1). We then had a little discussion about how her velocity related to her movements. For example, as Julia went from -8 to 2, her velocity was positive, and was increasing then decreasing. As Julia traveled from 2 to -1, her velocity was negative, and was decreasing then increasing. The homework packet has a nice fill-in-the-blank section that explains this a little better, and Mr. O'Brien says that the concept of velocity in relation to displacement and distance can be a little hard to grasp at first. Overall, it was a productive class!
Here are a few links to help with today's lesson:
http://www.schooltube.com/video/4eae3449edc2899b3d00/Calculating-the-distance-between-two-points-on-a-number-line-by-Camille-Morton She makes understanding distances easy enough for a five-year-old to understand. It's cheesy, so be warned.
http://www.intmath.com/Differential-equations/1_Solving-DEs.php This breaks down differential equations into a manageable form, which I found helpful before the quiz Wednesday.
Homework: Packet Mr. O'Brien handed out, quiz on differential equations Wednesday, extra credit to anyone who can find a good visual representation of #6 in the slope fields packet.
UPDATE:
While looking for some practice problems for the midterm, I came across this link:
http://www.analyzemath.com/calculus.html
It has a bunch of problems for quite a few of the topics we've done so far and applets to help explain them!
In class today, we began by talking about #6 on the slope fields packet. We also talked about the hot tub word problem. Here are a few helpful hints to keep in mind when working on Diffy Q problems:
-If you have a hard time with slope fields, make a table of values!
-You only need to add "C" to one side of your anti-derivative equation, as if you add c to both sides then subtract from one side, you still have a constant.
-Just because you are taking the anti-derivative of a fraction doesn't mean the anti-derivative is necessarily the natural log function
We then talked about Julia moving around the room. This taught us about position, displacement, and distance. We learned that position was Julia's location at a particular moment (e.g. position was -8 when Julia was at -8). Displacement was how far Julia traveled from her start location to her ending location, or final position minus initial position (e.g. Julia started at -8 and ended at -1 so her displacement was 7). We then decided that distance is how far Julia traveled total, and it must always be positive, as you can't un-travel. Julia's distance was 13, as it was the total of the absolute value of each leg of her trip (she started at -8, went to 2, then went back to -1). We then had a little discussion about how her velocity related to her movements. For example, as Julia went from -8 to 2, her velocity was positive, and was increasing then decreasing. As Julia traveled from 2 to -1, her velocity was negative, and was decreasing then increasing. The homework packet has a nice fill-in-the-blank section that explains this a little better, and Mr. O'Brien says that the concept of velocity in relation to displacement and distance can be a little hard to grasp at first. Overall, it was a productive class!
Here are a few links to help with today's lesson:
http://www.schooltube.com/video/4eae3449edc2899b3d00/Calculating-the-distance-between-two-points-on-a-number-line-by-Camille-Morton She makes understanding distances easy enough for a five-year-old to understand. It's cheesy, so be warned.
http://www.intmath.com/Differential-equations/1_Solving-DEs.php This breaks down differential equations into a manageable form, which I found helpful before the quiz Wednesday.
Homework: Packet Mr. O'Brien handed out, quiz on differential equations Wednesday, extra credit to anyone who can find a good visual representation of #6 in the slope fields packet.
UPDATE:
While looking for some practice problems for the midterm, I came across this link:
http://www.analyzemath.com/calculus.html
It has a bunch of problems for quite a few of the topics we've done so far and applets to help explain them!
Thursday, January 6, 2011
Scribe Post 01/06/2011
Today, to start off class, we got our Unit 3 Supercorrection Follow-up Tests back and went over the problems in class. Apparently, everyone did fairly well on it, but we went through each problem anyway for review.
Then we went to the blog, to do the two problem sets for the day that took up the whole class period.
The first one, 7-3, we went through and Mr. O'Brien explained step-by-step. The second one, 7-2, was mostly for practice.
Here are the notes/solutions to the problem sets, which you can find a copy of in the previous blog post as 'in class problems'.
(By the way, on the first page of solutions, the cut off note on the left hand side is supposed to read 'not all given in the original word problem'.)
And here is link about differential equations just in case you need a quick refresher. Example 1 is probably the best thing to go over if necessary.
Finally, our homework for tonight was a handout packet about Slope Fields. See Mr. O'Brien if you need a copy. And we were also assigned several word problems which you can find linked in the previous post or here.
The next scribe will be Caitlin Throne.
Then we went to the blog, to do the two problem sets for the day that took up the whole class period.
The first one, 7-3, we went through and Mr. O'Brien explained step-by-step. The second one, 7-2, was mostly for practice.
Here are the notes/solutions to the problem sets, which you can find a copy of in the previous blog post as 'in class problems'.
(By the way, on the first page of solutions, the cut off note on the left hand side is supposed to read 'not all given in the original word problem'.)
And here is link about differential equations just in case you need a quick refresher. Example 1 is probably the best thing to go over if necessary.
Finally, our homework for tonight was a handout packet about Slope Fields. See Mr. O'Brien if you need a copy. And we were also assigned several word problems which you can find linked in the previous post or here.
The next scribe will be Caitlin Throne.
Tuesday, January 4, 2011
Scribe Post 1/4/2011
We started off class with the Supercorrection Follow-Up Test. There were seven problems, and we were allowed to use calculators on problems 1 + 2. This took us up to 10:10, at which point we began a game centered around slope fields. We had to match pictures of slope fields with their respective differential equations and a general description of their solution curves.
If after this game, anyone is still a little confused, this link offers some practice problems and solutions http://www.education.com/reference/article/slope-fields/
Additionally, check out this link -- you can generate your own slope fields, set display range / number of tick marks, set initial conditions, etc. http://www.math.lsa.umich.edu/courses/116/slopefields.html
Tonight’s homework is on matching differential equations and slope fields. Apologies for the abnormally short scribe post, but this was pretty much all that went on today.
UPDATE: Turns out that slope fields pertain to the more difficult differential equations we've been working on of late rather well, and have done a good job introducing us to the concept of using initial values provided in the wording of the problem to generate our own solution curves. This site encapsulates that fairly well, and also talks a little bit about Euler's Method, the simplest way of obtaining a numerical solution of a differential equation.
If after this game, anyone is still a little confused, this link offers some practice problems and solutions http://www.education.com/reference/article/slope-fields/
Additionally, check out this link -- you can generate your own slope fields, set display range / number of tick marks, set initial conditions, etc. http://www.math.lsa.umich.edu/courses/116/slopefields.html
Tonight’s homework is on matching differential equations and slope fields. Apologies for the abnormally short scribe post, but this was pretty much all that went on today.
UPDATE: Turns out that slope fields pertain to the more difficult differential equations we've been working on of late rather well, and have done a good job introducing us to the concept of using initial values provided in the wording of the problem to generate our own solution curves. This site encapsulates that fairly well, and also talks a little bit about Euler's Method, the simplest way of obtaining a numerical solution of a differential equation.
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