Blank copy
Use to help with Supercorrections!
Friday, December 21, 2012
Thursday, December 13, 2012
Monday, December 10, 2012
Scribe Post 12/7
We began a marvelous Friday with a fairly difficult quiz on optimization, related rates, linearization, and the like. After about 50 minutes, the quiz ended, the groans began, and class moved on.
After settling back into our seats, we took a look at question 4 from Mr. O'Brien's AP Calculus exam from 1990! Mr. O'B got all excited, recollecting his heroic score of a 6 on the AP exam, the only person to ever score higher than a perfect score. Who could have guessed that the question was all about related rates! An excited murmur traveled across the room, as we began to realize that it's not even halfway through the year, yet we have all the knowledge to fully answer a question on the terrifying AP exam. We began to do the three parts of question 4.
Given that the radius "r" of a sphere is increasing at a constant rate of 0.04 centimeters per second, part A asked us to find the rate of increase of the sphere's volume when its radius measured 10 centimeters. The ever-important and oft-forgoten equation for volume of a sphere,
was also given in the problem. We went to work, first differentiating this equation for the volume of the sphere.
It is crucial to realize that we have plugged NO values in yet, we simply differentiated the entire equation. You can think of it as implicit differentiation if you'd like to, for that is what it technically is. It is also important to note that the constant, (4/3)π, remains unchanged when you take the derivative. From here, we simplified the derivative a little more, and then plugged the given values in.
But wait, how did we get the units? This is a key part of related rates: since volume is measured in centimeters cubed and time is measured in seconds, the answer must be measured in centimeters cubed per second. It makes a lot of sense - just match up the units and you're good!
Part A is done, easy! On to part B. This asked us the rate of increase of the area of a cross section through the center of a sphere when the volume of the sphere is 36π. Let's start with a picture! That often makes things easier.
We're looking for the rate of increase of the area (of a section that is a circle), so we first must remember the equation for the area of a circle. Easy!
What do you know, it's differentiable! Let's take the derivative, in the same manner as we did in part A.
Hmm, that looks good but we're kinda stuck. We're trying to find the rate of increase of the area and we know the rate of increase of the radius, but we don't know the radius. But! The fact that the volume of the sphere was 36π was given at the beginning of the problem, so we can solve for the radius!
Easy enough! Since we've already differentiated the equation for the rate of change of the area, let's plug in values.
And we're done! Onto part C.
This part asked for the radius when the volume and radius of the sphere were increasing at the same rate. Let's set both of those things equal to each other.
Next, let's plug in the values we've calculated above.
Simplify a little further.
Finally, square root both sides of the equation to find r.
Remember, since r is in centimeters the answer will be as well. We're done! Not too bad, certainly not as scary as many of us thought an AP exam would be. To check these answers, or to see another person's working of this same problem, check out this site and type in "1990 AB4 Solution"in the "Find" menu.
Also, please remember that the finely crafted opportunity day is on THURSDAY, so keep up with your IW and come to class ready to REVIEW next week! For additional help on related rates problems, check out this awesome applet about a melting snowball. If worse comes to worst, you can always count on our buddy PatrickJMT!
P.S. Ever wondered what he really looks like?
Next scribe will be ?
Given that the radius "r" of a sphere is increasing at a constant rate of 0.04 centimeters per second, part A asked us to find the rate of increase of the sphere's volume when its radius measured 10 centimeters. The ever-important and oft-forgoten equation for volume of a sphere,
Part A is done, easy! On to part B. This asked us the rate of increase of the area of a cross section through the center of a sphere when the volume of the sphere is 36π. Let's start with a picture! That often makes things easier.
This part asked for the radius when the volume and radius of the sphere were increasing at the same rate. Let's set both of those things equal to each other.
Also, please remember that the finely crafted opportunity day is on THURSDAY, so keep up with your IW and come to class ready to REVIEW next week! For additional help on related rates problems, check out this awesome applet about a melting snowball. If worse comes to worst, you can always count on our buddy PatrickJMT!
P.S. Ever wondered what he really looks like?
Next scribe will be ?
Friday, December 7, 2012
Mr. O'B's AP Exam!!!
Mr. O'Brien graduated from Camden-Rockport High School in 1990. How would you do on his AP Calculus exam? Let's try Question 4!
Wednesday, December 5, 2012
Scribe Post 12/5
At the start of class, O'Brien told us that unit three was finally drawing to a close, and that our final topic, related rates, would occupy our final two classes before the test, and that we would be supercorrecting the test going into our Christmas -- Hanukka -- Quanza -- Saturnalia... HOLIDAY break. Surely that's the best gift of all: forgiveness for our failures on opportunity day.
But before class could begin, a scribe had to be chosen, for Eddie, awake until 1:30 the night before slaving away on his scribe post, had forgotten to name his successor. The class quickly divided into two factions pleading not to be chosen: the Phyzards and the Bio-Nerds, complaining about "supercorrections" and "bio outlines" respectively. Even Gabe and Duncan of no-supercorrections did not step forward. Eddie was overwhelmed, but just as O'Brien was about to choose randomly from the mob, Crockett rose above the masses, a beacon of hope (see fig. 1) and volunteered his evening for the benefit of Phyzards and Bio-Nerds alike. Crockett's hair flowed back in a sudden classroom breeze and women swooned at his courage and selflessness.
Given a circular cone of height 10cm and radius r, increasing at a rate of 1 cm/s, what is the rate of change of the volume when the radius is 24?
O'Brien gave us a sort of loose procedure for problems like these:
1. Write what you know.
in this case we know that:
r = 24 cm
dr/dt = 1 cm/s
h = 10 cm
2. Draw a picture.
Drawing pictures makes it easier to find relationships. A helpful tactic is to superimpose your drawing over a set of axes, then maybe write some equations for lines that you see.
3. Find relationships.
To do this problem, we have to know that the volume of a cone is equal to 1/3Bh, or 1/3πr^2h
O'Brien told us that if we have equations for our variables, it's not always best to immediately substitute those in. However, if the value is a constant, we should substitute in immediately in order to avoid messy product rules.
Because h is constant, we substituted that in, leaving us with:
V = 10π/3 * r^2
4. Do calculus.
This is the easy part. If you've done the rest of the problem in a
We're trying to find the time rate of change of V, so we have to differentiate the function we just found with respect to t. "But our equation for V is in terms of r," you might claim, but implicit differentiation allows us to take the derivative anyway:
V' = 10π/3 * 2r * dr/dt
What's that? We have a dr/dt term! But that's okay, because dr/dt and r are both given values:
V' = 10π/3 * 2(24) * 1
This video is a pretty good explanation of related rates problems:
There is a series of videos by PatrickJMT on related rates problems that you can find here:
ALEX CRANS WILL BE THE NEXT SCRIBE
But before class could begin, a scribe had to be chosen, for Eddie, awake until 1:30 the night before slaving away on his scribe post, had forgotten to name his successor. The class quickly divided into two factions pleading not to be chosen: the Phyzards and the Bio-Nerds, complaining about "supercorrections" and "bio outlines" respectively. Even Gabe and Duncan of no-supercorrections did not step forward. Eddie was overwhelmed, but just as O'Brien was about to choose randomly from the mob, Crockett rose above the masses, a beacon of hope (see fig. 1) and volunteered his evening for the benefit of Phyzards and Bio-Nerds alike. Crockett's hair flowed back in a sudden classroom breeze and women swooned at his courage and selflessness.
 | |
| fig. 1: Crockett emerges just as the Phyzards and Bio-Nerds are at each others throats. |
Related Rates
O'Brien then gave us a problem:Given a circular cone of height 10cm and radius r, increasing at a rate of 1 cm/s, what is the rate of change of the volume when the radius is 24?
O'Brien gave us a sort of loose procedure for problems like these:
1. Write what you know.
in this case we know that:
r = 24 cm
dr/dt = 1 cm/s
h = 10 cm
2. Draw a picture.
Drawing pictures makes it easier to find relationships. A helpful tactic is to superimpose your drawing over a set of axes, then maybe write some equations for lines that you see.
3. Find relationships.
To do this problem, we have to know that the volume of a cone is equal to 1/3Bh, or 1/3πr^2h
O'Brien told us that if we have equations for our variables, it's not always best to immediately substitute those in. However, if the value is a constant, we should substitute in immediately in order to avoid messy product rules.
Because h is constant, we substituted that in, leaving us with:
V = 10π/3 * r^2
4. Do calculus.
This is the easy part. If you've done the rest of the problem in a
We're trying to find the time rate of change of V, so we have to differentiate the function we just found with respect to t. "But our equation for V is in terms of r," you might claim, but implicit differentiation allows us to take the derivative anyway:
V' = 10π/3 * 2r * dr/dt
What's that? We have a dr/dt term! But that's okay, because dr/dt and r are both given values:
V' = 10π/3 * 2(24) * 1
This video is a pretty good explanation of related rates problems:
There is a series of videos by PatrickJMT on related rates problems that you can find here:
ALEX CRANS WILL BE THE NEXT SCRIBE
Tuesday, December 4, 2012
Scribe Post 12/3
We began our beautifully wonderful mathematics experience on this fine day by going over the Unit 3 Quiz #2. Before we looked at the quiz, however, O'Brien informed our class that many of his Calculus students (our class especially) have been putting off the homework, and therefore many more people did less than excellently on the quiz. :( He also informed us that a mere one singular person passed in their IW 1-5 packet on time. Shame on us. :'(
After everybody stood up and saluted O'Brien, swearing on our sacred Calculus books that we would never again do an act so deplorable as that ever EVER again, we moved on to correcting the quiz.
Quiz problems went over in class:
1. nDeriv it! Why do more work than you have to? We're nothing but lazy high school students, and we need to act like it!
2. Double derive! At the points where y''=0 is where it changes from concave up to concave down.
3. Derive and conquer! *cough* nDeriv *cough*
4. Concavity occurs at points where p'' crosses the x-axis. function "p" is concave up when p'' is positive and it is concave down when p'' is negative.
5. Product rule and chain rule dat shindig. Once you get a super messy f '(x), then you can solve it just like O'Brien did like this:
Voilá! it's beautiful!
6. Use chain rule and power rule on dat conjunction junction, what's your function? 'cept you do it TWICE. #ohmygoggles. Once you get the second derivative, you'll see where y'' will be 0, and plug in that value for x to the y equation in order to find your answer.
7. Well, well, well. This was a tricky one. So tricky, in fact, that O'Brien declared that anybody who got all three parts of number 7 correct got a 100 on their quiz! WOWZERS! He also said that anybody else who got parts of it correct received bonus points, due to the trickiness of the question...in question.
This led to a discussion as to what makes the correct answers to these problems, and O'B presented us with these answers:
Crockett-Rockett Lalor, however, did not believe that these were very correct. He was sure that he had O'Brien fooled when O'Brien asked for any volunteers to come up to the board. Crockett went up and sketched his graph here:
Crockett claimed that the questions on the quiz were worded in a way that did not correctly define what was supposed to be graphed, and therefore this adorable little kitty cat, which O'Brien proclaimed had "conCATvity" was a legitimate answer to the question. O'Brien did not however, agree with Crockett, and proceeded to make Crockett fight hungry lions as punishment for his drawing of kittens on the board of the mathematics room. No, I'm just kidding, Crockett didn't fight lions, but he did have to sit down in his seat.
Now we move on to the fun stuff. #mathswag
O'Brien wrote three very strange words up on the board:
Linearization
Differential
OptimizationSo, as any normal class would do when confronted with strange words, we all asked in a harmonic symphony of voices, "Mr. O'Brien, what do these strange words mean?"
Mr. O'Brien then explained to us the meaning of these odd words as thus:
Linearization is the term used for when you find the equation of a tangent line at a point, which we've had extensive practice with. O'Brien said that we're pretty gosh-darn good at this already, so we accepted the compliment and moved on.
Differentials O'Brien explained with some fancy shmancy equations that he wrote up on the board:
There's also a little graphy graph there talking a bit about both linearization and differentiation. How very kind of Mr. O'Brien.
But what about that other word on the board? What is an Optimization?
Well, what's the best way for us to be taught? EXPLORATIONS! WOO!
O'Brien gave us an exploration to do called the "Maximal Cylindar in a Cone Problem" Now, this sounds complicated and hard, but it's really not! Everything that you need to know how to figure out is right on the sheet, and the sheet does a very nice job explaining it all.
After illustrating incrediferously how to do the exploration, O'Brien helped us out with a very tricky homework problem. This problem involved a person in a row boat who wishes to return to where she came from, however, she needs to decide what's a faster way, to row there at a speed of 2mph, or row to shore, then walk there at a speed of 5mph. The answer is somewhere in-between. Now, while we may struggle to figure out this problem, there is another species that already has us beaten, O'Brien told us. Dogs. Dogs know calculus. Dogs can calculate where the best place to jump into the water in order to retrieve a tennis ball is in order to get to the ball fastest. How does it feel, classmates, to know that the canine species is better and faster at Calculus than we are? All while sprinting after a ball. I can barely walk and talk at the same time, and dogs can sprint and do Calculus simultaneously. How is this possible? Well, my friends, the answer is quite simple: Magic. Back in the days of old, the magicians of the time were meddling in the affairs of the king, and were wondering just what is it that––
No. I'm getting off track. Back to math.
Here's the picture and working that O'Brien put on the board for us.
There we go. Easy as π
Homework was:
IW #7
* p. 231/7, 10, 37a
* p. 248/5, 17, 27, 41ab, 59, 60, 62
Apparently, Mr. O'Brien's magnanimity knows no bounds. The wonderful wizard of mathematics told us he would upload to the Even Answers section of the beautifully incredible answer booklet for chapter 5.4. The class applauded, Dr. I gave everybody the day off, the government came in with large bags of money, and the President even declared December 3rd as Mr. O'Brien Day, due to the graciousness shown by Mr. O'Brien.
Aaaaaand that's all, folks. I need to get out of here before I end up writing more paragraphs about magical wizards and dogs doing calculus.
Speaking of, if you want to read a little bit more about that topic, here's an article about aMagician Mathematician who's dog is a magical calculus dog!
Yay! Calculus dogs!
-Eddie McCluskey, faithful scribe
NEW SCRIBE: Crockett Lalor.
After everybody stood up and saluted O'Brien, swearing on our sacred Calculus books that we would never again do an act so deplorable as that ever EVER again, we moved on to correcting the quiz.
Quiz problems went over in class:
1. nDeriv it! Why do more work than you have to? We're nothing but lazy high school students, and we need to act like it!
2. Double derive! At the points where y''=0 is where it changes from concave up to concave down.
3. Derive and conquer! *cough* nDeriv *cough*
4. Concavity occurs at points where p'' crosses the x-axis. function "p" is concave up when p'' is positive and it is concave down when p'' is negative.
5. Product rule and chain rule dat shindig. Once you get a super messy f '(x), then you can solve it just like O'Brien did like this:
6. Use chain rule and power rule on dat conjunction junction, what's your function? 'cept you do it TWICE. #ohmygoggles. Once you get the second derivative, you'll see where y'' will be 0, and plug in that value for x to the y equation in order to find your answer.
7. Well, well, well. This was a tricky one. So tricky, in fact, that O'Brien declared that anybody who got all three parts of number 7 correct got a 100 on their quiz! WOWZERS! He also said that anybody else who got parts of it correct received bonus points, due to the trickiness of the question...in question.
This led to a discussion as to what makes the correct answers to these problems, and O'B presented us with these answers:
Crockett-Rockett Lalor, however, did not believe that these were very correct. He was sure that he had O'Brien fooled when O'Brien asked for any volunteers to come up to the board. Crockett went up and sketched his graph here:
Now we move on to the fun stuff. #mathswag
O'Brien wrote three very strange words up on the board:
Linearization
Differential
OptimizationSo, as any normal class would do when confronted with strange words, we all asked in a harmonic symphony of voices, "Mr. O'Brien, what do these strange words mean?"
Mr. O'Brien then explained to us the meaning of these odd words as thus:
Linearization is the term used for when you find the equation of a tangent line at a point, which we've had extensive practice with. O'Brien said that we're pretty gosh-darn good at this already, so we accepted the compliment and moved on.
Differentials O'Brien explained with some fancy shmancy equations that he wrote up on the board:
There's also a little graphy graph there talking a bit about both linearization and differentiation. How very kind of Mr. O'Brien.
But what about that other word on the board? What is an Optimization?
Well, what's the best way for us to be taught? EXPLORATIONS! WOO!
O'Brien gave us an exploration to do called the "Maximal Cylindar in a Cone Problem" Now, this sounds complicated and hard, but it's really not! Everything that you need to know how to figure out is right on the sheet, and the sheet does a very nice job explaining it all.
After illustrating incrediferously how to do the exploration, O'Brien helped us out with a very tricky homework problem. This problem involved a person in a row boat who wishes to return to where she came from, however, she needs to decide what's a faster way, to row there at a speed of 2mph, or row to shore, then walk there at a speed of 5mph. The answer is somewhere in-between. Now, while we may struggle to figure out this problem, there is another species that already has us beaten, O'Brien told us. Dogs. Dogs know calculus. Dogs can calculate where the best place to jump into the water in order to retrieve a tennis ball is in order to get to the ball fastest. How does it feel, classmates, to know that the canine species is better and faster at Calculus than we are? All while sprinting after a ball. I can barely walk and talk at the same time, and dogs can sprint and do Calculus simultaneously. How is this possible? Well, my friends, the answer is quite simple: Magic. Back in the days of old, the magicians of the time were meddling in the affairs of the king, and were wondering just what is it that––
No. I'm getting off track. Back to math.
Here's the picture and working that O'Brien put on the board for us.
There we go. Easy as π
Homework was:
IW #7
* p. 231/7, 10, 37a
* p. 248/5, 17, 27, 41ab, 59, 60, 62
Apparently, Mr. O'Brien's magnanimity knows no bounds. The wonderful wizard of mathematics told us he would upload to the Even Answers section of the beautifully incredible answer booklet for chapter 5.4. The class applauded, Dr. I gave everybody the day off, the government came in with large bags of money, and the President even declared December 3rd as Mr. O'Brien Day, due to the graciousness shown by Mr. O'Brien.
Aaaaaand that's all, folks. I need to get out of here before I end up writing more paragraphs about magical wizards and dogs doing calculus.
Speaking of, if you want to read a little bit more about that topic, here's an article about a
Yay! Calculus dogs!
-Eddie McCluskey, faithful scribe
NEW SCRIBE: Crockett Lalor.
Subscribe to:
Posts (Atom)