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Use to help with Supercorrections!
Friday, December 21, 2012
Thursday, December 13, 2012
Saturday, December 8, 2012
Scribe Post 12/7
For those of you who may not have been here to see it, we started off class today by looking at a video of a cute wombat. *Fun Fact* If you haven't already figured this out, OB's favorite animal is the wombat..
We were then subjected to a forty minute quiz on optimization problems, related rates, linearization, and differentials. Safe to say, it was a pretty tough quiz, seeing as there was minimal joking going on between us–something almost unheard of for period 4! While we were taking the quiz, an article of 15 Hidden Health Secrets of Lemons was on the Smartboard. Why, you may ask? Not sure, but if you want to know some more about lemons, feel free.
Once the quiz was over, O'Brien thought it'd be fun to compare our current knowledge of calculus to see how we would have done on the AP Exam that he took as a senior in 1990 (22 years ago..). According to OB, "back in the day" the exam was taken at the Congregational church (makes sense, since nothing goes more hand-in-hand than calculous and religion) and that you could not use a calculator on the free response questions. For this reason, back then, free response questions were more on "a type," whereas now, the concepts are more mixed and matched, making it important that we understand what actually going on in a problem.
So we looked at question 4, a related rate problem, which is about a sphere whose radius is increasing at a rate of 0.04 centimeters per second. Unfortunately, OB said that for this particular problem, a picture wouldn't really help you out that much. We should note, however, that as the rate of the radius increases, the volume will initially increase drastically and then begin to slow down over time. To help us visualize this, he used a balloon analogy: when you first blow air into a balloon, its volume increases rapidly, but as you continue to blow air into the balloon, you begin to notice the change in volume less and less. OB was immediately sassed out by Autumn for this analogy, perhaps rightfully so, seeing as this problem is CLEARLY about a sphere, not a balloon..
Okay, onto the problem. Seeing as you can all read what the question is asking for yourselves, I'm just going to dive in.
Part (a):
Given that: Find:
The key to related rates problems, is to use a relationship that relates the variables that you're given. For this problem, that relationship would be the volume of a sphere:
The next step is to take the derivative of this relationship with respect to t, the independent variable. Don't forget to use the chain rule and implicit differentiation when you do so! Once you do that, sub in your given values and solve!
Autumn then makes the comment that, thus far, this AP question is easier than the quiz we just took. OB tells Autumn not to "whinge".
Part (b):
Given that: Find:
This problem wants us to find the cross-sectional circle within the sphere, or the largest circle possible within the sphere. To do this, we'll use the equation for the area of a circle:
Just like before, we're going to differentiate with respect to t:
Hold on though! While we know the rate at which the radius is increasing, we DON'T know what the radius is. This is where the given volume comes into play!
Now that we have a value for our radius, we can finish solving the problem:
Part (c):
This part sounds confusing, but it's actually quite simple. The volume and radius are increasing at the same numerical rate, but the rates themselves are different because volume and radius have different units. Here's how to solve it:
Answers and solutions to the rest of the 1990 AP exam are found here.
Tips by OB on related rates problems:
1. Label everything nicely
2. Draw a picture to help you visualize
3. Have a personal "cheering section" in your mind... (is this what you do in nordic too OB?)
Still need help with related rates problems? Check out the videos on Autumn's scribe post.
IW for next class:
pg. 256/ 14, 16, 25, 30, 31, 33, 39, 41
The next scribe, chosen at random out of a hat, will be...
If you can't scribe on Tuesday, try to find someone else to switch with you. Otherwise, good luck!
Update!!
Here's a nice little song for everyone to get you into the calculous mood.
Seeing as midterms are coming up, I feel like a good update would be a refresher on some difficult related rates problems, particularly (for me, anyway) those involving cones. A good example can be found here by PatrickJMT.
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
The big thing in this scribe post is RELATED RATES, if that’s what you’re looking for, here it is!
We started off class with the question, “The side of a square is increasing at 2 cm/sec. How fast is the area increasing when the side is 4 cm long?” This brilliant question by O’Brien given with a wondrous animation on GeoGebra stumped our class. We kept asking questions like “at what time does the area start increasing? 0 seconds?” and “can’t you just tell us how to do it?” After a while of working on it our class decided on our answer: 8 cm^2/sec, which was wrong.
When O’Brien started actually helping us with this confusingly simple problem he threw our question right back at us: “Does it really matter at what number of seconds you start? 0 seconds? 1 seconds? 16 seconds?” He used a table (shown below) to prove that no, it doesn’t matter, no matter time you start at the volume is increasing at a steady rate.
We further proved this by looking at the rate of change at a moment in time instead of over 2 seconds. We did this by again, using a table (shown below) and looking at the times 1 second and 1.0001 seconds.
We noticed that for the change between the two times and between the two areas were both reeally close to zero, so we thought zero/zero = zero right? Or undefined? Then we thought, but wait, these two numbers aren’t equal to zero, they’re both just really close which means we’ll actually get a value here:
Once we did this we realized we just found the derivative by using the limit in indeterminate form. The derivative for this particular problem at this moment in time is about 16.0004.
After all this work O’Brien got his evil smile on to let us know there’s actually a much easier way to do all this. Thanks for letting us know before hand O’Brien. To use this easier way, we must first find the equation to this problem. By looking at our tables and the question asked in this problem it’s easy to see that our equation must be:
To solve for the rate of the area we must differentiate this equation with respect to t (time) (basically implicit differentiation):
Once again we proved that the rate or derivative at a moment in time is 16 cm^2/sec (above when we found the rate at a moment in time is was about 16). Important things to notice in the math above is that once we found the equation and differentiated it we plugged in the values we knew for s and ds/dt which were given in the original problem. It said “the side of a square is increasing at 2 cm/sec” which is ds/dt and that we are to solve for “when the side is 4 cm long” which is s.
Simple enough right?
The basic idea behind related rates that I’ve found O’Brien’s teachings and most of the sites and videos I’ve looked at to agree with is that each variable is now a function of time, or t, just like we did above. That means when we take the derivatives, we won’t be taking them in respect to each other like we used to, but instead with respect to t. This is important to know because some of the variables will not be changing with respect to t. These variables are like constants, and the derivative of them with respect to t is 0. When I say the derivative “with respect to t” I just mean that when you take the derivative of an equation you must multiple the variables in both sides by d/dt, for an example you can see we did this in the problem above.
And seeing as the name of the game is “related rates” it would be important to know if you didn’t already that something per something else, for instance cm/sec, would be a rate. So in an equation, whenever you see anything like that, it’s a rate and most likely important.
On the back of the exploration the Steps for every related rates problem: were given and I’ve found that almost every site or video on related rates uses basically the same steps so take note! They are important and helpful!
1. Label given information (Sketch where appropriate)
2. Note which quantities are variable, and which are constant
3. Identify rates d( )/dt which are given, and which are needed
4. Differentiate
5. Plug in date - ALWAYS AFTER YOU DIFFERENTIATE
The “beautiful man” (as Scotty would put it) Patrick JMT put it another way which I found helpful, he used an acronym: DREDS
Diagram
Rates
Equation
Derivative
Substitute specific info
He did that in this video, the rest was also helpful
Also for future reference Patrick JMT actually has a website and he has videos on there for every math concept imaginable so if you’re ever stuck look here! and *GASP* you can actually kind of see what he looks like!
And then while looking for other sites and videos on Related Rates which I thought would be helpful I’ve found this is actually quite the enjoyable experience because the people who make youtube videos on math problems are all so...interesting especially since most, unlike Patrick JMT who you only shows he’s hand in the videos, actually have their whole body in it. So there’s the expected nerdy, wonderfully awkward people trying to explain a math concept in a youtube video and then there’s this guy, Dr. Bob, who looks like he could be a body builder or a wrestling coach or something and is instead, on youtube being a math nerd, who, and I quote, said that he has a method for related rates problems that will “let us tear them down from start to finish.” Enjoy.
I hope I did a decent job explaining Related Rates and you all enjoyed my scribe post! The next scribe will be Sarah Mayberry, good luck!
We started off class with the question, “The side of a square is increasing at 2 cm/sec. How fast is the area increasing when the side is 4 cm long?” This brilliant question by O’Brien given with a wondrous animation on GeoGebra stumped our class. We kept asking questions like “at what time does the area start increasing? 0 seconds?” and “can’t you just tell us how to do it?” After a while of working on it our class decided on our answer: 8 cm^2/sec, which was wrong.
When O’Brien started actually helping us with this confusingly simple problem he threw our question right back at us: “Does it really matter at what number of seconds you start? 0 seconds? 1 seconds? 16 seconds?” He used a table (shown below) to prove that no, it doesn’t matter, no matter time you start at the volume is increasing at a steady rate.
We further proved this by looking at the rate of change at a moment in time instead of over 2 seconds. We did this by again, using a table (shown below) and looking at the times 1 second and 1.0001 seconds.
We noticed that for the change between the two times and between the two areas were both reeally close to zero, so we thought zero/zero = zero right? Or undefined? Then we thought, but wait, these two numbers aren’t equal to zero, they’re both just really close which means we’ll actually get a value here:
Once we did this we realized we just found the derivative by using the limit in indeterminate form. The derivative for this particular problem at this moment in time is about 16.0004.
After all this work O’Brien got his evil smile on to let us know there’s actually a much easier way to do all this. Thanks for letting us know before hand O’Brien. To use this easier way, we must first find the equation to this problem. By looking at our tables and the question asked in this problem it’s easy to see that our equation must be:
To solve for the rate of the area we must differentiate this equation with respect to t (time) (basically implicit differentiation):
Once again we proved that the rate or derivative at a moment in time is 16 cm^2/sec (above when we found the rate at a moment in time is was about 16). Important things to notice in the math above is that once we found the equation and differentiated it we plugged in the values we knew for s and ds/dt which were given in the original problem. It said “the side of a square is increasing at 2 cm/sec” which is ds/dt and that we are to solve for “when the side is 4 cm long” which is s.
Simple enough right?
The basic idea behind related rates that I’ve found O’Brien’s teachings and most of the sites and videos I’ve looked at to agree with is that each variable is now a function of time, or t, just like we did above. That means when we take the derivatives, we won’t be taking them in respect to each other like we used to, but instead with respect to t. This is important to know because some of the variables will not be changing with respect to t. These variables are like constants, and the derivative of them with respect to t is 0. When I say the derivative “with respect to t” I just mean that when you take the derivative of an equation you must multiple the variables in both sides by d/dt, for an example you can see we did this in the problem above.
And seeing as the name of the game is “related rates” it would be important to know if you didn’t already that something per something else, for instance cm/sec, would be a rate. So in an equation, whenever you see anything like that, it’s a rate and most likely important.
On the back of the exploration the Steps for every related rates problem: were given and I’ve found that almost every site or video on related rates uses basically the same steps so take note! They are important and helpful!
1. Label given information (Sketch where appropriate)
2. Note which quantities are variable, and which are constant
3. Identify rates d( )/dt which are given, and which are needed
4. Differentiate
5. Plug in date - ALWAYS AFTER YOU DIFFERENTIATE
The “beautiful man” (as Scotty would put it) Patrick JMT put it another way which I found helpful, he used an acronym: DREDS
Diagram
Rates
Equation
Derivative
Substitute specific info
He did that in this video, the rest was also helpful
Also for future reference Patrick JMT actually has a website and he has videos on there for every math concept imaginable so if you’re ever stuck look here! and *GASP* you can actually kind of see what he looks like!
And then while looking for other sites and videos on Related Rates which I thought would be helpful I’ve found this is actually quite the enjoyable experience because the people who make youtube videos on math problems are all so...interesting especially since most, unlike Patrick JMT who you only shows he’s hand in the videos, actually have their whole body in it. So there’s the expected nerdy, wonderfully awkward people trying to explain a math concept in a youtube video and then there’s this guy, Dr. Bob, who looks like he could be a body builder or a wrestling coach or something and is instead, on youtube being a math nerd, who, and I quote, said that he has a method for related rates problems that will “let us tear them down from start to finish.” Enjoy.
I hope I did a decent job explaining Related Rates and you all enjoyed my scribe post! The next scribe will be Sarah Mayberry, good luck!
Monday, December 3, 2012
Scribe Post 12/3
After going over the second quiz we went into today's lesson. We learned two words which we will start to see more frequently in the book and on the AP Exam.
Linearization: Another way of saying the tangent line.
An interesting applet to get an even better understanding of the tangent line:
http://calculusapplets.com/linearapprox.html
Differential:
We then worked on Exploration 8-3: Maximal Cylinder in a Cone Problem. For some reason we all struggled with this even though O'Brien said that it's really easy. Here are the answers:
1.
V(0) = 0
V(1) = 9π
V(2) = 24π
V(3) = 27π
V(4) = 0
2.
3.
4.
5.
6.
radius: 8/3
height: 4
volume: (128π/9)
7.
-find f'(x), solve for 0
-sub that value into f(x) and the function for height
8. You know what you learned better than I do.
Linearization: Another way of saying the tangent line.
An interesting applet to get an even better understanding of the tangent line:
http://calculusapplets.com/linearapprox.html
Differential:
find dy if x=1 and dx=0.01
We then worked on Exploration 8-3: Maximal Cylinder in a Cone Problem. For some reason we all struggled with this even though O'Brien said that it's really easy. Here are the answers:
1.
V(0) = 0
V(1) = 9π
V(2) = 24π
V(3) = 27π
V(4) = 0
2.
3.
4.
5.
6.
radius: 8/3
height: 4
volume: (128π/9)
7.
-find f'(x), solve for 0
-sub that value into f(x) and the function for height
8. You know what you learned better than I do.
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