Friday, November 30, 2007
Thursday, November 29, 2007
Optimization Problems
Well i am finally back to scribe, it has been a pretty long time since i had last scribed so here it is. Today in class, Mr. K introduced a new topic Optimization problems. Optimization problems are problems that deal with either finding the maximum or minimum of the function in the problem. although very rarely is the function given so you as a calculus student must use all your knowledge you've learnt up to now and create a formula/function that can be used to answer the question that is being asked. With that said Mr. K gave the class the 6 rules to live by when doing optimization problems, although do not really live by them, just follow them loosely.
6 Rules to Optimization Problems
1. Find what you are trying to either maximize or minimize.
2. Write an equation for it, make sure to use descriptive variables so you won't get confused when doing the problem.
3. Try to get the equation into a two variable form, where you are finding one of the variables, because you want to find the derivative.
4. Write the derivative of the equation in the two variable form.
5. Set the derivative to zero, and solve for the remaining values. (First Derivative Test or Second derivative Test)
6. If needed depending on what the question asks, Plug the value of that variable into the one or (two) equations to find all dimensions.
Now here is the explanation of the question done in class.
Well first we identified it was a maximum problems, because it asks for the box with dimensions with the largest volume.
Next the equation is made by using V=h*w*L, so h=x, w=16-2x, and L=21-2x. So the equation looks something like this, x(16-2x)(21-2x).
Then we multiply it out. 336x - 74x^2 + 4x^3
Then we take the derivative. Which is 336 - 148x + 12x^2
then by using the quadratic equation we find that x = 3 and 9.3, so we do a line graph and find it is positive before 3 and negative after 3 meaning that by the use of first derivative test, 3 is the maximum x can be.
Well, that was it for today's class. Tomorrow's Scribe will be M@rk.
6 Rules to Optimization Problems
1. Find what you are trying to either maximize or minimize.
2. Write an equation for it, make sure to use descriptive variables so you won't get confused when doing the problem.
3. Try to get the equation into a two variable form, where you are finding one of the variables, because you want to find the derivative.
4. Write the derivative of the equation in the two variable form.
5. Set the derivative to zero, and solve for the remaining values. (First Derivative Test or Second derivative Test)
6. If needed depending on what the question asks, Plug the value of that variable into the one or (two) equations to find all dimensions.
Now here is the explanation of the question done in class.
Well first we identified it was a maximum problems, because it asks for the box with dimensions with the largest volume.
Next the equation is made by using V=h*w*L, so h=x, w=16-2x, and L=21-2x. So the equation looks something like this, x(16-2x)(21-2x).
Then we multiply it out. 336x - 74x^2 + 4x^3
Then we take the derivative. Which is 336 - 148x + 12x^2
then by using the quadratic equation we find that x = 3 and 9.3, so we do a line graph and find it is positive before 3 and negative after 3 meaning that by the use of first derivative test, 3 is the maximum x can be.
Well, that was it for today's class. Tomorrow's Scribe will be M@rk.
Wednesday, November 28, 2007
Limits Involving Infinity
Hello all! It’s Robert here and I’m your scribe for today.
Today’s class was all about Limits Involving Infinity. Finding limits of infinity is all about finding the asymptotes of a graph.
Some things you need to remember when finding limits (finding asymptotes):
Today’s class was all about Limits Involving Infinity. Finding limits of infinity is all about finding the asymptotes of a graph.
Some things you need to remember when finding limits (finding asymptotes):
- When the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y=0.
- When the degree of the denominator is less than the degree of the numerator, there is no horizontal asymptote.
- When the degree of the numerator is equal to the degree of the denominator, you look to the leading coefficients of the term that has the highest exponent on x.
Well here are the slides for today's class. Take a look and see the different steps used in solving the problems.
That's it for today folks. The next scribe will be........Dino!!!
Tuesday, November 27, 2007
Scribe #5 out of A LOT!
Today's amazing class started off with a bit of side-tracking. Mr. K. introduced us to a couple anecdotes that will be helpful to everyone in their overall learning skills. Now of course Mr. K. got them from a source that I am quite lazy enough not to discover... but they were:
LEVELS OF LEARNING (Bloom's Taxonomy)
Basically describes the different ways we can learn. They are Remember, Understand, Apply, Analyze, Evaluate, and Create. When we begin to learn in elementary, we start at the beginning and eventually climb up the levels of learning. Currently we are at the levels of Analyze and Evaluate, which still require all the others, and use them in class everyday.
LEARNING PYRAMID
Shows the different media (or methods) from which one can learn and the percentage of that knowledge that one would retain from each method. These methods are (in order):Lectures, Reading, Audio-Visual, Demonstrations, Discussions, Practicing by Doing, and Teaching. It turns out we only retain about 5% of knowledge from Lectures and a whopping 90% from teaching... that's a huge gap!
So in conclusion, Mr. K. told us that when we do our DEV (Developing Expert Voices projects)
, we are combining the Create and Teaching methods. That is one heck of a lot of learning that we do!!!
But now I will jump into the actual lesson of the day...
Well, we first did a quick review of the First Derivative Test, which is:
Next, Mr. K. introduced, the Second Derivative Test, which goes as follows:
We then jumped into some sample questions using this test. These can be seen on the slides for today's class. As well, you can click on the links underneath the definition of the tests on the green pages for more practice problems.
Next Mr. K. showed us a little loop-hole... It is thought that when the second derivative (ƒ''(x)) is equal to 0 (zero), there is a point of inflection on the parent function right??? The thing is, it might not be. It may in fact be only a candidate for the point of inflection. Then we must test the concavity on either side of this point (whether ƒ'' is positive or negative). If the concavity is different on each siide, it is a point of inflection, but if it is the same on both sides, the point is not a point of inflection, it is a local extreme that is relatively flat in a certain interval. AN EXAMPLE OF THIS IS THE FUNCTION ƒ(x)=x^4
We then continued with practice problems (which can be seen on the slides; link is above) until the end of class.
Quick thing to add to our understanding of these tests:
First Derivative Test: Interval test; find the candidates for global extrema and check the intervals on each side of them to prove they are in fact extrema.
Second Derivative Test: Point test; find candidates of global extrema and find the concavity of the parent function at each candidate.
Mr. K. has said, that one is not necessarily better than the other, they are just DIFFERENT... meaning in different situations, different tests will be better to use. (I like the Second one better)
That just about wraps up my Scribe for the evening...
REMEMBER GUYS: 5.2 ODDS FOR HOMEWORK!!!
Next Scribe is.......
LEVELS OF LEARNING (Bloom's Taxonomy)
Basically describes the different ways we can learn. They are Remember, Understand, Apply, Analyze, Evaluate, and Create. When we begin to learn in elementary, we start at the beginning and eventually climb up the levels of learning. Currently we are at the levels of Analyze and Evaluate, which still require all the others, and use them in class everyday.
LEARNING PYRAMID
Shows the different media (or methods) from which one can learn and the percentage of that knowledge that one would retain from each method. These methods are (in order):Lectures, Reading, Audio-Visual, Demonstrations, Discussions, Practicing by Doing, and Teaching. It turns out we only retain about 5% of knowledge from Lectures and a whopping 90% from teaching... that's a huge gap!
So in conclusion, Mr. K. told us that when we do our DEV (Developing Expert Voices projects)
, we are combining the Create and Teaching methods. That is one heck of a lot of learning that we do!!!
But now I will jump into the actual lesson of the day...
Well, we first did a quick review of the First Derivative Test, which is:
Next, Mr. K. introduced, the Second Derivative Test, which goes as follows:
We then jumped into some sample questions using this test. These can be seen on the slides for today's class. As well, you can click on the links underneath the definition of the tests on the green pages for more practice problems.Next Mr. K. showed us a little loop-hole... It is thought that when the second derivative (ƒ''(x)) is equal to 0 (zero), there is a point of inflection on the parent function right??? The thing is, it might not be. It may in fact be only a candidate for the point of inflection. Then we must test the concavity on either side of this point (whether ƒ'' is positive or negative). If the concavity is different on each siide, it is a point of inflection, but if it is the same on both sides, the point is not a point of inflection, it is a local extreme that is relatively flat in a certain interval. AN EXAMPLE OF THIS IS THE FUNCTION ƒ(x)=x^4
We then continued with practice problems (which can be seen on the slides; link is above) until the end of class.
Quick thing to add to our understanding of these tests:
First Derivative Test: Interval test; find the candidates for global extrema and check the intervals on each side of them to prove they are in fact extrema.
Second Derivative Test: Point test; find candidates of global extrema and find the concavity of the parent function at each candidate.
Mr. K. has said, that one is not necessarily better than the other, they are just DIFFERENT... meaning in different situations, different tests will be better to use. (I like the Second one better)
That just about wraps up my Scribe for the evening...
REMEMBER GUYS: 5.2 ODDS FOR HOMEWORK!!!
Next Scribe is.......
Monday, November 26, 2007
Mondays post :)
First off before I begin, I don't know what this unit is, is it still the same? If you know put it in the chat so I can label this correctly. Now lets begin :)
The next scribe is I got no idea, GreyM your stuck tomorrow at lunch with me :) and that's it
Word of the day:
Gular- of or relating to the throat.
We watched a video in the beginning of class. I don't know the link or how to get it but it shows us about the graph of f(x) and a little about how f'(x) applies to it. Here's a pic of the graph in the video.
In class we only did three questions. Its all about finding the zeros of the derivative to find out is min/max critical points, and if they are local or global.
Here is the first question, its pretty straight forward. To find critical points find the root(s) of the derivative, which is 2. Put it on the number line to find out min or max, since its a restriction you have to find those too. Plug those into the the parent function to find their value and to determine global max since we already have the global min. (min = minimum) (max = maximum)
Next question. Same thing, all the detail is on the Pic :)
And last but not least the last question. Same thing and once again all details should be on it.
The next scribe is I got no idea, GreyM your stuck tomorrow at lunch with me :) and that's it
Saturday, November 24, 2007
BOB^4
Oh wow, for the first time in a couple of weeks... the blog actually loaded! Usually it takes a couple hours for the blog to fully load, but apparently not this time. But I don't have any time on the computer, I have to wake up early tomorrow and I pretty much have about 10 minutes left in me before I pass out on my bed due to exhaustion (I had to plan, move around objects to get ready for packing, and go around the city in search of supplies for the new house all day). )=. Thankfully, I remembered to BOB before the test (since the engineering outreach day overrided my obligation to take the test on thursday postponing it for Craig, Mark and I until monday) before it was really too late. I hope that everyone performed well on their test, and handled it with grace and ease (though it is a Mr. K test). Well in any case, it'd be best to get on with the post now.
I originally designated this unit to be one of my favorite units as we entered it approximately two weeks ago, and I still retain my previous position. I think that the overall rules involved with differentiation aren't terribly difficult, but applying them to certain functions such as related rates problems and with problems involving Newton's method or linear approximations might require more thought than many might have initially assumed. As with the application of any form of mathematics, problem repetition is definitely a profound factor in test preparation, but might be more required for this unit. As anyone in the class has probably noticed, practicing questions pertaining to this unit really help to adapt to the type of thoughts that are required to solve some of the aforementioned more challenging problems. I think that if I complete enough homework for these questions, I shouldn't find too much difficulty with the test. Throughout the progession of the unit, nothing really presented much difficulty or concerns to me except for the tedious process of reiterating calculations for Newton's method (though it can now be automated by our calculator using a custom program). I think that, especially considering the questions given on the pre-test, I'll have to recognize the exact context of the question to a fuller extent and really make sure that I have identified the entire scenario of the problem.
Now, I'm off to read some young students blogs and determine how I can comment on them in the most constructive manner possible, though it might not be so wise to do so with such a great lack in consciousness that I might fall asleep flat on my computer desk instantly =/. Nevertheless, I'll read some comments and maybe initiate some differentiation practice tonight and continue the arduous process of constant repetition tomorrow. I'll see everyone on monday, and once again, I hope everyone is confident with their performance on the test on Thursday. Though it might be too late for most, I'll still conclude the BOB post with the now cliche GOOD LUCK.
I originally designated this unit to be one of my favorite units as we entered it approximately two weeks ago, and I still retain my previous position. I think that the overall rules involved with differentiation aren't terribly difficult, but applying them to certain functions such as related rates problems and with problems involving Newton's method or linear approximations might require more thought than many might have initially assumed. As with the application of any form of mathematics, problem repetition is definitely a profound factor in test preparation, but might be more required for this unit. As anyone in the class has probably noticed, practicing questions pertaining to this unit really help to adapt to the type of thoughts that are required to solve some of the aforementioned more challenging problems. I think that if I complete enough homework for these questions, I shouldn't find too much difficulty with the test. Throughout the progession of the unit, nothing really presented much difficulty or concerns to me except for the tedious process of reiterating calculations for Newton's method (though it can now be automated by our calculator using a custom program). I think that, especially considering the questions given on the pre-test, I'll have to recognize the exact context of the question to a fuller extent and really make sure that I have identified the entire scenario of the problem.
Now, I'm off to read some young students blogs and determine how I can comment on them in the most constructive manner possible, though it might not be so wise to do so with such a great lack in consciousness that I might fall asleep flat on my computer desk instantly =/. Nevertheless, I'll read some comments and maybe initiate some differentiation practice tonight and continue the arduous process of constant repetition tomorrow. I'll see everyone on monday, and once again, I hope everyone is confident with their performance on the test on Thursday. Though it might be too late for most, I'll still conclude the BOB post with the now cliche GOOD LUCK.
Thursday, November 22, 2007
BOB... I hope?
Yikes, based on the fact that I wasn't writing the test today, I forgot that we had to BOB before today... well ladies and gentlemen, this will be short and sweet. This has been quite an interesting unit to say the least. I have enjoyed it (except the pre-test) that was intense!!! But yeah I like doing the problems from this unit because it feels really good when you finally get the solution to the most complicated related rates problem!!! But really, nothing I'm too worried about, I'll study a lot... and I guess it's a little late for "Good Luck", but I guess "Let the Force Be With You" would be appropriate for the rest of the year... =)
BOB
Well, this unit about differentiation rules has been a really tough one for me. This unit started off fairly easy, but it got a little challenging when the chain rule and related rates problems came up. I must admit that I'm still having troubles with those two concepts, but hopefully with the help of more studying, i can still get a decent mark. Thats it for my BOB, good luck everyone!
BoB
WOOT crazy episode of deal or no deal :)
first time bob ing so i hope this is right.
the unit was okay for me, better then previous experiences. I believe I'm starting to catch up and close the 2 year gap (last time I did math was in grade 10). I hope I do good on the test, as for fundementals I believe I got them all down.
peace, I love u GreyM :)
good luck all
first time bob ing so i hope this is right.
the unit was okay for me, better then previous experiences. I believe I'm starting to catch up and close the 2 year gap (last time I did math was in grade 10). I hope I do good on the test, as for fundementals I believe I got them all down.
peace, I love u GreyM :)
good luck all
My first BOB (keep forgetting)
I actually would've forgot again for this test, but, I looked at the blog, and was like, RIGHT, time to BOB.
This unit... was fairly easy to understand. I'm just having trouble understanding when to apply derivative rules at certain times. But for the most part, I understand what's going on. Yeah, I need to study more. I shall do that now. Laters.
This unit... was fairly easy to understand. I'm just having trouble understanding when to apply derivative rules at certain times. But for the most part, I understand what's going on. Yeah, I need to study more. I shall do that now. Laters.
Wednesday, November 21, 2007
BOB
This unit was not good for me. It was pretty difficult. Doing homework however definitely helps with the understanding. I have a feeling this test won't be one of the better ones but at least you learn from your mistakes. Great great. Well, good luck to everyone on the test tomorrow. Hopefully you guys studied! (y)
Bob
Well, I gotta say that this unit has been tough for me. Have to chalk that up to only getting to about 10 to 15 done in the homework, but it took me a long time and you just lose your will after a while. Well have started getting back on the horse the last couple of days and have finished up a portion of the homework I haven't done. Coupled with some hardcore studying tonight, tomorrow at my spare and lunch I figure I can pull an 80. I really do get the straight up rules for the derivatives, just identifying what I have to do in question is still my problem. Well good luck every1.
BOB 4
Well hello again doing my fourth bob for AP calculus. Well this unit was very challenging for me especially when we had gotten to the related rates problems. The chain rule gave me a little bit of problems in the very beginning although I was able to correct my mistakes and set myself on the right course. All in all I believe this test is going to be very difficult, in terms of the difficulty of the questions, especially the related rate problem that is probably going to be on the test. Well study hard and good luck to everyone for tomorrow's test.
Scribe five.
Today's class started off with Mr. K. showing us a new tool on the internet which we can possibly use for our DEVs. I cannot quite recall what that tool was called or what the site was but basically it is a tool where you can record your voice while explaining a video in the background where you can draw to highlight points and even fast forward or slow down the video. I will post a link on here later once I figure it out. Soon after that we got into our pretest.
We did our pretest and I think most of us found it difficult. Our pretest consisted of five [or fifteen lol] multiple choice questions and one long answer/free response question. So to get right on to it I think most of us got number one but basically you could put it into your calculator.
Once you realize that you can just simply plug that into your calculator.
The second question involved the quotient rule. Remember: the quotient rule-high d low minus low d high all over low low!
Basically, you must remember how to use the quotient to solve the question, so remember it! Number three was more difficult for a few of us. I didn't get it at first but I asked Mr. K. about it and hopefully I can do an all right job explaining it.
The next question involved the Pythagorean theorem.
uestion number five was also another difficult question we encountered. The last and final question was the long answer. Depending on what your slope was or how you drew your tangent line the answers varied among the class. Mr. K. said he wasn't looking for the correct answer in a question like this he was more interest in the process of how to go about the question. 
That was the last question. Do not forget to study or do a bob and that we don't have school on Friday! Everyone enjoy your extra long weekend! Good luck everyone!
The Next Scribe is...pending for now.
We did our pretest and I think most of us found it difficult. Our pretest consisted of five [or fifteen lol] multiple choice questions and one long answer/free response question. So to get right on to it I think most of us got number one but basically you could put it into your calculator.
Once you realize that you can just simply plug that into your calculator.The second question involved the quotient rule. Remember: the quotient rule-high d low minus low d high all over low low!
Basically, you must remember how to use the quotient to solve the question, so remember it! Number three was more difficult for a few of us. I didn't get it at first but I asked Mr. K. about it and hopefully I can do an all right job explaining it.
The next question involved the Pythagorean theorem.
uestion number five was also another difficult question we encountered. The last and final question was the long answer. Depending on what your slope was or how you drew your tangent line the answers varied among the class. Mr. K. said he wasn't looking for the correct answer in a question like this he was more interest in the process of how to go about the question. 
That was the last question. Do not forget to study or do a bob and that we don't have school on Friday! Everyone enjoy your extra long weekend! Good luck everyone!
The Next Scribe is...pending for now.13-0-13 V. 3
Hello this is my third BOB. I really found this unit difficult because of everything that goes into it. First we learned all of the rules and then we learned how to use those rules in more difficult questions like the related rates problems. When we first got introduced to related rates I was just absolutely lost but I kind of understand them and hopefully I won't do too bad on the test. We all have to do some studying (except for Chris maybe) because after that super hard pretest we were like =S ! There were questions that I didn't understand on the pretest but I asked Mr. K. about it and I think I get them now...even though he said one of those questions would not be on the test..it's still good to understand how to do them. I think I will try to get more help from Mr. K. because I find calculus extremely difficult and I'll understand it better. I think we should all do that if we are confused about calculus. Also, I am going to make a math dictionary for myself soon. I told myself I would before calculus started and I never got to it but I know Craig has made one for himself which is excellent and I think we should all invest in a math dictionary! Good luck to everyone ! I still have to do the scribe =S and study for math + English homework + Chem homework + study for world issues + bake cookies for bake sale...WOW ! okay good luck once again! =)
BOB - Differentiation rules
Well I better BOB because I no i'll need the mark. Well this unit is a real challenging one for me, especially when we got to the related rates problems. I really don't feel that confident especially after that pre-test! Well I guess I have a lot of studying to do when I get home from soccer. Good luck to everyone and do yourselves a favour and bob. Bye...Robert.
The Tuesday Scribe
Today, we had a substitute teacher, named Mr. Roseman. Boy, is he ever into football.
Off the Record with Mr. Roseman.
Anyways, we had to do supplementary problems in the textbook. Page 245, 1-41 all odds.
And, we kinda made a few jokes why Mr. K wasn't here today. Since the Mac that had all of Mr. K's lessons, crashed... and we got a new Mac. The smartboard wasn't as good as it used to be. We had the new software, and he said it was Old School, without all the mechanical things we used. So, we thought Mr. K lost his will to teach because the smartboard was his pride and joy. Haha, good class today.
Let's do newton's method to the a4! You do it Craig?
Anyways, pretest tomorrow and test on Thursday.
Next scribe is Aichelle!
Cya later...
Off the Record with Mr. Roseman.
Anyways, we had to do supplementary problems in the textbook. Page 245, 1-41 all odds.
And, we kinda made a few jokes why Mr. K wasn't here today. Since the Mac that had all of Mr. K's lessons, crashed... and we got a new Mac. The smartboard wasn't as good as it used to be. We had the new software, and he said it was Old School, without all the mechanical things we used. So, we thought Mr. K lost his will to teach because the smartboard was his pride and joy. Haha, good class today.
Let's do newton's method to the a4! You do it Craig?
Anyways, pretest tomorrow and test on Thursday.
Next scribe is Aichelle!
Cya later...
Monday, November 19, 2007
Tangent Approximation and Newton`s Method
Hi everyone! Its Mark here for today`s scribe.
We started today`s class by finding an approximate value of the square root of 37, without using our calculators to get the answer. The burning question is , how are we going to do that? We all know that its some where between the range of 6-6.1 since the square root of 36 is 6, but that is not a good approximation. Here comes the tangent approximation to the rescue. Our plan is to use the tangent line approximation of f(x)=square root of 36 @ x=36. First we need to graph the function, to see whether our approximate will be over or under the real value. If the function is concave up at that given point then our approximation will be over.If the function is concave down at that given point then our approximation will be under the real value. The graph of the function can be seen on the first slide. Second, find the derivative of the function @ x=36. After that create the tangent line at that given point. Then substitute x=37. We can see that our answer is approximately 0.006 larger than the actual value. The operations involve can be seen on the second slide.
We can therefore make a general rule that is true to every function to approximate any given value. The rule can be seen on the third slide.
The second thing we talked about in class is Newton's Method. We started our discussion about this method by going on a side story about Newton and Leibniz feud. After that, we went on describe Newton's method. His method is used for approximating zeros or roots of differentiable functions. Well, the big question is, how does it work? First, it starts with an initial guess, which is reasonably close to the true root, then the function is approximated by using the tangent line , and one computes the x-intercept of this tangent line . This x-intercept will typically be a better approximation to the function's root than the original guess. This process can be done over and over again until the root is found, this process is called iteration. In general, we can apply the formula that can be seen on the fifth slide to find a better approximation.
The third thing we talked about are the scenarios where Newton's method may fail. His method may fail if it produces a derivative of 0 or undefined. Also, it may fail if the steps just continues with the same set of values. Lastly, it wont work if the results diverge away from the root that you are interested in.
If you are feeling a little confused by this new concept, you can check this animation out.
That's it for today's scribe. Our home work for tonight is Exercise 4.7 (all the odd numbered questions). The next scribe will be Van.
We started today`s class by finding an approximate value of the square root of 37, without using our calculators to get the answer. The burning question is , how are we going to do that? We all know that its some where between the range of 6-6.1 since the square root of 36 is 6, but that is not a good approximation. Here comes the tangent approximation to the rescue. Our plan is to use the tangent line approximation of f(x)=square root of 36 @ x=36. First we need to graph the function, to see whether our approximate will be over or under the real value. If the function is concave up at that given point then our approximation will be over.If the function is concave down at that given point then our approximation will be under the real value. The graph of the function can be seen on the first slide. Second, find the derivative of the function @ x=36. After that create the tangent line at that given point. Then substitute x=37. We can see that our answer is approximately 0.006 larger than the actual value. The operations involve can be seen on the second slide.
We can therefore make a general rule that is true to every function to approximate any given value. The rule can be seen on the third slide.
The second thing we talked about in class is Newton's Method. We started our discussion about this method by going on a side story about Newton and Leibniz feud. After that, we went on describe Newton's method. His method is used for approximating zeros or roots of differentiable functions. Well, the big question is, how does it work? First, it starts with an initial guess, which is reasonably close to the true root, then the function is approximated by using the tangent line , and one computes the x-intercept of this tangent line . This x-intercept will typically be a better approximation to the function's root than the original guess. This process can be done over and over again until the root is found, this process is called iteration. In general, we can apply the formula that can be seen on the fifth slide to find a better approximation.
The third thing we talked about are the scenarios where Newton's method may fail. His method may fail if it produces a derivative of 0 or undefined. Also, it may fail if the steps just continues with the same set of values. Lastly, it wont work if the results diverge away from the root that you are interested in.
If you are feeling a little confused by this new concept, you can check this animation out.
That's it for today's scribe. Our home work for tonight is Exercise 4.7 (all the odd numbered questions). The next scribe will be Van.
Saturday, November 17, 2007
We're all related...implicitly!
Hello everybody! It's MrSiwWy here for Friday's scribe, though it might not quite as elegant as I previously planned to make this post, I think this format would be more applicable. I found that typing out the instructions alongside the solutions rather than incorporating each solution and explanation into a slide format would be easier to modify and edit in case any corrections are required. Also, in class Van was worried about copying everything on the boards down for his scribe post, though I thought I was going to be scribe (which I am) so I gave him my class notes to use for the blog. So most of what will be posted here will be from memory (though it doesn't really matter much) and the solutions for each of the example questions I had to finish just about 5 minutes ago. Well here I go, and I already can foresee this scribe to be a very long one (sorry everyone!), I tried to reduce the length by using a much smaller font, but nevertheless I hope that the amount of detail that I go into for each solution is sufficient for anyone to benefit from.
As class began, each student entered the class with the startling revelation that Mr. K was not in class (though he had already told us that he would be busy this class and might not be able to make it) but instead there was a delightful substitute -- Mrs. Adams. Also, we all slowly realized that the smart board was indeed not in full operation and we could not utilize it for Friday's class. With those two facts in mind, class continued and Mrs. Adams walked to the front of the class and told us what was to take place in class.
First, she presented herself to us, and noted that she had taught every level of mathematics in high school and some calculus as well (impressive!), she then told us that our homework was to complete exercise 4.6 in our textbooks (pages 235-237), all odd questions including question 22. I'll be reiterating this at the end of this scribe post to remind everyone of the homework. Anyway, she then showed us a booklet that we she was about to be handing out to the class, and read out an excerpt from the steps for solving related rates problems. The steps as outlined in the handout are as follows:
Steps for Solving Related Rates Problems
"1. Make a drawing of the situation if possible
2. Use letters to represent the variables involved in the situation e.g. x, y, etc.
3. Identify all rates of change given and those to be determined. Use calculus notation dx/dt, dy/dt, etc, to represent them.
4. Determine an equation that both
a. Involves the variables of step 2 and
b. Will involve the derivative of step 3, when differentiated
(You may need some geometrical formulas to do this)
5. Differentiate (by implicit differentiation) the equation found in step 4.
6. Substitute all known values into the differentiated equation
7. Use algebraic manipulation, if necessary, to solve for the desired unknown rate or quantity."
***At this point, I would also like to include the steps for solving these problems according to the textbook:
"1. State an equation, valid at any time, which relates the quantities that are changing.
2. Differentiate with respect to time to obtain a relationship between the rates of change (derivatives) of the quantities.
3. Substitute the specific values that the variables take on at the instant in question."
Using either of the similar algorithms for solving related rates problems are equally sufficient, though the latter might seem a bit more complicated. If these methods are followed carefully, not only might these problems become seemingly more simplistic but it could be easier to identify where mistakes have been made and one can learn from such mistakes in future endeavors to solve these problems.
Our class then split into three groups, since there are six questions in total, and each group was assigned to complete certain example questions from the handout and to post our solutions up on the whiteboards (it was subtly odd to travel back to the old "whiteboard" days!). Now here comes the long part of this scribe post, my detailed solutions for each of the example questions.
Example 1:
A spherical snowball melts in such a way that the instant at which its radius is 20 cm, its radius is decreasing at 3 cm/min. At what rate is the volume of the ball of snow changing at that instant?
Now this question doesn't actually seem too difficult. Though the main tribulation most students might face when they encounter a question similar to this is the fact that they must identify the equation relating the given rates, which might seem like too far back in our education to reach. So it's good to imagine the object's dimensions, and use geometric formula manipulation as much as necessary to determine a direct relationship between the variables. In this case, given directly by the diagram, it's obvious to see that the snowball is spherical in shape and that it is shrinking as it is melting.
You can see that I indicated what is known and what is not known, also, by drawing a diagram it's obviously easier to visualize and perceive exactly what is happening in the described scenario. Most importantly, I have identified an equation that relates the two given variables / rates, which are in this case V and r. Now, the next step to follow in this question would be to differentiate the above formula (step 5 in the first recipe, and step 2 in the second one) relating the two variables, in order to have the rates of each variable within the equation. This work appears as follows: 
It is quite obvious that I have designated that the hypotenuse of the triangle as the length of the ladder, y as the height the tip of the ladder against the wall relative to the ground, and x as distance the base of the ladder is from the wall. Now, as is visible by the diagram, the approach for this question to connect each of the variable is the Pythagorean theorem, stating that x2 + y2 = h2. But before differentiating the equation, it must noted that the value of y is not known, so since it will be utilized after the equation if differentiated (otherwise we would have two variables: the rate of y and also the value of y) it's value must be determined. Plugging in the x and h values into the above equation, y can be easily calculated.
Now that the height of the ladder against the wall is known, the equation can now be differentiated and solved for the rate of its descent against the wall.
"Given the solution above, it becomes apparent that the ladder is falling against the wall at a rate of 4 ft/sec."
b) For this portion of the question, I'll show not only the diagram and the variables and their corresponding values, but also the calculation for θ using trigonometry. I can do so at the same time since the situation is already familiar and the setup for the ladder against the wall has already been identified to create a right angled triangle. Also, I assigned the value for y as 20 ft, this being that since in part a, when x = 20 ft y was equal to 15 ft, now that x is equal to 15 ft y must be 20.
Now, to complete this question, an equation relating the angle to any other variable in the situation that is known must be created. One formula that can be used for this is the simple trigonometry equivalence for the sine function. Once this familiar equation is differentiated, the values can be plugged in to solve for the unknown, which is the rate of change for the angle formed between the wall and the ladder.
"Given the solution above, it becomes apparent that the angle formed between the wall and the ladder changes at a rate of 3/20 radians/sec."
Example 3:
A kite is flying 150 m high, where the wind causes it to move horizontally at the rate of 5 m per second. In order to maintain the kite at a height of 150 m, the person must allow more string to be let out. At what rate is the string being let out when the length of the string already out is 250m?
The diagram for this question is extremely important in understanding the mechanics of the situation, what variables are involved and how these variables are related altogether.
As visible in this diagram, the horizontal component of the kite from the individual holding the string to the kite is represented by x, the constant vertical distance (thus the 0 rate of change for the vertical distance) is represented by y and the string from the ground attached to the kite in the air is given by h. The horizontal distance of the kite from the controller of the kite when there is 250 m of string given out can be found by applying the Pythagorean theorem.
Now that we have the value for x, we can further use the Pythagorean theorem and differentiate it to then determine the rate that string is being let out as the kite fly's. We can do this since we know that this equation relates all three variables along with their rates, and also that the values for x, y, h and the rates for both x and y are known, meaning only the rate of h will remain unknown after we differentiate allowing us to solve for it.
"From this solution we can now deduce that the rate in which string is being let out when there is already 250 m of string out is 4 m /sec."
Example 4:
(a) Water runs into an inverted conical tank at the rate of 7 cubic feet per minute. The radius of the water's surface is always half the height of the water. How fast is the water level rising when the water is 2 feet deep?
(b) Suppose water is leaking out of the tank at a rate of 2 feet/sec in (a). How fast is the radius of surface of the water increasing when the water is 2 feet deep?
(a) This question comes off as very complicated, and in many ways it can be. It requires quite a bit of thought at first in order to construct a correct diagram to represent the situation. Since water is running into the tank shaped like a cone at a given rate, this must mean that this rate is actually the rate of change of the volume of water within the tank. Now, since the entire tank is conical in shape, and any liquid can only occupy a space within a container by taking the shape of that container, meaning the water within the tank will be conical in shape regardless of the amount. Also, since we know that both the tank and the amount of water in the tank are conical, we can now construct an appropriate diagram and also an equation relating the volume within the cone to the height of the water. I must note that since the question inquires about the rate of the increase in height of the water, and since the rate of volume increase is known, then it is possible to create a direct relationship between volume and the height of the water.
All that is left for this question is to differentiate the given equation and to determine the rate at which the water increases. The work for this is as follows: 
After differentiating the equation and substituting the height of the water and the rate of the volume's increase into the fresh derivative, the rate of the water's height increasing was determined. And as with any long answer question, it must end with a full sentence.
"The rate at which the water level is rising in the tank when the water is 2 feet deep is 7/4π ft/min."
(b) Now, the way that I perceived this question is that since water in being distributed to the tank at a rate of 7 ft3/min, but there is water leaking at a rate of -2ft3/sec, that must mean that the volume of the water within the tank is decreasing at a rate of -113 ft3/min (this is achieved by adding the rate of water entering to the rate of water leaving the tank, but warning take note of the units) Also, as I mentioned before, the question in part (a) was inquiring about the rate of increase of the height, but in this case it asks for the rate of increase for the radius. This means that we must search for a way to directly relate volume and radius, which can be done similarly as to how the volume was shown to relate directly to the height of the water by substituting in it's relationship with r, the same can be done here for the radius' relationship with height. This is detailed below, once again beside the accompanying diagram.
Now all that is left to this question is to determine what the radius of the water within the tank is when the water is 2 feet deep, meaning that using the fact that the radius is always half the height of the water, the radius is 1 foot. Using this information, it is now possible to solve for the rate of the radius' progression as water is leaking out at aforementioned rate of -113 ft3/min. 
"When water is leaking out of the tank at 2 ft3/sec, the radius of the water's surface will be decreasing at a rate of 113/8π ft/min"
Example 5:
An airplane, flying at 450 km/hr at a constant altitude of 5 km, is approaching a camera mounted on the ground. Let θ be the angle of elevation above the ground at which the camera is pointed. When θ = π/3, how fast does the camera have to rotate in order to keep the plane in view?
This question involves quite a bit of variables, since there is a vertical distance, a horizontal distance between the plane and the camera, a distance directly between the plane and the camera, and also an angle θ, all including corresponding rates. In order to formulate a connection between any of these variables however, it would be best to construct a diagram showing the travel of the plane and it's relation to the camera on the ground. Also, accompanying the diagram are calculations for both h and x, though they can be approached in different ways, I've put the method I used to calculate each (using trigonometry).
Since it states that the lighthouse is 3 km away from the point P on the shoreline, I took this to be the x component of a right angled triangle, where h the hypotenuse represents the beam of light from the lighthouse. Now since this beam of light is 1 km away from the point P, then this can be taken as the y component as depicted above. Now what the question is asking for is for the rate of change for the y component, since the y component indicates where the beam of light meets the shoreline, or in other words, the y component is indicates how the beam of light is moving along the shoreline. One trouble that arises when reading the question itself is the fact that the beam of light makes 4 revolutions per minute. What exactly does this mean?
This 4 revolutions per minute means that the beam of light makes four complete circles in one minute, meaning that the light travels 8π radians every minute. Therefore, the rate of growth for the angle θ is increasing at 8π radians per minute. Now, we also know that the distance the lighthouse is away from point P will always be 3 km, meaning that the rate of change in the x component will be 0. Also, the angle in question can be determined through the use of simple trigonometry, which would ideally be the tangent function in this case. Now, it would now be best to assemble these data accordingly and then determine a relationship that can be used to determine the rate of change for the y component of the triangle.
As class began, each student entered the class with the startling revelation that Mr. K was not in class (though he had already told us that he would be busy this class and might not be able to make it) but instead there was a delightful substitute -- Mrs. Adams. Also, we all slowly realized that the smart board was indeed not in full operation and we could not utilize it for Friday's class. With those two facts in mind, class continued and Mrs. Adams walked to the front of the class and told us what was to take place in class.
First, she presented herself to us, and noted that she had taught every level of mathematics in high school and some calculus as well (impressive!), she then told us that our homework was to complete exercise 4.6 in our textbooks (pages 235-237), all odd questions including question 22. I'll be reiterating this at the end of this scribe post to remind everyone of the homework. Anyway, she then showed us a booklet that we she was about to be handing out to the class, and read out an excerpt from the steps for solving related rates problems. The steps as outlined in the handout are as follows:
Steps for Solving Related Rates Problems
"1. Make a drawing of the situation if possible
2. Use letters to represent the variables involved in the situation e.g. x, y, etc.
3. Identify all rates of change given and those to be determined. Use calculus notation dx/dt, dy/dt, etc, to represent them.
4. Determine an equation that both
a. Involves the variables of step 2 and
b. Will involve the derivative of step 3, when differentiated
(You may need some geometrical formulas to do this)
5. Differentiate (by implicit differentiation) the equation found in step 4.
6. Substitute all known values into the differentiated equation
7. Use algebraic manipulation, if necessary, to solve for the desired unknown rate or quantity."
***At this point, I would also like to include the steps for solving these problems according to the textbook:
"1. State an equation, valid at any time, which relates the quantities that are changing.
2. Differentiate with respect to time to obtain a relationship between the rates of change (derivatives) of the quantities.
3. Substitute the specific values that the variables take on at the instant in question."
Using either of the similar algorithms for solving related rates problems are equally sufficient, though the latter might seem a bit more complicated. If these methods are followed carefully, not only might these problems become seemingly more simplistic but it could be easier to identify where mistakes have been made and one can learn from such mistakes in future endeavors to solve these problems.
Our class then split into three groups, since there are six questions in total, and each group was assigned to complete certain example questions from the handout and to post our solutions up on the whiteboards (it was subtly odd to travel back to the old "whiteboard" days!). Now here comes the long part of this scribe post, my detailed solutions for each of the example questions.
Example 1:
A spherical snowball melts in such a way that the instant at which its radius is 20 cm, its radius is decreasing at 3 cm/min. At what rate is the volume of the ball of snow changing at that instant?
Now this question doesn't actually seem too difficult. Though the main tribulation most students might face when they encounter a question similar to this is the fact that they must identify the equation relating the given rates, which might seem like too far back in our education to reach. So it's good to imagine the object's dimensions, and use geometric formula manipulation as much as necessary to determine a direct relationship between the variables. In this case, given directly by the diagram, it's obvious to see that the snowball is spherical in shape and that it is shrinking as it is melting.
You can see that I indicated what is known and what is not known, also, by drawing a diagram it's obviously easier to visualize and perceive exactly what is happening in the described scenario. Most importantly, I have identified an equation that relates the two given variables / rates, which are in this case V and r. Now, the next step to follow in this question would be to differentiate the above formula (step 5 in the first recipe, and step 2 in the second one) relating the two variables, in order to have the rates of each variable within the equation. This work appears as follows: Now, that is the answer to our question. Don't forget to complete each solution with a full sentence stating the value of the variable that was calculated in the context of the question. In this case, and example final statement for this solution could be:
"The volume of the snowball is decreasing at a rate of 4800π cm/min."
Example 2:
A ladder 25 feet long is leaning against the wall of the house, while the base of the ladder is pulled away from the wall at a rate of 3 feet per second.
a) How fast is the top moving down the wall when the base of the ladder is 20 feet away from the wall?
b) How fast is the angle between the ladder and the wall changing when the base of the ladder is 15 feet away from the wall?
a) Now for this question, it would be optimal to first identify what variables are present in the question and especially vital to construct a diagram to represent the situation given. The diagram and variables are detailed below.
It is quite obvious that I have designated that the hypotenuse of the triangle as the length of the ladder, y as the height the tip of the ladder against the wall relative to the ground, and x as distance the base of the ladder is from the wall. Now, as is visible by the diagram, the approach for this question to connect each of the variable is the Pythagorean theorem, stating that x2 + y2 = h2. But before differentiating the equation, it must noted that the value of y is not known, so since it will be utilized after the equation if differentiated (otherwise we would have two variables: the rate of y and also the value of y) it's value must be determined. Plugging in the x and h values into the above equation, y can be easily calculated.Now that the height of the ladder against the wall is known, the equation can now be differentiated and solved for the rate of its descent against the wall."Given the solution above, it becomes apparent that the ladder is falling against the wall at a rate of 4 ft/sec."b) For this portion of the question, I'll show not only the diagram and the variables and their corresponding values, but also the calculation for θ using trigonometry. I can do so at the same time since the situation is already familiar and the setup for the ladder against the wall has already been identified to create a right angled triangle. Also, I assigned the value for y as 20 ft, this being that since in part a, when x = 20 ft y was equal to 15 ft, now that x is equal to 15 ft y must be 20.
Now, to complete this question, an equation relating the angle to any other variable in the situation that is known must be created. One formula that can be used for this is the simple trigonometry equivalence for the sine function. Once this familiar equation is differentiated, the values can be plugged in to solve for the unknown, which is the rate of change for the angle formed between the wall and the ladder."Given the solution above, it becomes apparent that the angle formed between the wall and the ladder changes at a rate of 3/20 radians/sec."Example 3:
A kite is flying 150 m high, where the wind causes it to move horizontally at the rate of 5 m per second. In order to maintain the kite at a height of 150 m, the person must allow more string to be let out. At what rate is the string being let out when the length of the string already out is 250m?
The diagram for this question is extremely important in understanding the mechanics of the situation, what variables are involved and how these variables are related altogether.
As visible in this diagram, the horizontal component of the kite from the individual holding the string to the kite is represented by x, the constant vertical distance (thus the 0 rate of change for the vertical distance) is represented by y and the string from the ground attached to the kite in the air is given by h. The horizontal distance of the kite from the controller of the kite when there is 250 m of string given out can be found by applying the Pythagorean theorem. Now that we have the value for x, we can further use the Pythagorean theorem and differentiate it to then determine the rate that string is being let out as the kite fly's. We can do this since we know that this equation relates all three variables along with their rates, and also that the values for x, y, h and the rates for both x and y are known, meaning only the rate of h will remain unknown after we differentiate allowing us to solve for it."From this solution we can now deduce that the rate in which string is being let out when there is already 250 m of string out is 4 m /sec."Example 4:
(a) Water runs into an inverted conical tank at the rate of 7 cubic feet per minute. The radius of the water's surface is always half the height of the water. How fast is the water level rising when the water is 2 feet deep?
(b) Suppose water is leaking out of the tank at a rate of 2 feet/sec in (a). How fast is the radius of surface of the water increasing when the water is 2 feet deep?
(a) This question comes off as very complicated, and in many ways it can be. It requires quite a bit of thought at first in order to construct a correct diagram to represent the situation. Since water is running into the tank shaped like a cone at a given rate, this must mean that this rate is actually the rate of change of the volume of water within the tank. Now, since the entire tank is conical in shape, and any liquid can only occupy a space within a container by taking the shape of that container, meaning the water within the tank will be conical in shape regardless of the amount. Also, since we know that both the tank and the amount of water in the tank are conical, we can now construct an appropriate diagram and also an equation relating the volume within the cone to the height of the water. I must note that since the question inquires about the rate of the increase in height of the water, and since the rate of volume increase is known, then it is possible to create a direct relationship between volume and the height of the water.
All that is left for this question is to differentiate the given equation and to determine the rate at which the water increases. The work for this is as follows: After differentiating the equation and substituting the height of the water and the rate of the volume's increase into the fresh derivative, the rate of the water's height increasing was determined. And as with any long answer question, it must end with a full sentence."The rate at which the water level is rising in the tank when the water is 2 feet deep is 7/4π ft/min."
(b) Now, the way that I perceived this question is that since water in being distributed to the tank at a rate of 7 ft3/min, but there is water leaking at a rate of -2ft3/sec, that must mean that the volume of the water within the tank is decreasing at a rate of -113 ft3/min (this is achieved by adding the rate of water entering to the rate of water leaving the tank, but warning take note of the units) Also, as I mentioned before, the question in part (a) was inquiring about the rate of increase of the height, but in this case it asks for the rate of increase for the radius. This means that we must search for a way to directly relate volume and radius, which can be done similarly as to how the volume was shown to relate directly to the height of the water by substituting in it's relationship with r, the same can be done here for the radius' relationship with height. This is detailed below, once again beside the accompanying diagram.
Now all that is left to this question is to determine what the radius of the water within the tank is when the water is 2 feet deep, meaning that using the fact that the radius is always half the height of the water, the radius is 1 foot. Using this information, it is now possible to solve for the rate of the radius' progression as water is leaking out at aforementioned rate of -113 ft3/min. "When water is leaking out of the tank at 2 ft3/sec, the radius of the water's surface will be decreasing at a rate of 113/8π ft/min" Example 5:
An airplane, flying at 450 km/hr at a constant altitude of 5 km, is approaching a camera mounted on the ground. Let θ be the angle of elevation above the ground at which the camera is pointed. When θ = π/3, how fast does the camera have to rotate in order to keep the plane in view?
This question involves quite a bit of variables, since there is a vertical distance, a horizontal distance between the plane and the camera, a distance directly between the plane and the camera, and also an angle θ, all including corresponding rates. In order to formulate a connection between any of these variables however, it would be best to construct a diagram showing the travel of the plane and it's relation to the camera on the ground. Also, accompanying the diagram are calculations for both h and x, though they can be approached in different ways, I've put the method I used to calculate each (using trigonometry).
*Note that the rate for the altitude of the plane is not changing, this is because in the context of the question, the plane is flying at a constant 5 km meaning that it does not change at any time during it's flight. Also, the rate of change for the distance between the camera and the plane is negative, this is because the plane is moving closer to the camera, meaning that the distance between the two objects is in fact decreasing.
Now notice how I actually have two theta locations. That's because, due to our knowledge of transversal's way back in grade 7, opposite interior angles are equal, so we can in this way use the values for the lower triangle for the upper triangle (the plane's path) to determine the angle made between the plane and the camera. Using this knowledge and the above diagram, an equation must be found in order to determine the rate at which θ is increasing, which can easily be deduced by using trigonometry. But we must use tangent to do this, since only x and y are known by both their static and their dynamic values, meaning that we know their rates of changes and their static values simultaneously. Now, since we know that the tangent function will be put to use in order to determine the rate of θ changing, we must differentiate the equation appropriately and fully substitute the values into the equation. Be careful though, because this can get complicated. 
Note that this solution can work equally using either the upper triangle or the lower triangle since they retain the same dimensions. It's good to practice problems like these since not only is practicing mathematics problems beneficial in itself, but adapting to the procedure and the logic required for these problems is equally vital. Now to conclude this question, a full statement must follow.
Note that this solution can work equally using either the upper triangle or the lower triangle since they retain the same dimensions. It's good to practice problems like these since not only is practicing mathematics problems beneficial in itself, but adapting to the procedure and the logic required for these problems is equally vital. Now to conclude this question, a full statement must follow."The angle between the plane and the camera must change at a rate of -135/2 radians per minute in order to stay on the plane"
Example 6:
A lighthouse is on a small island 3 km away from the nearest point P on a straight shoreline and its light makes 4 revolutions per minute. How fast is the beam of light moving along the shoreline when it is 1 km from P?
This is a very complicated question, so I'll try to detail it as vividly as possible. First I'll portray my diagram before I go into an explanation.
Example 6:
A lighthouse is on a small island 3 km away from the nearest point P on a straight shoreline and its light makes 4 revolutions per minute. How fast is the beam of light moving along the shoreline when it is 1 km from P?
This is a very complicated question, so I'll try to detail it as vividly as possible. First I'll portray my diagram before I go into an explanation.
Since it states that the lighthouse is 3 km away from the point P on the shoreline, I took this to be the x component of a right angled triangle, where h the hypotenuse represents the beam of light from the lighthouse. Now since this beam of light is 1 km away from the point P, then this can be taken as the y component as depicted above. Now what the question is asking for is for the rate of change for the y component, since the y component indicates where the beam of light meets the shoreline, or in other words, the y component is indicates how the beam of light is moving along the shoreline. One trouble that arises when reading the question itself is the fact that the beam of light makes 4 revolutions per minute. What exactly does this mean?This 4 revolutions per minute means that the beam of light makes four complete circles in one minute, meaning that the light travels 8π radians every minute. Therefore, the rate of growth for the angle θ is increasing at 8π radians per minute. Now, we also know that the distance the lighthouse is away from point P will always be 3 km, meaning that the rate of change in the x component will be 0. Also, the angle in question can be determined through the use of simple trigonometry, which would ideally be the tangent function in this case. Now, it would now be best to assemble these data accordingly and then determine a relationship that can be used to determine the rate of change for the y component of the triangle.
Since we only know the two rates for the x component and the angle θ, we must find a way to relate the two with and also with y, so that the rate for y can be calculated. Using trigonometry once again, the tangent function can be used to achieve such a relationship. 
"The beam of light emitted from the lighthouse is moving at 8π/3 km/min along the shoreline"
Now that was one extremely long scribe, I'm sorry if this scribe being so long inconvenienced anyone, but the main purpose of this scribe's incredible length is mainly for the amount of detail I wanted to put into the explanations for each example. And as I mentioned earlier in my scribe post, I found a great site to use to review the geometric formulas that we might need for differentiating certain scenario's here.
That's all for my scribe for Friday's class, don't forget that our homework for the weekend is Exercise 4.6 complete all odds plus question number 22 (I think it's 22), well good night everyone have a great weekend. If anyone has any questions, complaints, inquiries, suggestions or corrections just comment or just contact me somehow. Well, I'm expecting a lot of questions and corrections on this post since I bet I made quite a few small mistakes, or maybe a lot of large mistakes in my work for some of the examples.
Oh and also, the next scribe will be MARK
"The beam of light emitted from the lighthouse is moving at 8π/3 km/min along the shoreline"Now that was one extremely long scribe, I'm sorry if this scribe being so long inconvenienced anyone, but the main purpose of this scribe's incredible length is mainly for the amount of detail I wanted to put into the explanations for each example. And as I mentioned earlier in my scribe post, I found a great site to use to review the geometric formulas that we might need for differentiating certain scenario's here.
That's all for my scribe for Friday's class, don't forget that our homework for the weekend is Exercise 4.6 complete all odds plus question number 22 (I think it's 22), well good night everyone have a great weekend. If anyone has any questions, complaints, inquiries, suggestions or corrections just comment or just contact me somehow. Well, I'm expecting a lot of questions and corrections on this post since I bet I made quite a few small mistakes, or maybe a lot of large mistakes in my work for some of the examples.
Oh and also, the next scribe will be MARK
Thursday, November 15, 2007
Scribe #4
Okay! .. So during today's class Mr. K gave us some websites that might help us with our DEV Project. A lot of the class was dedicated to talking about the DEV project and things that we can do to make it better. We then went onto some practice questions. It was a really basic class.
The next scribe will be Mrsiwwy!
The next scribe will be Mrsiwwy!
Wednesday, November 14, 2007
Today's Post OF Wednesday
Umm first i hate hump day so I'm not the happiest, second I was kinda counting on the slides from today so I will not do the best job ever. Sorry guys. Oh and if you want to tutor/mentor then you should go see teach man right away. It sounds cool with good experience and can be used as a reference to your skill in the future. And it might possible be worth a credit (check me on this.)
Hate to say this but I never wrote down the questions (just the work, bad habit), Il try to get them tomorrow if anyone wants.
ps. I don't use computer much for this stuff so I don't know how to get fancy so some stuff is in brackets.
Why surface area = 4(pie)r(squared) of any sphere
If you think of a baseball if you take it apart at the seems, it makes 4 circles.
Better detail down below. Sorry i felt like experimenting.
Anyways that's all I got. Have a good night. Peace :)
Hate to say this but I never wrote down the questions (just the work, bad habit), Il try to get them tomorrow if anyone wants.
ps. I don't use computer much for this stuff so I don't know how to get fancy so some stuff is in brackets.
Why surface area = 4(pie)r(squared) of any sphere
If you think of a baseball if you take it apart at the seems, it makes 4 circles.
Better detail down below. Sorry i felt like experimenting.
Anyways that's all I got. Have a good night. Peace :)
Tuesday, November 13, 2007
SCRIBE ( ^_^)
Hello! I hope everyone is having a great evening! I'm known as Tim-Math-Y and I shall be your scribe for today's classroom activities!
Today Mr. Kuropatwa was away and so, we had a substitute teacher. The instructions that were left for our class were to read through pages 231-234 and work/complete Exercise 4.6 numbers 1-10.
This chapter is known as Related Rates of Change. Here is the intro to this chapter: "Frequently, in trying to solve a physical problem we must establish a connection between rates of change (Derivatives). This connection is usually made by using implicit defferentiation and the Chain Rule.
The questions involve a series of rates where we must find the relationship that connect them. Your solutions may include the use of basic differentiation, product rule, quotient rule, chain rule, and more.
Good luck and Have fun!
Tomorrow's scribe: MrSiwwy? I really don't know if that scribe list is updated or not =), if not, we shall find a solution tomorrow =)
Have a Great night people!
Today Mr. Kuropatwa was away and so, we had a substitute teacher. The instructions that were left for our class were to read through pages 231-234 and work/complete Exercise 4.6 numbers 1-10.
This chapter is known as Related Rates of Change. Here is the intro to this chapter: "Frequently, in trying to solve a physical problem we must establish a connection between rates of change (Derivatives). This connection is usually made by using implicit defferentiation and the Chain Rule.
The questions involve a series of rates where we must find the relationship that connect them. Your solutions may include the use of basic differentiation, product rule, quotient rule, chain rule, and more.
Good luck and Have fun!
Tomorrow's scribe: MrSiwwy? I really don't know if that scribe list is updated or not =), if not, we shall find a solution tomorrow =)
Have a Great night people!
Scribe Episode 4: The Return of the Jedi
Oh My Goodness!!! It is quite late, but the Scribe is finally up, I believe most of you will be reading this tomorrow, but that's okay, it will be a great review!!! And BTW, how late is it if it was posted before the slides were??? Haha, anyway, here it is... Please tell me if there's anything wrong =) (Oh DARN I already found one spelling mistake, please ignored the "shoed" in the second slide -_-
Sunday, November 11, 2007
Scribe 4 - Differentiation Rules Quiz 1
Hey everyone! Once again I am scribe and this is what happened on Friday’s class.
We started off the class today by watching a post on our blog that was posted by Lani. The question is: Are there lessons to be learned from geese? (Sticking together as a “Flock” benefits us all) If you take a look at it you can learn all sorts of things about geese.
After talking about that for a little bit, we moved on to a quiz. The quiz was written in pre-test formation.
Quiz 1 on Differentiation Rules
1. a) To find the mortgage interest rate on January 1, 1985, you basically just plug 5 in for t using the first function given to you. You are given a specific time range so that is how you know to use the rate as a function of t formula.
1. b) Similarly, you are finding how many issued per year so you plug in the rate.
1. c) This question deals with composite functions. It is a bit tricky but all you do is find N'(r(t))*r'(t) after plugging the correct numbers into that formula.
2. The answer is (D) 4.3 because all you are finding is the area under the function between the interval 0-3.
We started off the class today by watching a post on our blog that was posted by Lani. The question is: Are there lessons to be learned from geese? (Sticking together as a “Flock” benefits us all) If you take a look at it you can learn all sorts of things about geese.
After talking about that for a little bit, we moved on to a quiz. The quiz was written in pre-test formation.
Quiz 1 on Differentiation Rules
1. a) To find the mortgage interest rate on January 1, 1985, you basically just plug 5 in for t using the first function given to you. You are given a specific time range so that is how you know to use the rate as a function of t formula.
1. b) Similarly, you are finding how many issued per year so you plug in the rate.
1. c) This question deals with composite functions. It is a bit tricky but all you do is find N'(r(t))*r'(t) after plugging the correct numbers into that formula.
2. The answer is (D) 4.3 because all you are finding is the area under the function between the interval 0-3.
Friday, November 9, 2007
Thursday, November 8, 2007
ebircS-Scribe
Hello everyone and welcome to my scribe! Enjoy (the angled words and other formatting came out a bit funky, but it's still easily legible except for maybe that second slide)
The next lucky recipient of the scribe will be Robert since he'll be back tomorrow, I think, if he isn't it's Craig, if Craig somehow doesn't get to class, it's Van, if not Van then it will be Sandy. Think I've covered the bases there.
G'night
The next lucky recipient of the scribe will be Robert since he'll be back tomorrow, I think, if he isn't it's Craig, if Craig somehow doesn't get to class, it's Van, if not Van then it will be Sandy. Think I've covered the bases there.
G'night
Wednesday, November 7, 2007
Scribe number four.
Hello, hello again.
I am filling in for Robert today because he does not have enough time to do one but don't worry, he can scribe at a later date! As you all may know or not know we have started a new cycle, which means anyone can be picked for scribe! *applause and cheering* So today's class started off with everyone writing their signatures on the smart board. However, to everyone's surprise it got eraseD!
Mr. K. started off by telling us that we'd be learning about derivatives of trigonometric functions and we also took a look at the first slide which can be found in the previous post. The picture on the slide was taken by a former student of Mr. K's for a flickR assignment.
Shortly after that we had all received our marks. We received two marks; raw and transformed. The raw mark is how we did in the term and the transformed mark is based on a curve.
Basically, Mr. K. took the average of our precal40S marks, which was 77.5 and based it on that. The reason why we have the transformation is because Mr. K. said it is to ensure you do not get penalized for taking a university course in high school. For a more thorough explanation of all this please see Mr. Siwwy's scribe below or by simply viewing the labels under his name or scribe posts. We also had all of our quizzes, pretests and tests back. We had a short discussion on a few of the answers because we were unsure of them. Also, see slide two to view how the curve looks like; blue = pc40s curve.
After that we finally got to our lesson. First we had a mini review and solved a problems on the board which can be found on slide three.
Then we put the equation sin(x) into our calculators and we found the derivative of that, which can be found on slide four. 
We found it to be cos(x). We also dealt with cos(x) and found the derivative of that to be -sin(x) . However, Mr. K. said that this does not tell us anything because it is not a proof. We took a look at the derivative rules and the Squeeze theorem-if two different functions are approaching the same point and there is a function sandwiched in between the two functions then it must also be approaching that point. [If I have that wrong, please inform me.] Next, we visited Visual Calculus, the website to view the proof for the derivative rules of a trigonometric function. To see the step by step proof on the site you may click this link or view the slides and click the little circle for the link on slide five.
After that we found the derivatives of secant(x) and cosecant(x). We found those by the proofs that have just been proved and the quotient rule [see Monday's scribe or slides for a review]. Please view slide six.
Slide seven explains how a reciprocal function's derivative can be found by using the quotient rule. 
Slide eight contains the rest of the trig. function's derivative. However, they are not complete.
They are for homework as well as the rest of the trig. questions we left out on the previous exercise [4.3].
That was basically our class. Hopefully you found this scribe post somewhat interesting and useful. If I have failed to mention something or if I misinterpreted anything, please let me know. Thanks! =)
Since Robert will be away tomorrow and he will not be able to scribe I am going to pick Graeme and I guess since Robert was supposed to scribe today he can scribe for Friday?.
I am filling in for Robert today because he does not have enough time to do one but don't worry, he can scribe at a later date! As you all may know or not know we have started a new cycle, which means anyone can be picked for scribe! *applause and cheering* So today's class started off with everyone writing their signatures on the smart board. However, to everyone's surprise it got eraseD!
Mr. K. started off by telling us that we'd be learning about derivatives of trigonometric functions and we also took a look at the first slide which can be found in the previous post. The picture on the slide was taken by a former student of Mr. K's for a flickR assignment.
Shortly after that we had all received our marks. We received two marks; raw and transformed. The raw mark is how we did in the term and the transformed mark is based on a curve.
Basically, Mr. K. took the average of our precal40S marks, which was 77.5 and based it on that. The reason why we have the transformation is because Mr. K. said it is to ensure you do not get penalized for taking a university course in high school. For a more thorough explanation of all this please see Mr. Siwwy's scribe below or by simply viewing the labels under his name or scribe posts. We also had all of our quizzes, pretests and tests back. We had a short discussion on a few of the answers because we were unsure of them. Also, see slide two to view how the curve looks like; blue = pc40s curve.After that we finally got to our lesson. First we had a mini review and solved a problems on the board which can be found on slide three.
Then we put the equation sin(x) into our calculators and we found the derivative of that, which can be found on slide four. 
We found it to be cos(x). We also dealt with cos(x) and found the derivative of that to be -sin(x) . However, Mr. K. said that this does not tell us anything because it is not a proof. We took a look at the derivative rules and the Squeeze theorem-if two different functions are approaching the same point and there is a function sandwiched in between the two functions then it must also be approaching that point. [If I have that wrong, please inform me.] Next, we visited Visual Calculus, the website to view the proof for the derivative rules of a trigonometric function. To see the step by step proof on the site you may click this link or view the slides and click the little circle for the link on slide five.After that we found the derivatives of secant(x) and cosecant(x). We found those by the proofs that have just been proved and the quotient rule [see Monday's scribe or slides for a review]. Please view slide six.
Slide seven explains how a reciprocal function's derivative can be found by using the quotient rule. 
Slide eight contains the rest of the trig. function's derivative. However, they are not complete.
They are for homework as well as the rest of the trig. questions we left out on the previous exercise [4.3].That was basically our class. Hopefully you found this scribe post somewhat interesting and useful. If I have failed to mention something or if I misinterpreted anything, please let me know. Thanks! =)
Since Robert will be away tomorrow and he will not be able to scribe I am going to pick Graeme and I guess since Robert was supposed to scribe today he can scribe for Friday?.
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