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Monday, December 15, 2008
Sunday, December 14, 2008
The Mean Value of BOBs
Once again, BOBing... I'm already up, so what's the point, I might as well get this one in while I'm taking a break from English and my lab write up. Looks to be another all-nighter...
Anyways, I really didn't have fun with the Optimization questions. I will have to do more questions to start seeing a pattern into doing those questions but as of right now, my mind is muddier then mud when it comes to Optimization. The key is getting your information straight and getting what your looking for straight. Other wise, your whole answer is going to be wrong.
The mean value theorem was simple enough and I enjoyed learning it, however, I need a bit more practice as I came to a few stumps when solving some of the questions and ended up not getting the right answer.
I'll be so-so for this test, but I don't feel too comfortable yet, but that's life.
~Rence, Out
Anyways, I really didn't have fun with the Optimization questions. I will have to do more questions to start seeing a pattern into doing those questions but as of right now, my mind is muddier then mud when it comes to Optimization. The key is getting your information straight and getting what your looking for straight. Other wise, your whole answer is going to be wrong.
The mean value theorem was simple enough and I enjoyed learning it, however, I need a bit more practice as I came to a few stumps when solving some of the questions and ended up not getting the right answer.
I'll be so-so for this test, but I don't feel too comfortable yet, but that's life.
~Rence, Out
The Gauntlet Is Thrown Down
The AP Calculus Commercial Challenge has been launched. What will you do with your 30 seconds?
Calculus Commercial Seed Video
Remember, your videos have to be posted to YouTube as replies to this one.
Calculus Commercial Seed Video
Remember, your videos have to be posted to YouTube as replies to this one.
BOB Version 5: Applications of the Derivative
My muddiest point, I'm sure everyone else is too, is learning about the optimization problems. I'm sure that the more practice I get in doing these problems, the better I'll be in solving those problems. "Practice makes perfect."
Limits involving infinity wasn't that hard as I first thought about it, since I missed the lesson on that because I was at a business field trip. Understanding that 1/x, as x approaches infinity, equals zero is the thing that underlies that topic.
Understanding the Mean Value Theorem came easy to me, until Rolley came along with his theorem and took things to the next level. I don't think Rolley will show up on the test since he isn't mentioned in our textbook.
An important lesson that I've learned from Mr. K is that if we learned how to do something, we must also learn how to undo that something to fully understand our actions or something like that. This was his introduction to antiderivatives, and I liked it. But we haven't yet finished learning about Auntie Derivatives and her family of functions (I like that title) but I'm sure it won't be as hard as optimization problems.
Good luck to those who are writing the test and pre-tests!
Limits involving infinity wasn't that hard as I first thought about it, since I missed the lesson on that because I was at a business field trip. Understanding that 1/x, as x approaches infinity, equals zero is the thing that underlies that topic.
Understanding the Mean Value Theorem came easy to me, until Rolley came along with his theorem and took things to the next level. I don't think Rolley will show up on the test since he isn't mentioned in our textbook.
An important lesson that I've learned from Mr. K is that if we learned how to do something, we must also learn how to undo that something to fully understand our actions or something like that. This was his introduction to antiderivatives, and I liked it. But we haven't yet finished learning about Auntie Derivatives and her family of functions (I like that title) but I'm sure it won't be as hard as optimization problems.
Good luck to those who are writing the test and pre-tests!
BOB for Applications of Derivatives
Yay!!! I'm the first BOBber. Anyways, this unit was quite interesting. Quite a bit of the stuff we did in this unit was already taught to us by Ms Karras about a month or two ago so it wasn't that tough.
My clear-as-crystal parts were with the First and Second Derivative Tests to find critical points and inflection points. I already knew about this back at Chapter 2 when I was experimenting with the derivatives.
The more muddier parts came when we had to apply those tests to the optimization questions. It took me a while to get well at it. I started understanding these questions when I had a moment of clarity. I did not have one in the shower, but I did have one when I was working on the Optimization homework. I started see how each formula (area, volume, etc) were related to each other and can be found quite easily through some algebraic massaging and gymnastics. Once I saw what was going on I began solving the questions at a much faster pace and with less and less mistakes. The important thing with word problems is the time it takes to solve the questions. Sure, some might be long but we have to see in which part of the solution can we change or remove to decrease the amount of time it takes in getting to the correct answer. This sounds like a possible optimization question for DEV. I CALL DIBS ON IT FOR MY GROUP. NO ONE TAKE THE IDEA!!!
I noticed that the big thing with the questions in this unit was the way that was required to answer the questions. We had to specifically state which test(s) were used in obtaining the answer. The way that we used the variables were quite important as well.
The Mean Value Theorem was quite simple as well so I didn't have much worries with that area.
I noticed something when we were working with the Antiderivatives the other day. It isn't in the chapter so does that mean we won't have to worry about it for the test?
Good luck everyone and brace yourselves for Mr. K's homework check. I got to get back to finishing my Chem Lab Write-Up.
Bye bye =D
My clear-as-crystal parts were with the First and Second Derivative Tests to find critical points and inflection points. I already knew about this back at Chapter 2 when I was experimenting with the derivatives.
The more muddier parts came when we had to apply those tests to the optimization questions. It took me a while to get well at it. I started understanding these questions when I had a moment of clarity. I did not have one in the shower, but I did have one when I was working on the Optimization homework. I started see how each formula (area, volume, etc) were related to each other and can be found quite easily through some algebraic massaging and gymnastics. Once I saw what was going on I began solving the questions at a much faster pace and with less and less mistakes. The important thing with word problems is the time it takes to solve the questions. Sure, some might be long but we have to see in which part of the solution can we change or remove to decrease the amount of time it takes in getting to the correct answer. This sounds like a possible optimization question for DEV. I CALL DIBS ON IT FOR MY GROUP. NO ONE TAKE THE IDEA!!!
I noticed that the big thing with the questions in this unit was the way that was required to answer the questions. We had to specifically state which test(s) were used in obtaining the answer. The way that we used the variables were quite important as well.
The Mean Value Theorem was quite simple as well so I didn't have much worries with that area.
I noticed something when we were working with the Antiderivatives the other day. It isn't in the chapter so does that mean we won't have to worry about it for the test?
Good luck everyone and brace yourselves for Mr. K's homework check. I got to get back to finishing my Chem Lab Write-Up.
Bye bye =D
Friday, December 12, 2008
Commercial Project
Because I have no other way to reach the other people in my commercial group, I'm posting this. If you're not Shelly or Yi-nan, ignore this!
Yi-nan and Shelly, if you're reading this send me an email at commercialproject@mailinator.com please (is a throwaway email account).
Antiderivatives with Auntie Derivative
Hi everyone and I'll be your scribe for today.
We started today's class with some practice with the Mean Value Theorem. The following picture show the questions and the solution. I'll explain the work afterward, don't worry :)
Okay so for all Mean Value Questions similar to the ones seen in the homework and on the image above, you must check to see if the function is continuous and differentiable in the given interval. In other words, make sure the function doesn't jump values, go to infinity (vertical asymptote), or have any cusps or corners. One example of a function that may not satisfy the Mean Value Theorem would be:
This function is differentiable as explained during class, but one thing may prevent us from using the Mean Value Theorem on it. That one thing is the given interval in the question.
The red function is the graph of f(x)=|x|. The Mean Value Theorem says that the function must be continuous and differentiable within the interval. But as you can see the derivative of the function (blue) is not exist when x=0. So there for if the interval given in the question includes 0 in it, the absolute function does not satisfy the Mean Value Theorem.
Okay back to the questions. So let's look at the domain of the functions and the interval given in the function. The first question involves the natural logarithm function. From Grade 12 Pre-Calculus we know that the argument of any logarithm function cannot be less than or equal to zero because any base, except zero, raised to any exponent cannot equal 0 or negative value so since the argument of any logarithmic function is a power, the power cannot be zero or negative. So let's look at the permissable values of the argument.
As you can see the argument is a quadratic function. Now lets try to imagine the graph of that quadratic. Do you see it? Well let me help you out. Well lets forget about the constant (4) and complete the square with the remaining terms. You get:
Now let's imagine this graph for now. It is a parabola, opening up, but shifted to the left 1 unit. When the function is factored out we see that the quadratic in the argument of the ln has a constant of 4 so then that means that the parabola shifts above the x-axis. We know from Grade 11 Pre-Calculus that if a parabola opens upwards and has a vertex above the x-axis, it has no roots nor can it produce negative outputs. So we don't have to worry about the argument of the logarithm function from the question on the slide being negative.
Now that we've gotten that out of the way, the fun starts. So let's plug and chug the value's into the formula of the Mean Value Theorem and find where on the function does it's derivative match the Secant line connecting the two endpoints of the function in the interval. If you don't remember the formula of the Mean Value Theorem it is:
Once f'(c) is found we then make that equal to the derivative of the original function which can be found on the image but I'll make an image of it anyways (Because I like Sitmo):
So what you can do to solve for the variable is shown on the image with the question on it. You can, one, punch the derivative function into your calculator in Y1 and in Y2 insert your value f'(c) that you calculated earlier and find where the two lines intersect. Or you can bringing over f'(c) to the side with the derivative function as seen in the image with the question on it and graphing that, then find the roots. Now you have the answer.
For the next question, just follow the same process. Find out if the function satisfies the Mean Value Theorem within the given interval, apply the formula of the Mean Value Theorem, and find at what other input of the function will it have an output with the same derivative as the Secant line connecting the two endpoints of the interval. So the big thing to watch out for are the permissable values of the function.
Okay once that was over we began a new part of derivatives, the Antiderivative. Most of the time in our lives, when we learn to do something, we want to learn how to undo it. Let us take multiplication as an example. To undo multiplication you use division. In Mathematics that is called applying the inverse. If you remember from previous classes in Pre-Calculus, the inverse of any function undoes what the original function does. In Derivatives, when you want to undo "differentiating" the function, you "antidifferentiate" it. Let me bring out the image:
On this slide, we were asked to differentiate the functions above. All of their derivatives were the same, they were 2x. Okay cool, now lets go backwards, can you find the parent function when given it's derivative. Sure you can, you just apply the power rule in reverse. But there is one problem, how do you if there is a constant or not. It could be 1, 4, 34873, or pi to the power of 3i. When you differentiate the parent function, the constant disappears because of the constant rule of differentiation. The derivative of any constant is zero. So in general, we just add in c where c is a constant, as seen on the image. Okay lets look at the constant of any function, lets take x2. When we add a constant to it, it causes the parabola to move up and down. All of those values may change the look of the parent function but they are all related in way because the derivatives are the same. So these functions are all related like a family.
We can check to see if our work was right by taking the derivative of the solution we made. If it's derivative was the same as the derivative given in the function, our answer is correct.
We were then asked to find the parent function if the derivative is x. Well the first guess was 1/x but when we applied the power rule to it, we didn't get the same derivative as in the question. Another guess was made and it was correct. (Good Job Jamie). So Mr. K gave began explaining the rule to finding the antiderivative, but the bell rang and class was over.
Okay let's recap. When working with the Mean Value Theorem, make sure there are not conflicts with the permissable values of the arguments of the functions and make sure the function is continuous and differentiable within the given interval.
We started the Antiderivative section of the Derivative Chapter and we'll continue it on Monday.
Oh Yeah, Remember Mr.K may do a sudden homework check when he feels like it. So get caught up with your homework or it's a zero for you. Just thought I'd remind you =D
Remember the test for Chapter 5 is on Wednesday, so that means the Pre-Test will be on Tuesday. We'll be starting the new Integrals unit on Thursday, and continue after the Holidays.
Well the next scribe will be my friend Jamie.
Good night and have a great weekend.
We started today's class with some practice with the Mean Value Theorem. The following picture show the questions and the solution. I'll explain the work afterward, don't worry :)
Okay so for all Mean Value Questions similar to the ones seen in the homework and on the image above, you must check to see if the function is continuous and differentiable in the given interval. In other words, make sure the function doesn't jump values, go to infinity (vertical asymptote), or have any cusps or corners. One example of a function that may not satisfy the Mean Value Theorem would be:
This function is differentiable as explained during class, but one thing may prevent us from using the Mean Value Theorem on it. That one thing is the given interval in the question.
The red function is the graph of f(x)=|x|. The Mean Value Theorem says that the function must be continuous and differentiable within the interval. But as you can see the derivative of the function (blue) is not exist when x=0. So there for if the interval given in the question includes 0 in it, the absolute function does not satisfy the Mean Value Theorem.
Okay back to the questions. So let's look at the domain of the functions and the interval given in the function. The first question involves the natural logarithm function. From Grade 12 Pre-Calculus we know that the argument of any logarithm function cannot be less than or equal to zero because any base, except zero, raised to any exponent cannot equal 0 or negative value so since the argument of any logarithmic function is a power, the power cannot be zero or negative. So let's look at the permissable values of the argument.
As you can see the argument is a quadratic function. Now lets try to imagine the graph of that quadratic. Do you see it? Well let me help you out. Well lets forget about the constant (4) and complete the square with the remaining terms. You get:
Now let's imagine this graph for now. It is a parabola, opening up, but shifted to the left 1 unit. When the function is factored out we see that the quadratic in the argument of the ln has a constant of 4 so then that means that the parabola shifts above the x-axis. We know from Grade 11 Pre-Calculus that if a parabola opens upwards and has a vertex above the x-axis, it has no roots nor can it produce negative outputs. So we don't have to worry about the argument of the logarithm function from the question on the slide being negative.
Now that we've gotten that out of the way, the fun starts. So let's plug and chug the value's into the formula of the Mean Value Theorem and find where on the function does it's derivative match the Secant line connecting the two endpoints of the function in the interval. If you don't remember the formula of the Mean Value Theorem it is:
Once f'(c) is found we then make that equal to the derivative of the original function which can be found on the image but I'll make an image of it anyways (Because I like Sitmo):
So what you can do to solve for the variable is shown on the image with the question on it. You can, one, punch the derivative function into your calculator in Y1 and in Y2 insert your value f'(c) that you calculated earlier and find where the two lines intersect. Or you can bringing over f'(c) to the side with the derivative function as seen in the image with the question on it and graphing that, then find the roots. Now you have the answer.
For the next question, just follow the same process. Find out if the function satisfies the Mean Value Theorem within the given interval, apply the formula of the Mean Value Theorem, and find at what other input of the function will it have an output with the same derivative as the Secant line connecting the two endpoints of the interval. So the big thing to watch out for are the permissable values of the function.
Okay once that was over we began a new part of derivatives, the Antiderivative. Most of the time in our lives, when we learn to do something, we want to learn how to undo it. Let us take multiplication as an example. To undo multiplication you use division. In Mathematics that is called applying the inverse. If you remember from previous classes in Pre-Calculus, the inverse of any function undoes what the original function does. In Derivatives, when you want to undo "differentiating" the function, you "antidifferentiate" it. Let me bring out the image:
On this slide, we were asked to differentiate the functions above. All of their derivatives were the same, they were 2x. Okay cool, now lets go backwards, can you find the parent function when given it's derivative. Sure you can, you just apply the power rule in reverse. But there is one problem, how do you if there is a constant or not. It could be 1, 4, 34873, or pi to the power of 3i. When you differentiate the parent function, the constant disappears because of the constant rule of differentiation. The derivative of any constant is zero. So in general, we just add in c where c is a constant, as seen on the image. Okay lets look at the constant of any function, lets take x2. When we add a constant to it, it causes the parabola to move up and down. All of those values may change the look of the parent function but they are all related in way because the derivatives are the same. So these functions are all related like a family.
We can check to see if our work was right by taking the derivative of the solution we made. If it's derivative was the same as the derivative given in the function, our answer is correct.
We were then asked to find the parent function if the derivative is x. Well the first guess was 1/x but when we applied the power rule to it, we didn't get the same derivative as in the question. Another guess was made and it was correct. (Good Job Jamie). So Mr. K gave began explaining the rule to finding the antiderivative, but the bell rang and class was over.
Okay let's recap. When working with the Mean Value Theorem, make sure there are not conflicts with the permissable values of the arguments of the functions and make sure the function is continuous and differentiable within the given interval.
We started the Antiderivative section of the Derivative Chapter and we'll continue it on Monday.
Oh Yeah, Remember Mr.K may do a sudden homework check when he feels like it. So get caught up with your homework or it's a zero for you. Just thought I'd remind you =D
Remember the test for Chapter 5 is on Wednesday, so that means the Pre-Test will be on Tuesday. We'll be starting the new Integrals unit on Thursday, and continue after the Holidays.
Well the next scribe will be my friend Jamie.
Good night and have a great weekend.
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