Thursday, March 27, 2008
Tuesday, March 25, 2008
Sunday, March 23, 2008
SCRiBE: Last Of Content
Hello everyone! I'm known as Tim-math-y and I'm the scribe for Thursday's lessons. On Thursday, our topic was: Solving Differential Equations Symbolically and Newton's Law of Cooling.
Introduction:
Solving Differential Equations Symbolically and Newton's Law of Cooling:
On Thursday, we learned more about differential equations, this time using the symbolic (algebraic) approach. This topic tied into Newton's Law of Cooling. According to this law, a hot object cools at a rate proportional to the difference between its own temperature and that of its environment.
By applying this concept, we solved questions that dealt with cooling.
Content and Lessons:
First we began by deconstructing the equation of "dy/dx = ky." Mr. K mentioned in the past that normally, you are not allowed to pull apart the differential operator. However, without going through the complexities of why it is possible, we pulled it apart.
By seperating the two variables, including both 'dy' and 'dx', we found a familiar term that could be antidifferentiated.
Through algebraic play, we found the equation that defined the parent function of cooling.
*Note: the variable "c" and "C" are different in value but generally both represent a constant.
Next we looked at an example question.
By using the information given (two pieces of data that showed us the temperature at a specific point in time) we substituted the values in and solved for the final equation that represented the temperature at any given time of this situation.
At this point, we considerred both questions solved.
Next, we looked at another example question (Don't be overwhelmed by the length of the question as most of it is just a background story that Mr. K conjured up: The information required is located in paragraph 3).
By applying the same steps from the previous question, we found the solution to the problem. Again, our variable assignments were the same. The only difference is that the coldest temperature in this situation was 25 degrees. We found the equation that still included the constant value of C.
By taking this a step further, we inputted the coordinates given (temperatures at given times) and solved for the value of C to in turn, solve for the final equation.
Finally, by inputting the temperature of the cola at an unknown point in time, we solved for the uknown value of 't'.
Conlusion:
We went through two problems that dealt with Newton's Law of Cooling, solving differential equations symbolically.
******
Homework is on the end of the slides posted in the previous post! Answer is on the following slide.
Tomorrow's Scribe is... Craig!
Happy Easter? Haha, good night.
Introduction:
Solving Differential Equations Symbolically and Newton's Law of Cooling:
On Thursday, we learned more about differential equations, this time using the symbolic (algebraic) approach. This topic tied into Newton's Law of Cooling. According to this law, a hot object cools at a rate proportional to the difference between its own temperature and that of its environment.
By applying this concept, we solved questions that dealt with cooling.
Content and Lessons:
First we began by deconstructing the equation of "dy/dx = ky." Mr. K mentioned in the past that normally, you are not allowed to pull apart the differential operator. However, without going through the complexities of why it is possible, we pulled it apart.By seperating the two variables, including both 'dy' and 'dx', we found a familiar term that could be antidifferentiated.
Through algebraic play, we found the equation that defined the parent function of cooling.
*Note: the variable "c" and "C" are different in value but generally both represent a constant.
Next we looked at an example question.- First we assigned variables, F = Temperature in Degrees of F and t = T time in Hours
- Second we applied the variables to "dy/dx=ky" to produce "dF/dt=k(F-20)" where 2o is the lowest temperature the object can attain in an environment
- By applying the same steps to seperate 'dF' and 'dt', we antidifferentiated and worked the out an equation that could be used to find 'F', the temperature, at any given time. The only unknown left is the constant that we have yet to solve.
By using the information given (two pieces of data that showed us the temperature at a specific point in time) we substituted the values in and solved for the final equation that represented the temperature at any given time of this situation.At this point, we considerred both questions solved.
Next, we looked at another example question (Don't be overwhelmed by the length of the question as most of it is just a background story that Mr. K conjured up: The information required is located in paragraph 3).
By applying the same steps from the previous question, we found the solution to the problem. Again, our variable assignments were the same. The only difference is that the coldest temperature in this situation was 25 degrees. We found the equation that still included the constant value of C.By taking this a step further, we inputted the coordinates given (temperatures at given times) and solved for the value of C to in turn, solve for the final equation.
Finally, by inputting the temperature of the cola at an unknown point in time, we solved for the uknown value of 't'.
Conlusion:
We went through two problems that dealt with Newton's Law of Cooling, solving differential equations symbolically.
******
Homework is on the end of the slides posted in the previous post! Answer is on the following slide.
Tomorrow's Scribe is... Craig!
Happy Easter? Haha, good night.
Thursday, March 20, 2008
Wednesday, March 19, 2008
The Edmonton Eulers Method
Hey everyone, it's MrSiwWy here as the scribe for March 18's class on Euler's method. I know I started this scribe post pretty late, but I just got home from the fashion show at school, which was definitely an enjoyable occasion. I really found this topic quite interesting, so I'll very much enjoy explaining his method as intricately as possible. Now with classes only every second day, I didn't think scribe duties would remain so frequent, though the class is indeed quite minuscule in comparison to the class size first semester. Oh yes, and before I begin, I think i must note that all explanations will be accompanied by an example furthering my explanation in a different text colour. Text in black will be initial explanations, while text in green will pertain to explanations using an example for the first part of the post and text in blue will pertain to explanations using an example for the second part of the post. Hope it works out well. Enjoy!
Scribe
~~~~~~
I can't remember exactly how class started, though I do recall it initiated with the usual abstract chatter and various technological utility exposures by Mr. K. But what I do remember, is beginning the actual lesson with a very fundamentally elegant and quite remarkable equation; Euler's identity.
Though we didn't go in depth into what Euler's identity is and how it was formulated, but we did vaguely discuss the implications of the identity. The sheer elegance of the equation derives from the fact that it contains and also intertwines the destiny of five of the most important constants in mathematics: 0, 1, (pi), i, and e. Euler's identity is as follows:
Now we transitioned into the basis of the day's topic: Euler's method. I think it's quite a useful technique, and is a very innovative technique with easy implementation (at our level at least). We started off with a demonstration of Euler's method without any formal introduction quite yet. This demonstration can be found here. If you follow the link, it might ease the subsequent explanations tremendously.
- First off, imagine that you are given an initial value problem. Recall that an initial value problem is a problem in which you are given a differential equation to solve as usual, but you are also given a point that exists on the parent function/solution to the differential equation.
*In this case, as used by the aforementioned demonstration, take the differential equation to be y'(t) = 1 - t + 4y(t) and the initial point to be y(0) = 1. This means that at t = 0, y(t) = 1. Following the above link will drastically clarify this example. Also, take note that we know the point (0,1) on the differential equation solution, and therefore have an exact solution and not just a general one. A general solution only represents the family of functions that could fit the solution for the differential equation. An exact one means that it's only specific function, and not a whole family.
- Now, there is an important idea that I must stress in order for Euler's method to be utilized easily, so pay attention in case you missed it in class. We must find a way to get another point on the parent function, which can be easily determined by plugging in the initial value/point into the differential equation to solve for the slope of the line tangent to the function at that point. Using this slope we can use our classic definition of slope to yield the next point that exists on the function y(t) given a certain step.
*The point (0,1) is known to lie on the function y(t), but how are we possibly going to get another point on this function? Well, as I stated earlier, since we know the differential equation, we can plug in the point (0,1) into the differential equation (which is y'(t) = 1 - t + 4y(t) in this case) and determine the slope of the line tangent to the function at that point. This is because the differential equation will solve for y'(t), which is the derivative of y(t) or rather the instantaneous rate of change of y(t) at a given t. This is particularly useful since using our rather classic definition of "slope" (rise over run) will yield our next point. By plugging in the initial point into the differential equation as follows, we can determine the required slope:
y'(0) = 1 - (0) + 4(1)
y'(0) = 5
Using this slope with the "old-school" rise over run definition of slope, we can easily determine the next point. But be careful, you must pay careful attention to the steps or the scale of each axis to correctly apply this definition. Since the steps we will be using in this case are 0.1, that means that instead of the function increasing upwards by 5 and increasing rightwards by 1, the function will increase upwards by 0.5 as it travels rightwards by 0.1. This means that the next point will be (0.1, 1.5). Using the next button on the demonstration page will automatically graph the next point on the function.
- Now that we have two points on the graph, we can easily repeat the above process to determine any subsequent points and find an accurate portrayal of the solution. The only problem is that you must use a sufficient amount of steps by decreasing the amount of change along the x-axis (or in this case t) so that each point will be closer to each other and each subsequent derivative will be more accurate than with little points. I advise using this website or this website to help grasp this concept. Review one of these sites to further your understanding of what I have just said. Also, concerning this idea, I tend to think of Euler's method as being quite similar to how an integral works. In an integral, you must let the dx values (or dt) get as close to 0 as possible, so as to increase how accurate the solution truly is or how well the integral fits the actual shape of the function. This is exactly how Euler's method works.
*Each time that you press the next button on the above demonstration, the graph will further the use of Euler's method and graph the next point on the parent function, thereby creating an approximate graph of said function. But notice how choppy each segment looks, even though as a calculus student(and others =p) you should recognize the function to be curved.
Once we were introduced to Euler's method using a bevy of demonstrations and tools to ease the idea of his method to us, we were asked to apply what we had just learned to a problem. This can be seen on the following slide:
In the above slide, all that we really did was find a way (basically on our own) to efficiently repeat the process I detailed above. Though there were some alterations in this problem.
*Basically, when you were given the initial point, you could plug the point into the differential equation (which is y' = y - x this time) to find the derivative of the function at that point. This derivative will be m, or the slope of the tangent line at that point, which can be used by rearranging the m = Δy/Δx equation into Δy = m * Δx. Since m has just been calculated, and Δx (which is the step as I mentioned above) is given in the question as 0.25. Calculating Δy will give us the change in y from the first point to the next point given a certain step, which means we'll know the next points x-coordinate (the first point x-coordinate plus the step) and y-coordinate (the first point's y-coordinate plus the Δy calculated). Thus, this "next" point will be our "initial point" in the next part of the solution. If you use the above slide to help with understanding this idea, then each row will be one iteration of determining the solution to the differential equation. An iteration means a complete round of doing something, such as the process that is being repeated within a loop (which can be taken as the process I detailed above in the first detailed section of this scribe post).
But how can we make this process simpler and reduce the irksomeness of the entire process, especially for cases with a lot more repetitions? One way that we worked out during class involved using the store function in our calculators. This method can actually have several approaches, but basically you can utilize your calculator's store functions in many ways to achieve the same thing. Here's basically what you do:
- Store the initial y value into any variable in your calculator, say A.
- Input this variable into the differential equation on your calculator, and multiply this new expression by the step, and finally add A to this whole thing and store the new answer back into A. It might be hard at first to type this all in within one shot, but it is doable if you do it enough. Below is an example of how it should look in your calculator using the above question.
*Applying this step to the slide above, here's what it would look like in your calculator:
(((A - 0)0.25) + A) -> A
- Simply repeat this line using the [2nd] then [enter] function in your calculator, each time changing the x value according to how the point has changed.
*Again applying this step to the slide above, here's what the next line would look like:
(((A - 0.25)0.25) + A) -> A
Using this method yielded the answers shown in the above slide, which could be taken as solved since the last point on the interval asked for (which was [0,1]) was determined, the question was answered. Though, Mr. K wrote down the exact answer to the problem at the bottom of the slide, which wasn't that close to the answer we arrived at.
So then we tried using more segments to approximate a solution that is much closer than achieved above. We did this by using our new found method of numerically solving initial value problems with Euler's method, we attempted another problem. Though it may look quite empty, all of the work was quickly repeated on our calculators using a method roughly equivalent to the one above, and we all arrived at the same answers as shown in the following slide. I suggest trying it out on your own for some practice. (note: The question is basically the exact same as the first, but it uses far more segments and a much smaller step).
Just before class ended, Mr. K distributed the "EULER" calculator program for us to use to quickly solve problems involving Euler's method.
I will post the algorithms and full code for the program here when I get the full version again, since someone accidentally erased a line in the program and I can't remember what that line was.
Okay, that's all for my scribe for now. I still have some stuff to edit, but it's getting pretty late now and I don't want to be late for Chemistry again tomorrow (I bet people in Chem would doubt that). Anyway, the next scribe will be:
Tim-MATH-y!
Well good night everyone, see you all in class tomorrow! Please just talk to me or comment if there are any questions, complaints, anxieties, confusions, etc. =p Good bye all!
Scribe
~~~~~~
I can't remember exactly how class started, though I do recall it initiated with the usual abstract chatter and various technological utility exposures by Mr. K. But what I do remember, is beginning the actual lesson with a very fundamentally elegant and quite remarkable equation; Euler's identity.
Though we didn't go in depth into what Euler's identity is and how it was formulated, but we did vaguely discuss the implications of the identity. The sheer elegance of the equation derives from the fact that it contains and also intertwines the destiny of five of the most important constants in mathematics: 0, 1, (pi), i, and e. Euler's identity is as follows:
Now we transitioned into the basis of the day's topic: Euler's method. I think it's quite a useful technique, and is a very innovative technique with easy implementation (at our level at least). We started off with a demonstration of Euler's method without any formal introduction quite yet. This demonstration can be found here. If you follow the link, it might ease the subsequent explanations tremendously.
- First off, imagine that you are given an initial value problem. Recall that an initial value problem is a problem in which you are given a differential equation to solve as usual, but you are also given a point that exists on the parent function/solution to the differential equation.
*In this case, as used by the aforementioned demonstration, take the differential equation to be y'(t) = 1 - t + 4y(t) and the initial point to be y(0) = 1. This means that at t = 0, y(t) = 1. Following the above link will drastically clarify this example. Also, take note that we know the point (0,1) on the differential equation solution, and therefore have an exact solution and not just a general one. A general solution only represents the family of functions that could fit the solution for the differential equation. An exact one means that it's only specific function, and not a whole family.
- Now, there is an important idea that I must stress in order for Euler's method to be utilized easily, so pay attention in case you missed it in class. We must find a way to get another point on the parent function, which can be easily determined by plugging in the initial value/point into the differential equation to solve for the slope of the line tangent to the function at that point. Using this slope we can use our classic definition of slope to yield the next point that exists on the function y(t) given a certain step.
*The point (0,1) is known to lie on the function y(t), but how are we possibly going to get another point on this function? Well, as I stated earlier, since we know the differential equation, we can plug in the point (0,1) into the differential equation (which is y'(t) = 1 - t + 4y(t) in this case) and determine the slope of the line tangent to the function at that point. This is because the differential equation will solve for y'(t), which is the derivative of y(t) or rather the instantaneous rate of change of y(t) at a given t. This is particularly useful since using our rather classic definition of "slope" (rise over run) will yield our next point. By plugging in the initial point into the differential equation as follows, we can determine the required slope:
y'(0) = 1 - (0) + 4(1)
y'(0) = 5
Using this slope with the "old-school" rise over run definition of slope, we can easily determine the next point. But be careful, you must pay careful attention to the steps or the scale of each axis to correctly apply this definition. Since the steps we will be using in this case are 0.1, that means that instead of the function increasing upwards by 5 and increasing rightwards by 1, the function will increase upwards by 0.5 as it travels rightwards by 0.1. This means that the next point will be (0.1, 1.5). Using the next button on the demonstration page will automatically graph the next point on the function.
- Now that we have two points on the graph, we can easily repeat the above process to determine any subsequent points and find an accurate portrayal of the solution. The only problem is that you must use a sufficient amount of steps by decreasing the amount of change along the x-axis (or in this case t) so that each point will be closer to each other and each subsequent derivative will be more accurate than with little points. I advise using this website or this website to help grasp this concept. Review one of these sites to further your understanding of what I have just said. Also, concerning this idea, I tend to think of Euler's method as being quite similar to how an integral works. In an integral, you must let the dx values (or dt) get as close to 0 as possible, so as to increase how accurate the solution truly is or how well the integral fits the actual shape of the function. This is exactly how Euler's method works.
*Each time that you press the next button on the above demonstration, the graph will further the use of Euler's method and graph the next point on the parent function, thereby creating an approximate graph of said function. But notice how choppy each segment looks, even though as a calculus student(and others =p) you should recognize the function to be curved.
Once we were introduced to Euler's method using a bevy of demonstrations and tools to ease the idea of his method to us, we were asked to apply what we had just learned to a problem. This can be seen on the following slide:
In the above slide, all that we really did was find a way (basically on our own) to efficiently repeat the process I detailed above. Though there were some alterations in this problem.
*Basically, when you were given the initial point, you could plug the point into the differential equation (which is y' = y - x this time) to find the derivative of the function at that point. This derivative will be m, or the slope of the tangent line at that point, which can be used by rearranging the m = Δy/Δx equation into Δy = m * Δx. Since m has just been calculated, and Δx (which is the step as I mentioned above) is given in the question as 0.25. Calculating Δy will give us the change in y from the first point to the next point given a certain step, which means we'll know the next points x-coordinate (the first point x-coordinate plus the step) and y-coordinate (the first point's y-coordinate plus the Δy calculated). Thus, this "next" point will be our "initial point" in the next part of the solution. If you use the above slide to help with understanding this idea, then each row will be one iteration of determining the solution to the differential equation. An iteration means a complete round of doing something, such as the process that is being repeated within a loop (which can be taken as the process I detailed above in the first detailed section of this scribe post).
But how can we make this process simpler and reduce the irksomeness of the entire process, especially for cases with a lot more repetitions? One way that we worked out during class involved using the store function in our calculators. This method can actually have several approaches, but basically you can utilize your calculator's store functions in many ways to achieve the same thing. Here's basically what you do:
- Store the initial y value into any variable in your calculator, say A.
- Input this variable into the differential equation on your calculator, and multiply this new expression by the step, and finally add A to this whole thing and store the new answer back into A. It might be hard at first to type this all in within one shot, but it is doable if you do it enough. Below is an example of how it should look in your calculator using the above question.
*Applying this step to the slide above, here's what it would look like in your calculator:
(((A - 0)0.25) + A) -> A
- Simply repeat this line using the [2nd] then [enter] function in your calculator, each time changing the x value according to how the point has changed.
*Again applying this step to the slide above, here's what the next line would look like:
(((A - 0.25)0.25) + A) -> A
Using this method yielded the answers shown in the above slide, which could be taken as solved since the last point on the interval asked for (which was [0,1]) was determined, the question was answered. Though, Mr. K wrote down the exact answer to the problem at the bottom of the slide, which wasn't that close to the answer we arrived at.
So then we tried using more segments to approximate a solution that is much closer than achieved above. We did this by using our new found method of numerically solving initial value problems with Euler's method, we attempted another problem. Though it may look quite empty, all of the work was quickly repeated on our calculators using a method roughly equivalent to the one above, and we all arrived at the same answers as shown in the following slide. I suggest trying it out on your own for some practice. (note: The question is basically the exact same as the first, but it uses far more segments and a much smaller step).
Just before class ended, Mr. K distributed the "EULER" calculator program for us to use to quickly solve problems involving Euler's method.
I will post the algorithms and full code for the program here when I get the full version again, since someone accidentally erased a line in the program and I can't remember what that line was.
Okay, that's all for my scribe for now. I still have some stuff to edit, but it's getting pretty late now and I don't want to be late for Chemistry again tomorrow (I bet people in Chem would doubt that). Anyway, the next scribe will be:
Tim-MATH-y!
Well good night everyone, see you all in class tomorrow! Please just talk to me or comment if there are any questions, complaints, anxieties, confusions, etc. =p Good bye all!
Tuesday, March 18, 2008
Monday, March 17, 2008
Article 13, Al Upton, and the minilegends
If you'd like to leave a comment to Al Upton and the minilegends in Australia click on the picture below to get to their blog.
You can also read more about Article 13 and the Convention on the Rights of the Child here. This movie illustrates what it's all about:
You can also read more about Article 13 and the Convention on the Rights of the Child here. This movie illustrates what it's all about:
The Pi day scribe
Well, a certainly enjoyable pi day it was. Lots of treats and Friday!. Doesn't get much better than that.
With all the celebrating, we didn't do very much.
We learned, the slope fields were the graphs of the parent function. Just moved all over the place.
Due to my internet moving unbelievably slow, I will add more "pi" pictures when it starts to pick up. (Litterally, 3 minutes to load a single page)
Next scribe.
Mr.SiwWy-chan
With all the celebrating, we didn't do very much.
We learned, the slope fields were the graphs of the parent function. Just moved all over the place.
Due to my internet moving unbelievably slow, I will add more "pi" pictures when it starts to pick up. (Litterally, 3 minutes to load a single page)
Next scribe.
Mr.SiwWy-chan
Friday, March 14, 2008
Absolutely Late Scribe..... Sorry guys.
WOW! Very, very sorry about the tardiness of this scribe... I worked both last and this evening and today was very busy.
So Wednesday's class started with a very unusual period in which we were given our tests back and allowed 10 minutes to continue them since we didn't get a chance last Thursday.
Then we briefly discussed the Developing Expert Voices projects as Mr. K. handed back the proposals.
Following that was a little conversation about "π day" (which is today actually). I'm bringing either a Strawberry-Mango or Blueberry-Peach pie =D.
Finally we started the lesson for the day. The first thing we did was a quick review of Differential Problems.
It dealt with a set amount of oil in a reserve, states that a function A(t) as the rate of consumption, and provides the derivative of the rate [ A'(t) ]. How long until the reserve is completely emptied.
Basically, to solve it, find the antiderivative, A(t)+C, of the the given function and then plug in t=0 to find the constant (C).
From there it is as simple as finding the positive root of that function and that will be the answer.
Next we moved onto the new content.
SLOPE FIELDS
these are kind of tough to explain... I'll let some pictures speak some of my words.
This is the FIELD in which we plot the SLOPES. Easy enough right?
So we first tried it with the expression ( ∂y/∂x = y ). This means that the derivative of any point in the function is the value of the function at that point (y-coordinate).
•Let's say we want to plot the point (1,1). The derivative (SLOPE) of the function at that point is 1, so we draw a small line with slope 1 at that point.
•Next, take the point (3,2). The derivative at that point is 2, so there's another small line, but with a slope 2 at the point (3,2).
Now, you understand how they are constructed, this is how the SLOPE FIELD of the function(s) that have ( ∂y/∂x = y ).
From here, you can solve for the point (1,1).
To do this, draw a curve that follows the slope of the line in either direction until the next whole number is reached. At that point, change the curve's slope to that of the new poitn and continue until the next whole number and so on... WHAT FUNCTION DOES THE CURVE LOOK LIKE???
Correct! It is easy to see that the function that has a derivative equal to the value of the function and passes through the point (1,1) is y=e^x.
However, you can also see that there are many other functions with a similar shape that could have been drawn, depending on the starting point.
This is because the SLOPE FIELD shows the entire family of functions [ ∫ƒ(x)∂x + C ].
So, by solving for a point, you can find the exact one function from that family.
We also did SLOPE FIELDS for ∂y/∂x = 2x and ∂y/∂x = -x/y... can you guess what functions they are???
This was the end of the lesson as we realized that any function can be determined by plotting it's derivative in a SLOPE FIELD and we can find which function out of a family given a point on the specific function.
This concludes my scribe post, once again, sorry it was so late.
The scribe for PI (π) Day will be...
VAN!!!!!!!!!!
So Wednesday's class started with a very unusual period in which we were given our tests back and allowed 10 minutes to continue them since we didn't get a chance last Thursday.
Then we briefly discussed the Developing Expert Voices projects as Mr. K. handed back the proposals.
Following that was a little conversation about "π day" (which is today actually). I'm bringing either a Strawberry-Mango or Blueberry-Peach pie =D.
Finally we started the lesson for the day. The first thing we did was a quick review of Differential Problems.
It dealt with a set amount of oil in a reserve, states that a function A(t) as the rate of consumption, and provides the derivative of the rate [ A'(t) ]. How long until the reserve is completely emptied.
Basically, to solve it, find the antiderivative, A(t)+C, of the the given function and then plug in t=0 to find the constant (C).
From there it is as simple as finding the positive root of that function and that will be the answer.
Next we moved onto the new content.
SLOPE FIELDS
these are kind of tough to explain... I'll let some pictures speak some of my words.
This is the FIELD in which we plot the SLOPES. Easy enough right?So we first tried it with the expression ( ∂y/∂x = y ). This means that the derivative of any point in the function is the value of the function at that point (y-coordinate).
•Let's say we want to plot the point (1,1). The derivative (SLOPE) of the function at that point is 1, so we draw a small line with slope 1 at that point.
•Next, take the point (3,2). The derivative at that point is 2, so there's another small line, but with a slope 2 at the point (3,2).
Now, you understand how they are constructed, this is how the SLOPE FIELD of the function(s) that have ( ∂y/∂x = y ).
From here, you can solve for the point (1,1).
To do this, draw a curve that follows the slope of the line in either direction until the next whole number is reached. At that point, change the curve's slope to that of the new poitn and continue until the next whole number and so on... WHAT FUNCTION DOES THE CURVE LOOK LIKE???
Correct! It is easy to see that the function that has a derivative equal to the value of the function and passes through the point (1,1) is y=e^x.However, you can also see that there are many other functions with a similar shape that could have been drawn, depending on the starting point.
This is because the SLOPE FIELD shows the entire family of functions [ ∫ƒ(x)∂x + C ].
So, by solving for a point, you can find the exact one function from that family.
We also did SLOPE FIELDS for ∂y/∂x = 2x and ∂y/∂x = -x/y... can you guess what functions they are???
This was the end of the lesson as we realized that any function can be determined by plotting it's derivative in a SLOPE FIELD and we can find which function out of a family given a point on the specific function.
This concludes my scribe post, once again, sorry it was so late.
The scribe for PI (π) Day will be...
VAN!!!!!!!!!!
Thursday, March 13, 2008
Wednesday, March 12, 2008
Tuesday, March 11, 2008
Monday, March 10, 2008
Wednesday, March 5, 2008
bob
this unit was just the same as the rest. im still trying to catch back up. thanks to GreyM I think i will be able to do just that. Im putting more effort into cathing backup now so i hope i do good.
The Integral BOB
I have to say I feel more confident going into this test then I did going into the other test earlier today. I get how these problems work most of the time, just using the unit logic and the idea of taking one of the parameters and chopping it up into little bits. I grok the average value of a function idea. I waver a bit on the solids but really they are not terribly hard. I think I will do fairly well tomorrow. G'night.
BOB ^ 8
Well, I went on to check out the calculus blog to BOB and surprisingly enough, there aren't many BOB's up for this test yet, especially considering what time is it already. I hope that people didn't forget =/.
Anyways, this unit, applications of integrals, was definitely not a cake walk. I think there's quite a bit of material left for me to review, mostly due to the fact that I seem to be lost in a haze since we only have class every second day now and I'm usually too busy with other things to finish all of the homework. But all in all, a lot of the applications of integrals we have covered during this unit are quite similar. Such as the similarity between evaluating the volumes of various solids revolved about the x or y-axis or a particular line like y = -1 and using density functions and such to find a mass or population (the last section of this unit).
Since I mentioned the last section of this unit anyway, I think that's probably the only place where I might run into some trouble on the test tomorrow. The main reason for this is because the process of solving such problems aren't straight forward in the least, though once you repeat the logic undertaken each time you attempt a problem of this sort (where unit analysis incredibly simplifies the matter) the approach for each question becomes more apparent each time.
Besides the aforementioned troublesome questions, I don't think that there is anything else I might find excruciatingly difficult on the test. The only things we really have to know for tomorrow are:
- The difference between displacement and distance when integrating a velocity/speed function.
- Determining definite integrals to represent the volume of a solid, and evaluating Riemann sums and these self-generated definite integrals.
- How to use integrals to determine the average value of a function. (Just think of finding the average normally, ex. adding up all your marks (an integral) over the total amount of marks (the size of the interval)).
- How to use integrals for differing scenarios, such as oil density or population density of cities. I suggest reviewing chapter 8.5 in the textbook for elaboration and some practice on this topic.
Well that's it, I haven't completed such a long BOB in quite a while. Well, I hope everyone does great on the test tomorrow, don't forget to go over all your notes and practice some problems tonight. Study hard everyone! I know I will.
I wish everyone good luck on tomorrow's test and a good night!
Anyways, this unit, applications of integrals, was definitely not a cake walk. I think there's quite a bit of material left for me to review, mostly due to the fact that I seem to be lost in a haze since we only have class every second day now and I'm usually too busy with other things to finish all of the homework. But all in all, a lot of the applications of integrals we have covered during this unit are quite similar. Such as the similarity between evaluating the volumes of various solids revolved about the x or y-axis or a particular line like y = -1 and using density functions and such to find a mass or population (the last section of this unit).
Since I mentioned the last section of this unit anyway, I think that's probably the only place where I might run into some trouble on the test tomorrow. The main reason for this is because the process of solving such problems aren't straight forward in the least, though once you repeat the logic undertaken each time you attempt a problem of this sort (where unit analysis incredibly simplifies the matter) the approach for each question becomes more apparent each time.
Besides the aforementioned troublesome questions, I don't think that there is anything else I might find excruciatingly difficult on the test. The only things we really have to know for tomorrow are:
- The difference between displacement and distance when integrating a velocity/speed function.
- Determining definite integrals to represent the volume of a solid, and evaluating Riemann sums and these self-generated definite integrals.
- How to use integrals to determine the average value of a function. (Just think of finding the average normally, ex. adding up all your marks (an integral) over the total amount of marks (the size of the interval)).
- How to use integrals for differing scenarios, such as oil density or population density of cities. I suggest reviewing chapter 8.5 in the textbook for elaboration and some practice on this topic.
Well that's it, I haven't completed such a long BOB in quite a while. Well, I hope everyone does great on the test tomorrow, don't forget to go over all your notes and practice some problems tonight. Study hard everyone! I know I will.
I wish everyone good luck on tomorrow's test and a good night!
Pre-Test blog (Gearing Up)
what do you know I made it back for class :)
Sorry but i forgot about scribe until tonight, but all Im going over is the Pre-test. Nothing else really happened...I believe.
Personally i needed the pre-test because they was somethings I didnt even realize I didnt know and now I have the chance to look over it again.
Now on to the pre-test.
It was made up of 6 questions. Ill be going over each of the questions.
Heres the questions. All the work is done on the pages itself.
Most of these are for me to rework them so sorry if something seems really obvious.
Tuesday, March 4, 2008
BoB!
Ah! Once again there's an upcoming test! I am confident in saying that I am prepared, but not quite.
There are still a couple of things that I'm worried about. However I can't seem to place it. It's just that feeling in the gut. Hopefully some more practice will clear out any doubts.
Confidence may be my only hope! Good luck to everyone else on the test :)
There are still a couple of things that I'm worried about. However I can't seem to place it. It's just that feeling in the gut. Hopefully some more practice will clear out any doubts.
Confidence may be my only hope! Good luck to everyone else on the test :)
Monday, March 3, 2008
Craig's BOB
Well, just here to do a quick little B.O.B. The unit of Applications of Integrals has been a pretty interesting one because of how close it gets to real time problems. I have enjoyed solving these problems and even though at times they seemed complicated, once you leave all the words behind and put it into mathematical equations it became a lot easier.
As well, it gave a bit more practice for solving integrals a bit quicker and allowed us to use a lot of previous techniques/skills to do them.
I would have to say that the hardest part of the unit was the last bit with the radial density because of how much translation from real circumstances to mathematical equations and diagrams. However, once again as soon as it was translated, it became quite easy to solve.
So after the pre-test, we will see what kinds of questions to expect on the test, and what areas we need to brush up on a bit.
See you all tomorrow and good luck on the pre-test and test!
As well, it gave a bit more practice for solving integrals a bit quicker and allowed us to use a lot of previous techniques/skills to do them.
I would have to say that the hardest part of the unit was the last bit with the radial density because of how much translation from real circumstances to mathematical equations and diagrams. However, once again as soon as it was translated, it became quite easy to solve.
So after the pre-test, we will see what kinds of questions to expect on the test, and what areas we need to brush up on a bit.
See you all tomorrow and good luck on the pre-test and test!
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