Differential Equations BOB

When we first started this unit, it was quite easy. We learned a lot of stuff in this unit from our previous units. The Euler's Method when done by hand was tedious, but not hard. Slope fields were pretty cool to draw.

We went into the velocity and acceleration problems next. It's much easier to follow the question by writing down everything that's given in the question, and from the initial information, finding hidden information by doing a bit of work.

Newton's Law of Cooling was not bad in class, but looking back I can't remember much about it. I still have trouble on when to use certain equations and using the calculator to solve certain problems, so I'm still sort of worried. I just wish we had a class today for review or maybe a formula sheet, I suppose this would help by refreshing my memory on the different equations. Good luck on the test everyone.

April 13.

Today in class we were finally introduced to our Wiki Solutions Manual, a little project where we thoroughly solve one question from a set of ten given. We post the solutions on the blog in the first week, and in the second week we find a mistake or simply make another person's solution more elegant.

Easy enough, and this will prepare us for the DEV projects that are soon arriving deadline. We will use LaTeX code to make the equations a lot easier to write out, on the page itself on the right side bar, there is a link that gives all the codes right below the "sitmo" box.

The goal of this project is to come up with perfect solutions to the calculus questions posted, through various minds working and correcting a single problem. You may correct an already corrected problem if you believe something has to be corrected. You may check if a question has been edited by simply clicking "page history" on the post itself.

If you're still not sure on how to do this project, there is an explanation and example on the wiki page itself. Remember, first come, first serve as far as the questions go. The tab for the Wiki Solutions Manual is found at the top of the blog by the banner.

We will have a pre-test tomorrow, on our unit of Differential Equations.

We had a quiz today and we were allowed to use calculators.


Find what x is equal to, when y is equal to one, by inputting the y-value into the given equation. Find the derivative of that equation, and bring the y-derivative value to one side, then input the points that you found from the first step. We do this because the derivative on a point on a graph is the slope.

Question 2.



The area under a graph is found my integrating it over a certain interval. We're trying to find what interval that is. If the area under both graphs is equal at a certain interval, we equate them and find the integral. Then we solve for k.

Question 3.



Integrate it. Use chain rule. The interval given is 1 to 2x.

Question 4.



We found the area from the x-axis to the graph for each part. I`m not quite sure how we did this. But I`ll find out.

We also did question 3 of our worksheet we got on Wednesday when there was a substitute teacher. We didn`t quite finish in class, but I`m sure we will go over the question in tomorrows class.

Good-luck on the pre-test everyone and the next scribe is Jamie.

It's all about Euler.

Today in class we spoke mostly of a man named Euler. A famous mathematician that made many discoveries in the field of calculus. He was responsible for the number "e", the symbol of pi and the use of "i" to represent an imaginary number. Euler had an identity and a method.



Euler also worked with many brilliant people, one such person is Goldbach, who stated that any even number larger than 2 can be written with the sum of any 2 prime numbers, if you feel like being famous some day, find an exception to this statement. Euler was also a father and he claimed to have made many mathematical discoveries while holding a baby in his hands, and as his other children played around his feet, now that is some fatherly love. Mr. K thought this was quite the favorable trait.

Now that we know a bit about Euler we can move on to his method. Which was the main focus in today's class. If we have a derivative that tells the slope of any line and you want to find a specific function, and not just a family of functions. You would need a slope and point, then you can get the equation of the line from the slope field. Calculate the equation of the line by moving a known point on one of the slope field lines by 0.1 units, find the slope at that point of the line then move the point 0.1 units on the new line and find a slope line at that new point, and repeat. Using the tangent lines gives a close approximation of the equation of the line. A great example is found on slide 2.


You can decrease the amount of error by using smaller intervals, such as 0.001 units instead of 0.01, but this is a very tedious process when done by hand. This method generates a table of values.

Another example we used was using local linear approximation to find 1.001¹⁰⁰.


Find the derivative of the function by replacing 1.001 with x to make it easier. Then replace x with a number close to 1.001 such as 1, this will be the x-value. By solving the original equation the y-value given would be 1. Solving the derivative equation would give the slope, which is 100. Inputting this into the equation of a line and then replacing 1 with 1.001 to find a close approximation of the original value of 1.001¹⁰⁰ which would be 1.1

Our homework is found on slides 6. We are given the initial point, the differential equation, the amount of steps between the given interval and the change in the x-value. The first few steps are done for us. All the equations we need to use are given at the top, and we find the slope by inputting the points in the differential equation. The change of x remains constant. Have fun.


and slide 7 is also homework.

Also, great job to Jamie for her performance in the talent show! Good luck on your homework and the next scribe will be Benofschool.

BOB on Applications of Integrals

At first I thought this unit was going to be quite easy going, but to my surprise, or my demise it became quite confusing and difficult.

We started out by just finding simple distance and displacement given a specific scenario or problem, sure this was easy but then we moved to revolving 2d graphs.

In my opinion these graphs have no business revolving whatsoever. This was just confusing. We revolved these graphs on various points all over the plane, and certain points caused the shape to be a washer or cylindrical shell which required a few more calculations.

Good luck on the test everyone. I'm off to bed.

Todays Scribe Post: Cross-Sections.

Before we started with our lesson, we talked about some other things. For example we talked about how Obama wrote a book on how he will reform the education system in the United States of America. I wouldn't mind reading this, but then again I wouldn't have to know about this due to us being Canadian and all. This discussion led to the awesome education system of Canada, and everything awesome about Canada. Except for our phone companies, which are pretty much trash.

All the randomness aside, we actually continued with our lesson on applications of integrals, we learned something new. We were supposed to learn this before rotations of 2d graphs and so forth, but we did not. We pretty much learned the hard stuff first that mainly dealt with the rotation of 2d objects which would make up a 3d object, which we had to visualize. This is cross-sections not rotations. A good example of cross-sections is like slices of bread from a bread loaf, and my favorite example is sushi sliced from a sushi roll. It looks like a rectangle from Birdseye view but when sliced, each cross-section is a circle! Amazing?! I know, just like magic.

Our first problem looked like this:




As the question states, we have to find the volume of the 3d shape the graph makes between the graph itself, the x-axis, x=0 and x=2. The formula used is the formula for the area of a circle, as shown in the slide. The only dimension we needed was the radius, and that's already given by the graph which is x². So we input this into the equation, and integrate it between the interval 0 to 2. Easy enough.

For our 2nd slide we had something a bit harder, but pretty much the same concept, but now we're dealing with different shapes than circles, such as semi-circles, equilateral triangles and squares, so our area equations would be different for each one.





The width of each cross-section would be dx in the integral equation. The length or radius would depend on the given function of the graph, in this case we have x².
Now the length of each cross-section when considering the semi-circle would be the diameter, and we would divide this by 2 to get the radius, r = x²/2. The equation of the area would be A = pi*r²/2 and we input r = x²/2 for the radius, and solve over the given interval, simple enough.
The square would be easy solve as the equation for the area is A = s² and s = x².
The equilateral triangle would have a base of x² and the height would have to be found using a bit of calculations. Imagine cutting the triangle in 2 equal parts right down the middle, splitting the base in two. The line used to cut would be the height. This would make a right angle triangle, and knowing an equilateral triangle all the angles are pi/3. Using tan(pi/3) = h/(b/2) we find that height is:
The area of a triangle is A = bh/2. We know b = x² and h. Throw all that together and we get this:We integrate it over the interval 0 to 2, and voila. As long as we know the shape of the cross-section, the corresponding area equation for this shape and the values for the dimensions to input into the equation then we know the anwser, its simple as pie. Pi day is almost here. Our homework was a worksheet handed to us entitled "Volumes of Revolution Excerises" and this should be worked on for continued success. The next scribe will be Lawrence.

Guess who's back? BOB

Today in class we had a pretty simple and straight forward pre-test. It was 4 mulitple choice and 1 free response. The pre-test questions ranged from definite integrals, such as finding the area under a graph in a certain interval, to integration by parts and substitution to find the antiderivative of a certain functions.

For question 1 of multiple choice I used the rsum program on my calculator to find the actual value of the function through the interval [1,3] and then I found the area under the graph on the interval [1,3] and there was a difference of 0.0416667 which is approximately 1/24, which was anwser B.

For question 2 I used trial and error on my calculator finding the area under the curve of the graph. Whichever equaled 6, would be the correct anwser, which was option D.

Question 3 was also the same concept as question 2. Where ln2 was equal to approximately 0.693, and I used trial and error to find the area under the curve that equaled ln2.

Question 4 was easily solved, just use integration by parts on the function y = xcos(x) and you should end up with option A.

Open response was a bit harder but not by much. For part a) we used substituion to find the antiderivative and then we found what it would equal at each endpoint and the table made it easy by giving us the values. Part b) was the same deal, but we used integration by parts.

Overall this pre-test was by far the easiest, in my perspective. Good luck everyone. It's time to BOB.

I was never really worried about this unit, because all of it seemed really straight forward. I was really only worried about when to use what, such as substitution and integration by parts to find an antiderivative. I know I won't have trouble solving either of them, as long as I make the right decision in the first place. The midpoint sum, trapezoid sum and Simpson sum seem to be quite easy, but I do believe we could have spent maybe a class or two more in this section. I didn't see these type of questions in the pre-test so this is maybe why it seemed brief in class, because it won't be a significant part in the test. I feel quite comfortable going into the test, as long as we're allowed to use our calculators. I'd be nothing without the graph and rsum programs!

The next scribe will be! Joyce.

Finding anti derivatives!

First class of the "finding anti derivatives" unit was today, and it wasn't so bad. We also talked about the test we had yesterday, and by far, the free responses were a lot harder then expected, so we went over the answers in class. The test seemed to be quite long, but in fact, it was about 2/3 its original size! Now that's a fun fact. The answers to the free response questions are as follows, and I will try my best to explain them.
Okay so this here is the first free response questions, and is the easiest of the two we went over in class. The answer would be to sketch the graph y = (Sinx)/x. Where x cannot equal zero which means its a discontinuity at x = 0, but it also says that the graph exists at 1 where x = 0, so it pretty much fills in this discontinuity. Confusing? Yeah, I know. Part b asks for what values of x does A have a local minimum, and A is the parent function of the graph we drew, so the local minimum of the parent function would be the point where it the x values decrease then increase, and on the derivative, it would be where the values are less than zero, to when they begin to be greater than 0, this would be at 2pi, and it would repeat every cycle. Part c asks for the coordinates of the first inflection point of the parent function right of the origin, this would be at 3pi/2 where the derivative graph decreases then increases. Finally, part d asks to solve the following equation accurate to one decimal point.
This is quite simple, because we just use our calculators, and keep integrating the graph until the area we get is 1. It's pretty much trial and error, and the answer should be 1.1
I honestly don't know how to explain the answer to the 2nd free response question, so I'm not going to try, but Mr. K gave us an alternate solution, which we will learn about later on in this unit. Here it is:
Part b asks us to find where k = 2 and this is simple enough, just plug in the value 2 where k is supposed to be, the answer is 4.

Part c is here:
It's pretty much a rates problem, where we are given the rate of k in respect to time, and we have to find the rate of A(x) (parent function) in respect to time. We know when k = 2, A =4 from part b, and we have the area of A as a function of k from part a. We pretty much derived the equation of A and multiplied it by the rate given, and plugged in 2 for k, and the answer is 4.

That was it for the test questions. Mr. K also introduced us to a new site entitled Mathway which pretty much solves all our questions on calculus, of course you should use this for help, and not answers for upcoming exercises. Now onto our actual unit, "finding anti derivatives" we went over many derivative rules, and we anti derived them and don't forget to add +C, when anti deriving an equation, because the C stands for a constant, and this constant disappears when it is derived, so don't forget! We also anti derived some questions toward the end of class, and we also looked at a list of partial anti-derivatives rules. Here it is! Our next scribe will be Rence.

BOB!

First off this unit was not bad, not bad at all. I feel relatively confident about this unit, although there were a couple confusing sections. I'm sure they won't be so bad, especially since I'm studying them right now. Hopefully it's enough! Today the pre-test wasn't so bad. I didn't do so well, but after a quick revising, I realized how simple it really was, and hopefully I will remember this by the time the test comes along. I get confused about the accumulation functions, and how they're related to derivatives, and the derivative of certain functions, but I think overall, I'm going to be alright. Good luck everyone.

Team FLJ Timeline

This is FLJ's Timeline as of today. Timeline is subject to change.

TIMELINE:

JANUARY 9 - Record Copied Examples of Questions for Project Use -Complete-

• Six examples are to be copied from Calculus textbook to be used as models for questions to be created.
  

JANUARY 23 - Create Six Questions from Copied Questions for Project Use

• From the six examples taken from the book, they will then be used to create questions that are going to be displayed in the project.
  

FEBRUARY 13 - Solve and Annotate the Six Created Questions

• The six created questions are to be solved, corrected, edited and annotated as to be properly displayed in the project
  

MARCH 6 - Begin Filming Gameplay

• Gameplay filming will begin via Xbox 360 and Halo 3. Filming will begin as soon as a HDPVR is acquired. Many long nights of filming will occur.
  

MARCH 31 - Complete all Gameplay Film

• It is expected that by this date, filming will be completed
  

APRIL 1 - Begin Editing Gameplay Film

• Post-production of the project will begin. Special effects, voice acting and video editing will take place. Many long nights will again, also occur.

APRIL 13 - Begin DEV Blog Construction

• A separate blog will be created for the DEV to be published to. Will consist of posts for Solved questions, Reflections, Sources and Credits.

MAY 1 - DEV Revision

• Prior to "handing the project in", the blog will be revised that it is in correct layout, questions will be revised for errors, and links and videos will be tested for optimal viewing.
  

MAY 3 - 100% DEV Project Completion

• Film will be completed and fully uploaded. Link to DEV blog will be revealed on the Expert Voices blog. Digital copies will be created and will be published via DEV blog. Project will be "handed-in" for evaluation.

This project is a sequel and will most likely require you to view the previous project : MITSOP.

This Project is Rated EG for Educational Guidance

This film contains scenes of Intense Math, Frightening Equations, and the Use of Patterns

BOB, Applications on Derivatives

After break, everything just BROKE. No, I'm joking, but I did forget a lot of stuff about this unit. It's awesome how I'm one of the last BOB's so I can read other peoples BOB's and have a sort of mini review on everything, especially Benofschool's BOB. I was chatting with him earlier, and I asked him "whats up?" and you know what he says? He says "I'm helping people study" Wow, can that kid get an award or something? Thanks bud!

Applications of derivatives, where should I start? How can I start? I don't remember a lot, but I did read other people BOB's! I know about the 1st and 2nd derivative tests, and what they're used for, pretty much, but I don't know if I will be able to graph them, and I have a feeling that's going to be a big chunk of the test. Optimization problems seem quite blurry to me, but I did look at some questions on the internet, such as examples, so hopefully I can just understand them by reading a couple more. The mean value theorm was easy enough to remember, atleast the theorm part of it, but if there are any questions on it, I'm not so sure I can make it through alive! That's pretty much it for BOB and I. Good luck everyone.

First day back.

Today in class, we didn't do much. Our main focus though, was the pre-test. It wasn't worth any marks but it was valuable. The answers can be found on today's class slides. The pre-test was actually quite hard, but maybe that's because it was our first day back, and we just forgot about calculus over the holidays. I know I'm not alone here.

We found out that our DEV will have a final due date on May 3rd, which means we have 4 months to complete it, which is plenty of time, so don't stress, but don't procrastinate either. We have to copy and not create questions similar to what we will be doing by the end of this week for our DEV. 4 questions for a soloist, 5 questions if it's you and a partner, and 6 questions if you got a group of three.

Mr. K also showed us a video reply to our calculus ad "What is a derivative?" which was made by someone else, it was pretty cool and pretty flashy. Our next integrals commercial will have a time cap of 60 secs, or one whole minute, and will be due on February 1st.

Time to study everyone! Exam week is the last week of January, but from the looks of things our class doesn't have to worry much, because it'll be pretty lightweight on us calculus heavyweights.

That was pretty much what we went over in today's calculus class. That next scribe will now be Kristina.

Inifinite Limits

Today in class we spent the majority of our time learning, and expanding our understanding of infinite limits and asymptotes, of both the horizontal and vertical type. First off, I'd suggest you watch the video found on the slides of today's slide show, because it's very informative and useful when it comes to better understanding this unit, which will help you understand this scribe post (hopefully). We started out about stating how infinite limits come in 2 "flavours" when a limit of "x" goes to some value, and that equals infinity, or approaches infinity, since infinity isn't a number. The other "flavour" is when the value of x approaches infinity and equals to some value. Those are the 2 main rules.

We also mentioned sequences and series. Where sequences are a series of numbers, that follow some rule, and are related to each other by this rule, and a series is the sum of the numbers in a given sequence. I'm not entirely sure why we mentioned this, but I think we brought it up to further understand the value of infinity, or the not-value of infinity, since infinity isn't a value at all, if that makes sense.

Okay so found in slide 2 of the slide show from today's class, we can see the equation that looks something like this : f(x) = (x-a)(x-b)/(x-c)(x-d), this is pretty much the general equation we used for the whole class. With this equation, we looked at how to find asymptotes, of the horizontal and vertical type. A little note about asymptotes: the graph doesn't necessarily leave these sections "untouched" it depends on what section of the graphed function you're looking at, because there can be a section of the graph, that might actually touch or cross where the graph isn't shown.

In a rational function when the denominator equals 0, we have a vertical asymptote. With this said, the denominator determines the vertical asymptotes, and the numerator determines the roots of the graph. Vertical asymptotes are related to limits when x approaches some value and the result is infinity, and horizontal asymptotes are related to limits when x approaches infinity and the result is some value.

Horizontal asymptotes are found on slide 6 of today's slide show. To find the asymptotes we lower the our highest given degree of a polynomial to 0 (the power on x). We multiply the numerator and denominator by the reciprocal of the given polynomial, as found in slide 6. After doing this, we should be left with the coefficients of those polynomials we reduced. All the other values should be over "x". When infinity is substituted into these x-values, then those values won't exist, and your just left with the coefficients, which will be your horizontal asymptote.

Doing these steps are important because if you substitute infinity into the x-values of the original equation, then most likely the result will be infinity divided by infinity, which does not equal 1! This is called an indeterminate form, other indeterminate forms are 0/infinity, infinity/o and o/o, in university if you decide to take calculus, you will further learn about these indeterminate forms. If you get a indeterminate value of 2/0, when trying to find the horizontal asymptote, then there is no horizontal asymptote, but instead might be a slanted asymptote. Also, if trying to find the horizontal asymptote, and the value is 0/1 then the horizontal asymptote will be the x-axis, regardless.

That pretty much sums up our class. All fun stuff aside. Hopefully you guys watched that video like 20 times like me.The homework was exercise 5.3, all the odd questions including 6 and 20. The next scribe will be Kristina, because I guess she's first on the list.

BOB, it's a test.

Well, its that time again, when we have a test and we have to call upon BOB. Today I didn't think the pre-test was going to be all that difficult, but it was. I thought it was gonna be a breeze because I felt that I knew what was going on about those word problems Mr. K kept giving to us in class, but the pre-test wasn't that simple. I totally forgot about the other stuff that we learned about before Mr. K arrived. Now I need to study pretty hard to try and remember the problems and rules and stuff. I also need some practice on those word problems. I'm sort of worried, but I'll just look back at what I have and try my best. Good luck everyone.

5-foot Girl and her Shadow...

Okay so today in class was a pretty fun one. We started with a problem: A light is at the top of a 16 foot pole. A girl, 5 feet tall walks away from the pole at a rate of 4 ft/sec. At what rate is the tip of her shadow moving when she is 18 ft. from the pole? At what rate is the length of shadow increasing?

Okay so we tackled this question together. First we wrote out everything we had, and everything we need, thanks to the question we already had a rate to work from. Which was the rate at which the girl walked away from the pole, it was 4 ft./sec. We had the height of the pole: 16ft, and we had the girls height: 5ft. We need to find the rate when the shadow is moving and the rate of the length of the shadow as it increases. Let's draw out a diagram!

As you can see, that is a triangle, not just a regular triangle, it's a right angle triangle. Okay, but there are 2 triangles, and they happen to be similar, because they are proportional, just one is smaller, with smaller sides. Okay, we'll get back to that later, but for now we should write down our rates.







Our rates are in relation to time, because if there was no time, there would be no movement going on, and well we wouldn't have a moving shadow or girl without time.. right? We're given the rate of the distance covered by the girl which is 4ft./sec, so our first rate is equal to 4. Our second rate is the rate at which the shadow is moving and our final rate is the rate of the length of her shadow as it increases. Now, all the letters such as "d", "s" and "a" are given by our diagram. From our diagram, we got "a" from adding together d and s (d + s = a, pay attention to this equation, we're going to use it later on). We should find some ratios, because the triangles in the diagram are similar (refer to diagram), we know that side "s" is to 5 ft. as side "a" is to 16 ft. s/5 = a/16, from this we know that 16/5 = a/s. We know that a is equal to d+s, now we just throw it in.
First things first, substitute a for d+s and rewrite the equation, then put like values on one side and solve by finding the lowest common denominator, etc. Substitute dd/dt with 4, because that's what we were originally given. simplify 4/16 to 1/4. Divide out 11/80 to finally find ds/dt, which is the answer to one of out questions! The rate at which the tip of the shadow is moving when the girl is at 18ft. away from the lamp post.

Now that we have ds/dt and dd/dt, we can solve for da/dt.

That was the answer to our last question, the rate the length of her shadow is increasing.

We also learned a little jig for the Quotient Rule, we rehearsed it a few times so feel free to practice. It goes like this: lo - di- hi minus hi - di - lo all over lo - lo. Where hi is the numerator of the given function, lo is the denominator and di is short for derivative.

Last thing we did in class was similar to the question we got in the beginning, but I'm not sure what it was about, because we didn't have much time to work on it. I'm sure we'll go over it thoroughly next class though. But for now, my post is done! Next scribe will be Joyce.

Where we at?! BOB

Okay well, I was in the middle of watching James Bond, then I realized that I had to BOB, James BOB. It actually kinda ruined the movie, haha. First off all, welcome back Mr. K, I'm excited to learn!

Our first unit about graphing was quite simple, it was pretty much a review of our previous years of graphing functions and all that. Not much to say here folks. The second unit about derivtives was pretty tough, the most I remember is about limits, I got the hand of those pretty fast, everything else was kinda blurry.. to say the least especially graphing a derivitive of a funtion from nother graph of a function, talk about confusing. The 3rd unit, was about definite integrals, was this unit short?, because I hardly remember anything about it, I remember rieman sums, but not fully.. actually not much at all only that I can use my calculator for this. This current unit was pretty cool, I felt like I did a lot in this unit because of the product rules and chain rules, and such and such. The chain rule isn't crystal clear, but I'm sure I'll get there. ESPECIALLY SINCE MR. K IS BACK! CAN I GET A HIP HIP!? HURRAY!

October 22. The Definite Integral

I guess I'm scribe again since I've been caught by Mr. K. That's alright though, no big deal. At least he's still around. :D

Today we started off my watching a youtube video entitled "Lobachevsky - Tom Lehrer" The video is about a mathematician named Lobachevsky who supposedly plagiarized from a fellow mathematician named Gauss.

Well back to the subject. We started out by graphing some values.


The red bars represent the velocity or speed at the given time intervals as it changes gradually. The orange and green bars represent pretty much another graph, where the change in velocity was instant. Whether the change in speed was gradual or instant, we can't determine so. The curve in between the orange and green is the distance travelled, and this is what we will be trying to find.

We then try to find the lower limit by finding area of the lower limits (found in red bars).

Lower limit = sum of area of least possible #'s.

We multiply each value by the time interval (1).
Lower limit = 1.4(1) + 2.7(1) + 3.5(1) + 4.5(1) + 5(1) = 17.1 ft.
Don't add the last value, because this is the upper limit and it can't get any higher than this (t=5).
Now we calculate the upper limit, which includes the orange and green bars.
Upper limit = 2.7(1) + 3.5(1) + 4.5(1) + 5(1) + 5.7(1) = 21.4 ft.
Now estimate the distance by taking the average of these two areas. (17.1 + 21.4)/2 = 19.3 ft.
If we're talking about limits, the range would be 17.1 ft. to 21.4 ft.
At this point we find the ambivalent area (area of uncertainty) which is pretty much the orange-green bars. Ambivalent area = 1.3 + 0.8 + 1 + 0.5+ 0.2 = 4.3 ft.
We have an estimated area of 4.3 ft. under the curve which doesn't exactly match the previous results we estimated.
Now we decrease the time intervals to be more precise, and to see what happens.

Let's find the lower limit, or left edge of this graph. Left edge = 0.7 + 1.35 + 1.5 + 1.75 + 2.05 + 2.25 + 2.4 + 2.5 + 2.7 = 18.2 ft.

Now we find the upper edge which is pretty much the same, but we exclude the number of least value and include the number of most value. It should equal 20.35 ft.

The distance will be between 18.2 ft. and 20.35 ft.
The mean distance is (18.2 + 20.35) / 2 = 19.3 ft.

The average distance remains the same as our last graph.

Therefore by decreasing the size of the time intervals we have brought both the upper and lower limits' ranges lower.
At 1 sec. interval range = 17.1 ft. and 21.4 ft.
At 0.5 sec. interval range = 18.2 ft. and 20.35 ft.
The mean stayed the same at 19.3 ft.

That was all we did in class, our homework is Exercise 3.1 questions 1, 3, 6, 7 and 9. Hope you enjoyed, if any issues, be sure to tell me and I'll be sure to correct them. The next scribe will be well, not Richard, but Not Paul.

Friday's class of a couple Theroms.

Today in class we learnt about a couple new theorems. The first one is the "Intermediate Value Theorem". This theorem pretty much states that according to slide 1, found on the graph, that if y= f(x) is continuous on an interval [a, b] and y = k in between points y=f(b) and y=f(a) then there is an x-value of c where f(c) = k. If a discontinuity is found, remove it. I believe that what Mrs. E means is if there is a discontinuity between the chosen interval of [a,b] then chose a new interval by moving it to a place on the graph where there is no discontinuity. Confusing? Yeah, just look at the graph and read the notes a couple times. That's what I did.

We also looked at how to find zeros on a continuous function without our calculators. We were given an example on slide 2. f(x) = x³-3x+1. To find the zero of this function, we chose points [0,1] for this example. We then found out where the x-values were either positive or negative. On the example the point 0 was input into the function and equaled to 1. Which was positive ( f(0) = 0-3(0) +1 = 1). At the point of 1, the function was a equal to a negative number ( f(1) = 1 -3(1) + 1 = -1). Now we know that the zero of the function will be in between those two points. Now we want to hone in on a smaller region, so we should cut that interval of 0 to 1, in half. So we find the f(1/2) and see if it's positive or negative, so we know where to look next. The function of f(1/2) turned out to be negative, as found on the slide. We then know to look between points [0 and 1/2] because the zero will be in between a negative and positive value. It's best to divide the section by half to cover greater amount of space between the points. The next point we used was 1/4. We input this into the function and found it to be positive. Now we can look between points [1/4 and 1/2]. We then used the point of 3/8 in our function, which is in between our given points. It turned out this was negative, so now we know our zero is between 1/4 and 3/8 which is small enough of an interval where our zero is found, so we stopped there.

The next theorem we used was entitled the "Extreme Value Theorem". This states that on a continuous function, it will always have a maximum and minimum and will be an extremum value from where all points on the function will either larger or smaller depending if its the maximum or minimum extremum. Lets say that on a continuous function on interval [a,b] there exists numbers c and d where all x-values in [a,b] f(c) ≤ f(x) ≤ f(d). d will be the maximum and all x-values will be less that or equal to d. c is the minimum and all x-values will be greater that c or equal.

That's all we did in today's Calculus class. If anything is wrong, talk to me and I'll be sure to change it. The next scribe will be Lawrence!