 Well, we are at the last day, day 59. I thought it would be somewhere close to 60, so didn't quite make it to 60. I think the task today is to kind of go through what you did on that first, and then after that task is accomplished. Then we'll go back and look at things that have made it to tests that you might see again on the final exam, and what have we covered since test four that might also have a chance of making it to the final exam. So I think we have addressed some of that already. What have we done since test four? We've done a little bit more with the binomial series, and hopefully I've connected those well enough to you that really the binomial series and the Taylor polynomial, yeah, you guys help yourselves there because I don't want to take any of them with me. They're really one in the same. The binomial series kind of was rooted or founded using the Taylor series, so when they talk about a Taylor expansion and you want to use a binomial, that's fair game. They're all the same. They're in the same category, but we have done more with the binomial series expansion than since test four, so that's potential final exam material. The applications I think we decided last class are probably too lengthy. They're good. They're valuable. I'm glad we got that far in the course, and we looked at them, connected some things maybe in physics that you have used before that kind of wondered why we had two different formulas and how we could get away with that. But those problems are too lengthy for a final exam. And we have, I think, did we finish the most recent test as far as going back through everything on test four? Do we get all the way through that one? Okay. Yeah, we didn't address the bonus. Actually the bonus, I mean it's appropriate, but probably would have been more appropriate for test three, because that was kind of the topic there in test three, but it is a couple of infinite geometric series. The ratio is 4.4, which absolute value of that is less than one, so it does converge. But just a simple one. The first one, it drops five feet, and then after that, the amount that it comes back up, it also drops back down. And theoretically this process of responding to a height of 40% of what it was prior to that, in theory that ball never does actually sit on the ground, right? Doesn't it kind of continue to respond to 40% of what it was before? We know eventually that it does just sit there on the ground, so it is a mathematical model. It does a great job, but the reality of it is not completely modeled by the infinite geometric series that we're trying to use. So I think I drew a diagram of that, and then the one five foot trip down is kind of off by itself, unless you want to add it to all the other trips down, and all the trips up, the first trip up is 40% of five feet. So I just split it apart into two different infinite geometric series, and then added them together. Did anybody get full credit on the bonus? Yeah, I think you did. What if you got the right answers, but you didn't do the work that you did? I don't know. I'd have to see something. What I don't want to do is reward, oh, I think it's 19 and two thirds. I think I'm going to write that down. And then you happen to get the right answer. But if you've got anything else, all right, so I think we're done with test four. So let's just kind of continue our way back. So let's look at test three. I am hoping that as we, more importantly, that as you look back at this stuff that you'll say, oh, yeah, we've been doing that a lot even since then. For example, on test three, question five, the first one, five A, you've got an alternating series. We've dealt a lot with alternating series. So we could probably, if all we need to do is determine convergence, use the alternating series test, but the additional directions, if it converges, find the value to which it converges, that's not going to get us that. So we've got to come up with something else. It is an infinite geometric series. Because it alternates, the ratio is negative, right? And it's a matter of figuring out what it is that we multiply through by in that particular series. Well, a dead giveaway, if the first term is one, isn't the next term the ratio. That's the only way you can get from one to something else is to multiply by that something else. So the ratio is negative 0.4. I think if I remember right, the most common error on this problem was that people had the ratio at 0.4. The ratio in this case is negative 0.4. And the first term is one. So the sum of an infinite geometric series that does, in fact, converge is what? What over what? A over 1 minus R. A over 1 minus R. So in this case, 1 over 1 minus negative 0.4. What does 5B look like to you? It's harmonic. There's a 5. You can pull the 5 out in front. It's not going to save it. If it's diverging, it's not going to cause it to diverge. If it's converging, it's just five times whatever we get with that what's left in the argument. So that is divergent. And that's enough of a reason that it's harmonic. We've used that pretty much consistently. What do you think of as a possible method when you look at 5C? And I don't really care right now if you're looking at your answers, just kind of looking at these problems as if you were looking at them for the first time. Partial fractions, decomposition. And when we did that on these kind of series, what type of series did we, I think, just about every time come up with telescoping. Now could you determine convergence without using that? Yes. You could kind of multiply the denominator out, 5 over n cubed plus some other, or n squared plus some other stuff. You could use a p-series, right, comparison test. You could determine convergence, but the best way to determine the value to which it converges is to decompose it into partial fractions and write out several terms of the telescoping series. See what remains, see what drops out. 5D, that may be the easiest problem on any test this semester. Some of you rejected that gift, that was a gift. What is the deal with 5D that makes it less than a minute to do that problem? It's the limit. The limit is one big. If the limit of the nth term is not zero, we know automatically it diverges, which is the converse, I mean the contrapositive of the theorem that's in the book which says if it converges, then the limit approaches zero. Take the contrapositive of that. If the limit is not approaching zero, then it does not converge. So I think they call it a test for divergence in the book. So 1n over 3n, the limit is one-third, which is not zero, therefore it diverges. That's the end of that problem. Integral test, if it doesn't say use the integral test, one of the ways you might want to look at it is this argument, if I had x's instead of n's, would that be integrable? If it is, then you could use the integral test. So you have an n squared or an x squared in the numerator, n cubed or x cubed in the denominator, looks like the numerator could potentially be the derivative of the denominator, so that'd be a natural log format. I think this ended up diverging, right? Is that correct? Because we had a natural log of some number and the number itself was getting larger and larger, therefore the result was not converging. What is your line of thought in 5f? When you look at that, what do you think of? Okay, comparison, maybe. I would recommend when you first look at that the plus two is not going to be very much of an issue, right, for large values of n. So it's really kind of 1 over 3n in a lot of ways, and what is 1 over 3n? That's harmonic. So we're thinking that this diverges. If we're going to use the comparison test, what would this argument have to be in relationship to just 1 over n or 1 over 3n in order for it to also diverge? It'd have to be greater, right? It's got a larger denominator, which makes the value of this argument smaller, so even though it kind of falls out of the comparison test, we still have a general idea that it is divergent. So the next kind of more powerful test that will give us a conclusion since the comparison test fails is the limit comparison test. And it doesn't matter which one you put in the denominator, which one you put in the numerator, if you get an answer and one of them diverges, then they both diverge. If you get some finite answer and one of them converges, the other one also converges. No, in other words, if you get, let's say, five-thirds and the one that we knew about was, in fact, convergent, well, either the one in question is either five-thirds times the one that was convergent or, if we put them in the other order, three-fifths. Either way, it's also convergent. It's either five-thirds as large or three-fifths as large, but in either case, it's still convergent. Isn't the answer that it converges to? No. That's correct. So we get a decision, but we don't get a value to which it converges, because this one does not converge, right? No, it does. So if it's... Can you do that one? Oh, we've got the... Yeah, you've got the solution. Yeah, okay. Go ahead. If you get an answer, then it's convergent. If the one that we're comparing it to in the limit comparison test is also convergent and you get two-thirds or 11-seventeenths or whatever you get, the one in question also converges. So it's not really... So you get the one in question to be related to the one that you know about by some numerical scaling factor. So if the one diverges that we know about, and we find that this one is two-thirds of that or three-halves of that, either way, it doesn't change the fact that it's still going to diverge. How would you never get a number? How would that... If it does not exist. Or really, I think we've got some potential problems if you get zero, because if you think about that zero times a divergent series, is that going to diverge or converge? I don't think we really know a whole lot there. But two-thirds, five-seventeenths, those are all fair game. Does not exist and zero are potential problem quotients. Obviously, the bonus there is a third order. You're not responsible to solve third order differential equations, but some of the same techniques you used in second order, which is the earlier part of the test. You should be prepared to, as you look back, do all three cases of second order differential equations. The three cases are where the characteristic equation had two distinct real roots. That could be with a homogeneous or non-homogeneous. Be prepared to do that. Find the two R values, decide on the nature of the solution. If it has a real root, but it's a double root, what does the solution look like there? And the other one is what? Two distinct real roots, a real root that is a double root, imaginary or complex root, and that's when we enter into the trigonometric type solution, the oscillatory solution. First one is homogeneous. Second one is also homogeneous with some boundary or initial conditions. And the second one, or the third one, is non-homogeneous, kind of left me there for a second. I haven't looked at it for a few days. The right side may not be, so you don't just look at this for review. The right side could be a polynomial, which this is. What else could appear on the right side? Some trigonometric, and really not any trigonometric, but sines, or cosines, or sines plus cosines is kind of the subset we looked at in here. And what else? Exponentials. Now if it happens to be a constant, a constant is really a polynomial, so that fits into that category. So you would need to find the homogeneous solution, the particular solution, the final solution is the sum of those two. And then we had a spring problem, an application problem, remember in a spring problem, Hooke's Law, the force required to stretch or compress a spring is directly proportional to the amount you have stretched it or compressed it. Any issues, questions about any topics that you see on test three? So that's going to be a pretty good review for you. With the exception of number three, you're going to have to look elsewhere to get an exponential equation on the right side and or trigonometric sine or cosine on the right side. Questions, issues, test three, content. Alright, test two. So I came home last night and my wife had the TV on, and I don't know what time it was, probably about 5.30, and this was on. This class was on, and she goes, she's just shaking her head when I walk in, she goes, I don't understand you. She goes, how do you get so excited about math? She goes, I have no idea what you're talking about, and you just really seem to be all into it and stuff. I said, well, no. You made the decision to kind of get married 27 years ago, so you know what you were getting into, some weird math person. So I do, I like it, I like this stuff, and obviously she doesn't. Really getting to be a marital problem. And she's from Carolina, so I don't know, it compounds the situation. Test two, orthogonal trajectories. We have a derivative, we want to, or we have an equation, I guess here, and we want to solve for the derivative. We want an orthogonal trajectory, orthogonal means perpendicular, we want to flip it and negate it, and the one thing that happens in this problem is that if you have a k in your derivative, dy over dx, you don't want that k to appear in the flipped and negated version, you want to be able to get rid of that k. What is k in relationship to the original x and y quantities from your original equation? So this is an example where you had a k in the derivative. You can flip it and negate it, that's fine, but don't try to anti-differentiate it in order to solve for the orthogonal trajectory, you need to get rid of the k. So k in this problem was replaced by y over e to the x, and the nice thing about that is that it drastically simplified the thing that you had to anti-differentiate. So instead of being this ugly thing with e to the negative x's in it and k's in it, it ends up being 1 over y, that's a lot easier to anti-differentiate than had you not eliminated k. Remember, when you do the anti-differentiation of both sides, kind of it's a separable differential equation format, is that you include a plus c on one side or the other. At that time in the solution, don't think you can recoup that at the end of the problem, just put a plus c or k on your last step because you're going to do things beyond adding the c or the k. In this case, doubling both sides and taking the square root of both sides. So that c or k could be under a radical, could be in the numerator, in the denominator depending on what algebra solved for y from the point in time where you anti-differentiated the cope. After you took the negative reciprocal, what did you do in the next step? That's a separable differential equation. So multiply both sides by dx and multiply both sides by y because you want to get the y's and the dy's on one side and the x's and the dx's on the other. And that would have to be a separable differential equation for us at that point in time. I don't think we had any other exact techniques that we did at this particular stage in a problem. Is that okay? Second one, center of mass. I see center of mass as a potential exam question. I mean there is some memorization there. If you had it memorized for test two, you might want to kind of refresh it. Rememorize it if that's the case. I think that's something you might see in another math or math related class, center of mass. Force due to liquid pressure or hydrostatic pressure, number three, that's potential. Memorization there, just some words, the density of the stuff, doing the pressing, the depth of each horizontal strip in terms of variables that we set up on our own axis system and then the area of each horizontal strip. And then the accumulation of all of the hydrostatic pressures on each of the strips of each of the kind of described areas, we don't have to worry about that, that is added up through the definite integral. Also be prepared on the exam for directions that say set up only. It is, I don't think time will be a factor, you have three hours to do the final exam. But it is a time saver and if I've found out if you know how to integrate something like this one in another problem, I don't need to necessarily find that out in this problem as well. For differential equation, but this is Euler's method, I think that's a method that we'll have some carryover beyond this class. Any iterative approximating technique like this? I know you'll revisit that, certainly if you go to math 341, which Sarah was talking about before class, she was looking forward to all the other math classes that she would be going to beyond this one, so you'll see that in 341, this Euler's method. Just remember that as you compute kind of the intermediate x and y values along the way, that you're using the prior information. If you're trying to find x3, y3, you're using x2, y2, so that's any time you need to sub in an x value, using all the previous information, none of the newly found information along the way. This is one, and I kind of like to ask the question this way, obviously I ask it on the test this way, so it might appear on the final exam this way. Here you've done this approximation in part a, using Euler's method, go ahead and solve this exactly using separable differential equation techniques, see if by chance our approximation is in fact related to the exact value, so it kind of serves as a little bit of an internal check, sometimes the incremental approach to Euler's method, you start to deviate from the exact curve, and it's kind of hard to get the answer back close again, but at least it kind of tells you if it's reasonable, so I like that method of asking that question and kind of get two birds with one stone there too. And everybody's favorite, the tank problem, so I have never given a Math 241 exam without a tank problem on it, and here's a hint, I'm not about to start now, I'm not ready to turn over a new leaf, so tank problems will be on the final exam, tank problem will be on the final exam, this isn't a literal tank, but the room here with a certain volume to the room is our tank for this problem, just remember to pay attention to the stuff that in the problem that we're trying to analyze, in this case it's carbon dioxide, so we need to pay attention to how much carbon dioxide's coming into the room or tank, and also how much is leaving the room, so it's the rate at which the stuff that we're paying attention to is coming in, minus the rate at which the stuff that we're paying attention to is going out. Rate in, minus rate out, certainly you want to pay attention to the concentration of the stuff coming in, as well as the rate of flow of the stuff coming in, same thing on the stuff going out, why is this a problem in this particular part of our course? Because once you get it set up, if you look at problem five, you should get a separable differential equation, right? And usually the, if you establish your variable, like in this problem, C is the amount of CO2 in the room at any time t, that will come into play somehow in your differential equation. Notice the concentration of the rate at which carbon dioxide is going out in the solution to this problem, how much is in the room, C, out of the total volume of the room, 16,000 cubic feet, wasn't it, yes, so that gives us a concentration. And these have some little tricky parts to them because you're going to have to exponentiate both sides and then put in the initial conditions in the problem. And we also had a problem six, you had your choice, Newton's law of cooling, and a logistic population problem, I guess in a sense each of them made it to the test, some of you chose one, some of you chose the other. So they have a chance, so if you did the first number six, you want to either make sure you study this or look elsewhere, back at your notes, back in the book for the logistic growth problem. Anything, any formulas that were given to you on the tests throughout the semester will also be given to you on the final exam. And in like fashion, anything that was not given to you that was expected for you to know on the tests will be expected of you on the final exam as well. You know, I always have to do this. So if we had a three by five card or a sheet of paper on the final exam, how many of you would like to have that? Just a little kind of formula sheet, three by five card piece of paper, okay, we're not going to do that. I was just kind of curious to see. But that's nice that you would like that. I appreciate knowing your opinion, but we're not going to do that. Sorry, you knew that was coming in here. You knew that was coming. It just kind of adds fuel to that fire that he's just a mean spirited person, sorry. So that's how I kind of raised my children. Honey, you want to go get some ice cream? You do? We're not going. What are you talking about? We're not going to get ice cream. I didn't do that, really. I didn't do that. We have to do that once each course, so we've got it on tape now. You know, I happen to watch the ones that happened in last semester. That you... It sounds worse on tape. Pretty horrible. It's so bad. That you were laughing at me? Yeah. Okay. Do we have any snorting? We don't have any snorters. All right. Test two. Anybody? Any issues? Any...that's kind of loaded. Test two is loaded with stuff. Applications. Yeah. Yes? Are you going to give us any evidence of which type of...all of the problems that we're going to have? You know, this lindrical oil tank, one like the pressure one, that's compared to the work problem, because there are like six different types of those and it's pretty cumbersome to... Yeah. It'll kind of be a work or hydrostatic pressure. Work is you're integrating force times distance, right? So it could be one of those where you're kind of...you've got this tank full of water and you slice it and then you figure out how much work is required to move. One of the slices to the top of the tank, one of those, force times distance. Or this one, density of the stuff, doing the pressing, take a look at horizontal strips parallel to the surface of the liquid, what is their depth and then what is the area of one of those strips, invariable form. I mean you're not...you don't want a number. So I'd probably come from one of those two. Anything else from test two? All right, test one, a lot of this was probably stuff that you had or at least potentially had, maybe kind of in a hurry at the end of math 141, trig integrals, powers of sine and cosine, decomposing into partial fractions, which we also then use later in this course with series. Trapezoidal rule or Simpson's rule, I think that was technically new. I don't think that was covered in 141. If I remember right, 141 ends at 5.8 in the text and trapezoidal and Simpson's rule are 5.9, so that's kind of new material. Table of integrals, subset of the table of integrals is possible. Match up the integrand after you've kind of made it match up exactly by correcting for coefficients that aren't there, that you need, kind of a comparison of what you have after you doctor it up to the table of integrals. That shouldn't be a difficult problem, but you may have to kind of doctor up the integrand to make a match. Improper integral, we had a variety of those. You may want to look at the one that kind of made it to the test and then look back in your notes or at the book at some other types of improper integrals, some other things that could happen that would make an integral improper. Area, I guess bounded between two curves is what it looks like to me. Number six, and if you took that area and revolved it around the x-axis or the y-axis, what would be the volume of that region? Both of those are, I think, important models in Calc 2 because they are accumulation models, that whatever it is you're talking about, you're talking about an area of a region, you slice it up into skinny little rectangles, you describe the area of one of those skinny little rectangles, let the calculus do the work by adding an infinite number of those skinny little rectangles together. Same thing with volume, you slice it up, perpendicular to the axis of revolution, they get excited about areas, I wish I could get them excited about volumes. Slice it up, perpendicular to the axis of revolution, describe the volumes, remember they might be solid discs, they might be washers. I think probably I'll steer clear of cylindrical shells. So solid disc or washer, when you slice it up, that's what each disc is going to be. And length of a curve, there's technically I guess three ways there for length of a curve starts out being 1 plus first derivative squared, right, is one of them, if you have y in terms of x and that works, it might also be 1 plus derivative of x with respect to y, you can integrate that with respect to y from a y value to another and I think this is parametric in it on the test, what's it look like, right, so it kind of depends on what you're handed in the problem, all of them are kind of rooted in the same thing, this is kind of a description of an individual hypotenuse and then we use the accumulation model to add up all these hypotenuses, if that's a word, from starting point to ending point. Any questions, issues with test one, that's a lot of stuff. And I honestly, I think final exams are helpful, I know you don't think that in the place where you are, try to convince yourself that they are because they kind of force you to look at all this stuff again, to hopefully see that this course, even though it is somewhat disconnected, it is all connected and some of it's connected that the first time we had a concept, let's say partial fractions, we didn't know how we were going to see that again in a sequence and series problem, but we did see it again, it's the same technique, it just helps us solve a completely different kind of problem. So the course is connected and I hope that you find that to be true as you prepare for the final exam. Now the date for this group of students for the final exam is Friday, May 1st from 8 to 11 and that is in this room. And I know that a couple of you have issues getting here at 10.15, make sure you set a backup alarm because 8 is very different from 10.15. Questions? I bet you, you folks, I sent a couple emails out to you, they were kind of harsh initially that if you weren't going to come to remember those emails, because I had some people in the 141 class that just kind of disappeared and it just looks weird when you look at the tapes and this person's never here and she's never here and that's why I was trying to make sure, plus I knew there were some people that wanted in this class that it was full by the time their registration window came open. So hopefully it didn't turn out quite as badly as you might have thought initially. I think that you have been a very pleasant group of students. I don't know if the camera affects that, it kind of affects me. I'd probably be a little bit looser, maybe a little bit less guarded in some things I say if the cameras weren't running, but hopefully it was a good course for you and you can finish strong with a strong final exam grade. If you are on the border, and that is not just for students in this class, but all my classes, somebody is on the fence between a B and a C and they really do the final exam right. I mean they get a 92 on the final exam and they're on the border between a B and a C. What is their final grade? They get a B. I mean they prove that they deserve that if they're, but you've got to be on the border. I'm not going to take somebody with a D that gets an A on the exam and say, well they get an A. I mean that's not representative of the body of work so to speak, but if you are on the fence, the final exam grade could in essence be more to you in terms of waiting than just the waiting that it's given as far as the percentage. So finish strong and I will see you on the first.