 Welcome to our review for Exam 4 for Math 1220, Calculus 2 for students at Southern Utah University. As usual, I'll be your professor today, Dr. Angela Misseldine. This is our fourth and last midterm exam for Calculus 2. There is one exam after this one, that is the final exam. It's a comprehensive exam and has some important differences that we'll see in the future. This video, of course, will focus on Exam 4. The topics for Exam 4 are going to be those sections we've covered in lectures 32 through 42. There's mainly two topics being covered on this exam. There's the topic of polar coordinates and polar functions. These were the topics we had discussed in lessons 32, 33, and 34, and then the rest of this topic, the rest of this unit is actually being on the topic of series. A series we had introduced in lesson 35, talking about sequences, then formerly we introduced series in lecture 36, and we continue that on through lecture 42 there. These are the main two topics that you're going to see on this exam, a little bit about polar functions, mostly about convergence of infinite series. I should make mention that power series, that is series that involve variables, will not be covered on this exam. It will show up on the final exam, as the final exam will go all the way to lecture 50. But for this test, Exam 4, we just go through series, nothing about power series. The time, place, and manner of this exam, it changes from semester to semester, so I'm not going to include that information in this video. Instead, go to the exam syllabus or to Canvas itself to find out some of those specific information, that information specific to the semester. This video is going to focus on the contents of the exams, that is the types of questions you're going to see, and structure of the exam as well. This exam is going to contain 11 questions. Honestly, most of them are going to be found in the multiple choice section. Seven of the questions on this exam are multiple choice. And as such, the multiple choice questions are going to be a little bit more heavy than they've been on other exams. Each multiple choice question will be worth six points. And as usual, multiple choice is all or nothing, so you want to be very careful to get those questions correct. This exam will have a very short, short response section, only two questions, but each of those questions will be worth 10 points. And then the free response section will also be quite short. There are two questions. The question 10, which will be worth 16 points. And question 11 will have two parts, which are 10 points each. So question 11 is worth 20 points total. So this exam is going to feel shorter than past exams we've seen, but each of the questions is worth a whole lot more points. So you do want to make sure that you do pay attention and get these questions right. And so without further ado, let's talk about the specific types of questions you should expect to see on this exam. So starting with the multiple choice section. Question number one, you'll be given a polar graph, something like this, for which you'll probably be given a lemason of some kind. So your function might look like R equals A plus B sine theta, R A plus B cosine theta, something like that. We've also done some of these roses and lilies, whatever you want to call them, R equals A times sine of A theta, B theta, something like that, or maybe cosine theta. That's probably what you're going to get, but I'm not going to say it's going to be necessarily one of those things. This question, you can try to memorize familiar parameters and formats of functions. But honestly, I really want you to treat this as like process of elimination as you're given this polar graph, which of these options have to be ludicrous? Only one will make sense. The others you can rule up by examining the graph. Like for example, when you look at this graph right here, when the angle theta is equal to zero, the radius is equal to two, right? That's something we can very easily see. We can also see that when the angle is pi halves that the radius here, what is that? One, two, three, four, five. So you can look at the polar coordinates and interpret the graph. And by that, you could eliminate possibilities until there's only one left. That's how I really want you to approach this one. Of course, again, if you want to memorize, like what's the basic format of a limousine or a cardioid or a lily, one of those flower drawings we've done, you can do that, but really process elimination is going to be your best tool on this one. This is just about polar graphs. And so you can go back to lesson 32 for some further practice on polar coordinates and their associated graphs of polar functions. Question number two is gonna ask you to compute the limit of a sequence. And that's exactly like we did in lesson 35. 35 introduced us to sequences, which are discrete functions. And in particular, we're interested with the limit as n goes to infinity. Be comfortable with those type of calculations. There's a lot of different variety you can see here. It is a multiple choice question, so you don't necessarily have to provide the argumentation. You just have to get the correct answer. Now, if the sequence is divergent, that is the limit doesn't exist, you would select option F here. But if it is convergent, you need to then select the correct limit, which will be one of the other five options. So use your limit calculations. A lot of these will come down to basically calculus one problems, but there are a few things like factorials that are purely these discrete functions. So you wanna be familiar with those going forward. Remember in factorials, the product of one times two times three times four all the way up to n. So something like that could come into play, all right? Question number three is gonna be a question about polar functions again. And in particular, it's gonna ask you essentially to calculate the derivative of a polar function. I hope I can spell that word correctly there, derivative. So in particular in this one, since we want you to find the slope of the tangent line and that situation, notice it's asking you to compute dy over dx. The slope of the tangent line is the change of y with respect to x, it needs to be rise over run. Now in the polar setting, we handle this just like we do with the parametric setting. You're gonna use the chain rule to count and so you calculate dy over d theta over dx over d theta. And you're using the parameterization that x equals r cosine theta and y equals r sine theta. And so this always is gonna turn out by the product rule. You're gonna get an r prime sine theta plus r cosine theta on the top. And in the denominator, you get an r prime cosine theta minus r sine theta like so. And then you can compute the derivative. So question number three, we'll probably have you compute the derivative and do something with this. But this one is we're looking for the slope of the tangent line. We probably need to plug in the point of tangency and go from there. Polar derivatives showed up in lesson 33. So go back to lesson 33 and the connected assignments to lesson 33 to see what would be appropriate questions. What are possible questions you could see here for number three. All right, kind of wrote over this one a little bit. So let me erase some of this. For question number four, you'll be asked to determine whether a series is convergent or divergent. If you believe the series to be divergent, you would select choice F. If the series is convergent, you actually need to evaluate what is the sum of the series. Now with that in mind, right? You have to find the exact sum. There's basically only two types of series that we are capable of finding the exact value as opposed to an approximate. And those situations are geometric series, which geometric series we introduced in lesson 36. The thing we've learned about geometric series is you take the sum where N equals zero to infinity here of a geometric series, A times R to the N. This is gonna add up to be A on top and then one minus R on the bottom. So the things to remember here is A is the first term of the series. So even if you don't start at one or start at zero, that's okay. A here is just the first term. So plug in whatever this number is into the sequence. You get a number that goes on top. And then on the bottom, you get one minus the ratio. A ratio is how is the geometric series progressing, right? This series, remember, will be convergent. It's convergent if and only if the absolute value of the common ratio is less than one. If your ratio of a geometric series is too big, it'll be divergence, you would select choice F. If it is a small ratio, then you can compute the series using this formula and you would select that option if you have a geometric series. The other family of series that we can get an exact value from were telescoping series. And we saw those in lesson 37. So with the telescoping series, because there's a bunch of cancellation of terms, you can actually come up with an exact form for the partial sums. So because after all, the infinite series by definition is equal to the limit of the partial sums where SN is the sum up to the number N at that point. Telescoping series, we can get an exact form for the partial sums because most of the terms cancel out. Take the limit as N goes to infinity and that would then give you the value. If that limit doesn't exist, that means the series was divergent. So you'd select divergent. If that limit does exist, then that would be the sum of the series and you would select that option. With the telescoping series, remember you have the head of the beast, but you also have the tail. If the tail vanishes, that will mean it's convergent, but sometimes the tail doesn't vanish. But even if it doesn't vanish, it still could be convergent. Go back to lesson 37 and the associated exercises for some more examples of these things. The last question on the first page, question number five will ask you to compute an integral involving polar functions of some kind. So this question number five will ask you to do some exercise associated to lesson 34. The two main topics that you should be prepared for in this one is to calculate the area of a polar region. So the area of a polar curve, that is the area inside of a polar curve will be the integral from alpha to beta. These are measurements of theta. You get one half R squared d theta. That's if you're finding the area inside of a polar curve like this question's asking you. It could ask you to find the area between two polar curves for which then if your outer radius is capital R squared and your inner radius is little R squared, the area function will then be a difference of squares, much like the washer method, it's a difference of squares. You can find the area there, in which case then alpha and beta would then be the intersections between those two polar curves. You would set the two Rs equal to each other and solve for it. That would give you those parameters alpha and beta. So most likely you're gonna be asked to find a polar area. Those are the type of questions we did the most often, but you could also be asked to arc length for which the arc length formula is the same basic thing we integrate the DS. Now in the situation of polar functions, the version of DS that you're gonna be interested in is gonna be the square root of, you're gonna get an R prime squared plus R squared d theta, where here R prime means dr d theta. So we take the derivative of theta, sorry, the derivative of R with respect to theta. That then gets us through the first page of this exam. There's two more multiple choice questions. Questions six and seven are gonna be similar. It's gonna give you six infinite series. And then you have to determine whether the series is convergent or divergent by a specific test, okay? So you do need to make sure you read the instructions. It could ask you which of the following series is divergent or it could ask you which of the following series is convergent, do pay attention to the directions. Amongst all six options, there always will be some convergent series and some divergent series. So do ask, do notice, is it asking for convergence or divergence because it can switch between the two questions here. And then it's gonna mention a specific convergence test. So number six, this version, it asks, is it convergent, or sorry, excuse me, which series is divergent by the divergence test? Number seven asks which of these series is convergent by the ultimate series test? You need to know the assumptions of each of these convergence tests and then what are their conclusions? With regard to the divergence test, the divergence test says if the sequence of terms in the series converges to some number L that's not equal to zero, then the series where you add together the A sub N, that would then be divergent, okay? That's what the divergence test shows. The divergence test cannot show that a series is convergent, it can only show that a series is divergent if the sequence of terms doesn't go to zero, right? So of these six options, you have to find the one series which is divergent by the divergence test. Now, clearly if any of these series are convergent, then you can throw them out because they're not divergent. But then you have to find a series that is divergent by the divergence test. So I wanna throw, I wanna mention the distractor right here. If you look at F, F is the harmonic series. We know that the harmonic series is divergent, but the reason it's divergent doesn't come from the divergence test. It actually comes from something else like the P test, the integral test is how we originally showed that this series was divergent because the sequence one over N does go towards zero, but nonetheless, even though the sequence converges to zero, the series is divergent. So even though it's divergent, it's not divergent by the divergence test. So the correct answer would not be F. So as you're working through number six and number seven, make sure that you pay attention to, well, is it convergent or divergent? And make sure it's convergent or divergent by the test that's provided. With the alternating series test, for example, the alternating series test cannot show that a series is divergent. Typically these two get coupled together, but as alternating series can show that a series is convergent, but it only applies to alternating series. Like the first one here isn't even an alternating series. So if it's convergent or not, it can't be proven convergent by the alternating series. For example, B here, this is a convergent P series because P equals two in that situation. So it's convergent, but it's not convergent by the alternating series test. The alternating series test would not apply to this one because it's not even an alternating series. For the alternating series test, what you want is you have to have an alternating series. So you should expect to have like a negative one to some power. And then if you look at the sequence, ignoring the negative one there. So if you look at its absolute sequence, you have to argue that that sequence is decreasing towards zero. That is it converges to zero and is eventually decreasing. It's a multiple choice question. So you don't have to show all those details, but be aware, it's six points. So if you make a goof, you lose all of those. So be meticulous on these ones. It has to be convergent by the alternating series test. Even though this one's convergent, the alternating series test doesn't apply here. This one's divergence, the harmonic series again. So you have to find the one and only one sequence which is convergent by the alternating series test. So what are you gonna see on questions six or seven? Well, you could see the divergence test. You could see the alternating series test. Those are options you could see there. I may mention to it a little bit, but you could also see the P test. For example, this series is convergent by the P test. This series is divergent by the P test. The other four options, whether they're convergent or not, doesn't matter because the P test doesn't apply to them. So the P test is an important one to look out for here. Also another test that you could see here is the test for absolute, absolute convergence. Because as we saw previously, that a series is absolutely convergent, then it's convergent. So sometimes you can prove that a series is absolutely convergent easier than you can show that it's convergent. That's, we've seen some examples of that. Like for example, this friend right here, right? This one, I mean, this one's not absolutely convergent, but if I put a two on the bottom, a sine of n over n squared, that would then be absolute convergent. And that's actually easier to show that it is convergent since absolute convergence implies convergence, you can use the test for absolute convergence to help you out in that situation. So let's remind ourselves where we can find some of these things that we want some more practice. If you're looking for the divergence test, that was actually introduced in lecture 36. You can go back there for some more examples and always go to the company assignments also. The tests for absolute convergence showed up in lesson 42, excuse me, 41. Excuse me, 41. 41's when we did the ratio test, which absolute convergence was introduced there. For the ultimate series test, that was given to us in lesson 41, excuse me, 40, 41 was absolute convergence there. And then the P test was given to us in lesson 38. So question six or seven is gonna give you two of these four convergence tests that it's gonna ask you, is it convergent or divergent by these tests? So you should be very comfortable with these tests. We have fairly simple checks like the P test. Oh boy, that's a very simple test to check. And so that's why it justifies as a multiple choice question. All of these tests for the most part are fairly simple checks. Perhaps the test for absolute convergence is the hardest one, but it's not so bad, which is why they're put as multiple choice questions. There's really not a lot you have to argue. And so a lot of this could be done as mental math, but feel free to write some notes in the margin if you need some more space. So that then ends the longest section of the exam, the multiple choice. Remember, seven questions, six points each. So there's 42 points eligible for the first response, excuse me for the multiple choice section, a little bit less than half of the points of the exam, out of 100 points, of course. We then get to the short response section. There are two questions here, each worth 10 points. So there's 20 points total. Question number eight, this one, unlike the previous ones, these questions in the short response section are still gonna ask you to determine the convergence or divergence of a series. Just so you're aware, three sections of this exam come from polar functions, and all of those questions were in the multiple choice section. We had one about polar graphs, about polar derivatives, about polar integrals, those were all in the multiple choice section. The short response and the free response will have no questions about polar functions. So there will be no more polar functions for the rest of this exam at this point. The short response, like I said, you'll be given two series for which, for formatting purposes, I might phrase in the following way, here's the sequence, and then determine the convergence of the series, like so. These short response questions do have a multiple choice part to it. So you have to decide, is this series convergent or divergent? If you believe it to be convergent, select that option. If you believe it to be divergent, select that option. So in that respect, it's kind of like a multiple choice, but unlike the multiple choice sections, you do have to provide your justification here. Now it's a short response question, so some of these things are already provided to you and you kind of have to fill in the blanks. So like with question number eight, the way that this is formatted, it's like, okay, here's a series, excuse me, here's a sequence, is the series convergent or divergent by the limit comparison test, okay? This one says limit comparison test, but randomizing this question, it could be the comparison test. Question number eight will ask you to either use the comparison test or the limit comparison test to determine the convergence or divergence of a series. So question number eight will cover an example from lesson 39 about the comparison test or the limit comparison test. The two tests are very similar, but they're not exactly the same thing. With regard to the comparison test, all right, let's remember what the comparison test tells us. So we have two series that are related to each other. So we have a series of A sub N and let's say that's less than or equal to the series of B sub N. Now if B sub N is convergent, that is it's a finite value, then that means that the smaller series will likewise be finite. And so therefore the convergence of the series B N will imply the convergence of the series A N. On the other hand, if the series of A N is divergent, that is it adds up to infinity because these do have to be positive terms of the series there to apply this. That actually would imply that B N is likewise divergent. So the comparison test goes in these directions. If the bigger series is convergent, the smaller series is convergent. If the smaller series is divergent, then the bigger series is divergent. When of course all of these are greater than or equal to zero. That's all that the comparison test can do. Now for the limit comparison test that you see on the screen right here, the limit comparison test says that if the limit as N goes to infinity of A N over B N is equal to one, then the convergence of the series of the A Ns will be the same as the convergence of the series of the B Ns, okay? Which of course this limit doesn't have to equal one. It just has to be a positive finite number. But for this one, I'm asking you to find the exact sequence which will make the limit of the ratios equal to one. So what's like the simplest perfect match? You're gonna write that in here. It's like B N equals yada, yada, yada. You tell me that. Now that series B N, if you're doing a comparison, whether you're doing the comparison test or the limit comparison test, that new sequence you're comparing to, B N should be relatively simple. It probably should be either a P series or a geometric series. Something super simple that we can very quickly find the convergence of that series. So simple and I even asked for it. Like if you came up with like, oh yeah, this sequence is going to be a P series. It's gonna be one over N to the fourth. I'm not saying that's the right answer, but if you think that the sequence that you're gonna compare to is one over B to the fourth, well then that series as a P series with P equals four would be convergent. So you'd say that, okay, this series is convergent because it's limit comparable to a P series which is convergent. So to get full credit here, you need to tell me is it convergent or divergent? And what's the series that you compare to? What's the sequence you compare to? By all means there's space here to provide work. If there's more that you need to show here. And that's a good idea of course. So that's question number eight. Question number nine will be a very similar format for which it'll give you a sequence and then you have to determine whether the series associated to that sequence is convergent or divergent. And it'll specifically tell you which convergence test you should be using. Now this one as you see on the screen is formatted using the integral test. The integral test tells you that the convergence of a series is identical to the convergence of an integral if the sequence of terms in the sum is positive is positive and decreasing towards zero. Sort of similar, similar assumptions to the alternate series test there. So then therefore if you want to know if you think this series is convergent it's because this integral equals a number like say it turned out to be two and it would be divergent if you think this is equal to infinity, right? This is a calculation here and for which you might wanna provide the details here much like with this one you might wanna provide the details of the limit here to justify this thing, right? Showing your work can get you partial credit on these questions if you make a mistake somewhere the partial credit can cover your base there. So that's what the space is provided for. Do show your work as necessary. Now this version gave us the integral test for which the integral test was first introduced in lesson 37, but really you wanna go to lesson 38. That's where all the good examples are but there's some stuff in 37. So look at both if you need some more practice with integral test. This also could involve either the root test or the ratio test. That's another question that can come up here which of course this was given to us in lesson 41 with the ratio test remember the ratio test you wanna take the limit as n goes to infinity where you're gonna take the absolute value of a n plus one over a n you wanna compute that limit the limit of ratios and consecutive terms you're gonna compute a limit here and you want that limit to be less than one. If this limits less than one then the series will be convergent if it's greater than one then it actually will be divergent and if it's equal to one then it's inconclusive but I won't give you that one using the root test what you wanna do is calculate a limit you're gonna get the limit as n goes to infinity here the nth root of a sub n here you get a limit you want that likewise to be less than one if it's less than one the root test will give you that it's a convergent series if it's greater than one then it'll be divergent if it's equal to one it's inconclusive and so you can imagine what would you have to do for a ratio test you'd have to actually calculate this limit I'd want you to put the answer on the line that's provided but it's probably not just something you can do super simple show all the work and that's how you can get full credit on these questions and that's why they're worth 10 points there are some limit calculations that'll be necessary for these problems all right now we move to the last page of the exam and to the free response the free response much like the show response is only two questions long but these questions are worth a lot more cause there's a lot more you have to show unlike the previous questions which there's a lot of scaffolding there basically there's a lot of hand holding like on the previous example I told you you have to use the integral test question number 10 this is going to be our street fighter question in fact this is street fighter two cause this is the second time we've had to do these street fighters we did with integrals before now we're doing a series with question number 10 it's worth 16 points that's a lot of points on this exam here you'll be given a random series and you have to determine whether it's absolutely convergent, conditionally convergent or divergent and in the blank provided you need to show all of your work any details that are omitted I will take away from these 16 points 16 means there's a lot of points possible and so omitting details means you can forfeit those points don't do that show all of your work there I need some explanations I need some calculations like if you're doing a limit comparison test what are you comparing to and what was the limit are you doing the ratio test well what was the limit associated to the ratio test much like we did on much like we said on some of those previous examples there also if you're using any convergence tests you need to specify what those convergence tests are and why it worked so let's say you're doing a limit comparison test you wanna do a limit comparison test on this it's like okay this thing's kinda complicated I'm gonna compare it to something else what's the new sequence you're comparing to tell me what that is what then would be the limit of the old sequence versus the new sequence there's some calculations there let's say you end up with a two alright it's like okay since this is a positive finite number this means the convergence of the series B sub N is the same as the series A sub N by the limit comparison test so you're gonna use the limit comparison test but why is this series this new series convergent maybe it's a maybe it's convergent by the P test because you end up with a P equal to three or something like that but that gives you convergence with it is it absolutely convergent or conditionally convergent we do have to distinguish between those two cases absolute convergence means that the absolute series converges the series of positive terms converges conditionally convergent means that it converges but the absolute series doesn't converge we saw some examples of this involving alternating series for example the series where you take the alternating harmonic series this is a conditionally convergent series it is convergent you can use the alternating series test to show that it's convergent but it's not absolutely convergent so you have to argue that this series converges again by the alternating series test but then you have to also show that the series the harmonic series is divergent and you can use the P test in that situation so to show that something that's conditionally convergent you have to show that the series converges but the absolute series diverges providing all the tests and reasons for that so there's a lot going on this problem which is why it's worth so many points I've seen too many students who've sacrificed points because they didn't do everything they needed to do don't do that don't give up your points there fight for every point there and that's what makes this a street fighting question anything could happen it could be a relatively short one maybe it's just a simple application of the ratio test maybe but it could be more convoluted anything goes on this one any tests you want to use you can use them just tell me what they are and you do need to do that for full credit then we get to the last question of this exam question 11 it does have two parts and each part is worth 10 points each and therefore collectively this problem is worth 20 points this problem is going to be about remainders and as such it's going to be about estimation we want to estimate the series because after all geometric series and telescoping series we can compute the exact value but for these other series we've done when we determine they're convergent we don't know what the value is but we can approximate it using partial sums so the infinite series where n equals zero to infinity of a sequence a sub n this will be approximately the same thing as the sum where n equals zero to some number let's say we stop at m a sub n right here this is the partial sum s sub m and the thing is this is a good approximation if the remainder n equals m plus one to infinity of a sub n so this is what we call the remainder rm this happens if this thing it's absolute value is small is less than small I'll be more precise so you'd be like it's less than epsilon something like that anyways so if you're trying to find a remainder using the integral test the key formula there is that the remainder rn is less than equal to the integral n to infinity of f of x dx so you'd be interested in calculating integral like this this gives you an error bound for the situation and once you have an error bound you can then calculate like how large should n be to guarantee that your error is smaller than a certain amount we've seen things like that before so that's how you take care of the integral test you use the integral that helps you determine the convergence of the series you just change the lower parameter to be a generic variable you can solve for that if you're out using this error bound this is if you use the integral test to define convergence here if you use the comparison test be aware in that situation this actually has a much easier bound with the comparison test if you're like oh the series a sub n is bounded above by the series b sub n so you know that b sub n was convergent in that situation when this situation you can then use that the error of the series a n is going to be bounded above by the error of the b n series and so then you just make this thing get small so typically what you're going to do is if you did the comparison test you compare this to a geometric series for which you can actually compute exactly what that t sub n is going to equal to or if it was like a p series well the p series is actually a special case of the integral test so you'd use the error bound from over here in that situation so be prepared for that the last possibility is then if you prove a series is convergent using the alternating series test so you have something like negative one to the n times a sub n where a sub n is a positive number if you know that's convergent by the alternating series test this is a nice little thing here because the remainder in that situation r sub n is actually bounded above by a n plus one so you just have to make the next term in the sequence be sufficiently small and that'll guarantee that the error is small so those are the three types of remainders that you're gonna have to be prepared for integral test, comparison test and alternating series test question 11 will ask us to it'll ask us to compute a remainder an error bound I should say using one of those three tests and then find out what is the largest n that would guarantee the error is sufficiently small these type of remainder calculations for the integral test happened in less than 38 for the comparison test we did in less than 39 and for the alternating series test we did it in less than 40 and that then brings us to the end of exam four thanks for watching hopefully this helps you as you prepare for the exam do of course look at the resources on Canvas like the exam syllabus, the practice exam for further practice if you have any other questions about what will be on this exam or not please let me know and I'll be glad to help you as best I can best of luck on this exam