 We have this midterm tonight, my goal for today is rather than doing this stuff, I'm just going to leave you. I would prefer if you have specific things you want me to know, or if you're asking about them. But I can just remind you of everything you've done if you want, although it's a little hard to sit back in an hour. Yeah. Yeah? You want me to do it by example or can you just give me some words? Okay, so I'm just going to make up what I haven't done before. We'll say theta times the cosine of three steps. You want to figure out what it looks like. Just like if you're plotting a rectangular graph. What do you do? You figure out masses and nins. You essentially plot points, but you plot points in an intelligent manner. You plot points by finding important points rather than just randomly picking things and saying, gee, where's this goal? I don't know. So what we want to do here is essentially plot points in a smart way. So let me just, for this one, just for simplicity, just say theta is only positive. If theta is negative, that also will affect it, but let's just start with theta positive. Okay, so important points here. So these will be places where the thing changes direction or where it's zero. Because zero is an easy place to find. So when it's zero, it's going to be different than when it's not zero. So when r is zero, well that will mean that theta times the cosine of three theta is zero. So that means theta is zero where the cosine of three theta is zero. So the cosine of three theta is zero when? Well that's the same thing at three theta with some multiple of pi over two because the cosine is zero whenever we're at pi over two we're negative pi over two. So that means that theta is some multiple of pi over six. So that already tells us a fair amount of information. Where should I start making the graph? Doesn't matter. So it tells us that at zero and every multiple of pi over six, the thing comes through the origin. And that might be enough for you to figure out what the graph looks like already. I don't know, is it clear to people what the graph looks like already? Yes, no, maybe? Okay, and so let's analyze a little further. Between these points when it's zero, it's going to be positive and then negative and then positive and then negative and so on. Right? So we can look for, so if you want, look for when the radius is at its maximum and its minimum and that will be, well so we could take the derivative or we could just realize by a little bit of analysis that we know what the cosine looks like and multiplying it by theta is not going to change it that much except stretching. So the maximum, so the maximum of cosine is going to occur when the sine is zero. Right? So the sine of three theta is zero. So this is halfway between these points. And multiplying it by theta is not going to change that very much. So here, just by thinking a little bit, or if you want, you can take the derivative and set it to zero and see when the derivative is zero. I don't want to do that because it will take longer. But the derivative is easy. Cosine three theta plus theta sine plus three theta sine two theta plus whatever. So let's start making the graph here. So we already see it's important to know that it starts here at the origin when theta is zero. And then as theta increases, r will increase until theta equals pi over six. And at pi over six, so I guess I have divided this into thirds. So this is pi over six. I'll call it two pi over six which is also pi over three. This is three pi over six which is also called pi over two and so on. In each one of these little sectors, what happens? The cosine increases and then it decreases. And when I multiply it by theta, since theta is a positive number, it increases a little less because theta is less than one but then a little more. So I'll get a loop like that when theta is pi over six just from the cosine going up and down again. Is this clear? So now in this next region here between pi over six and two pi over six, the cosine increases and decreases but it's negative. And when I multiply by theta which is a positive number, this will increase and decrease but it will be negative. So in some sense we want it to go here but it's negative so it goes back here. And it will be longer than this loop because I'm multiplying by theta which is bigger here than it is there. Yes? No? OK. And then we just continue in this way. Then when we go between two pi over six and three pi over six, the cosine is positive because that's a cosine between two pi over three and whatever it is. I lost that. Anyway, the cosine will be positive again and it will grow and shrink. So we'll get a slightly bigger loop. And then in this next region between here and here. So this is supposed to be tangent here. In this next region we divide this into thirds between here and here. Well the cosine is three times this angle is negative. I'm multiplying by theta which makes it bigger. So I will get a slightly bigger loop down here. And then I will get a bigger loop up here. And then I will get a negative loop here which is even bigger. And then I'll get a loop here which is bigger. And it just continues in this way with bigger and bigger loops. So it's like a spiral of loops. So let me just draw it a little more quickly. It goes bigger and bigger loops like that forever. This is a pretty crappy picture but I think you get the idea. And this process is the same for pretty much any polar graphs. The point, what I want you to take away from in terms of these polar graphs is not, well let's see, I know that cosine 2 theta is a four leaf rose and that sine theta is a circle and blah, blah, blah. That's not really useful information. I want you to be able to understand the relationship between the equation and the graphs. If this were a smaller class I might ask you to actually graph in polar coordinates. It's very difficult to grade student graphs. Student generated graphs or anybody, you know, it's a little hard for you to tell whether that's the right graph or not even if you know where the right graph looks like. So it would be like asking you to draw a picture. Some people are good at drawing, some people are not good at drawing. It's hard to tell whether you have the idea. So practically asking you to produce a graph can happen in a class like this. But you should understand if I show you a graph whether it's right and you should also be able to, if I give you an equation and ask you something about it, be able to answer in a sensible manner. So for example, if I ask you to find the area of the largest loop between theta equals zero and theta equals 2 pi, that requires you to understand the graph but not actually produce it. So I'm not going to ask you to draw graphs. But I will ask you questions that may require you to understand the graph. More polar issues? Yes. Okay. Do you want me to find the area of one of these loops or something else? Yeah. Okay. Which one do you want? The largest one. The largest one. Okay. So which one is the largest one? The last one. The last one. Right. So the last one will be from theta equals, well, the highest value will be 2 pi. I mean, there is no last one because bigger than 2 pi is just a bigger loop. But the last one between zero and 2 pi is a well-defined concept. So, okay. I could also ask you to find the area of the whole thing. But that would be mean because there's a whole bunch of separate calculation here. So, find area of largest loop. And maybe I can't even do this integral. Well, it's parts. Okay. From, so let me just write, let me just say instead of say, find area, write an integral for the area of the largest loop for r equals theta cosine 3 theta with theta between zero and 2 pi. Okay. So the largest one is the last one. So is for theta less than 2 pi and bigger than 2 pi minus pi over 6, whatever that is. 9 pi over, you know, whatever 2 pi minus pi over 6 minus, yeah. Right? So now I guess that's 11 pi over 6. So that's what the largest loop is. And so now we need to think about what represents the area. Let me draw the picture not correctly just so that it's bigger. So we have this loop that looks kind of like that. And I want to find the area as it goes from 11 pi over 6 to 2 pi. And our area consists of adding up a bunch of little wedges like this. So that means it's going to be an integral from 11, 11 pi over 6. It's hard to write 11 pi. You can't even cross the 11. 11 pi over 6 to 2 pi. And the area of one of these little wedges is related to pi r squared. You can either remember that the part that I want is r squared, 1 half r squared, or we can calculate it. So I mean you can remember that this is 1 half f of theta. Well, f of theta is right there. So my r is theta cosine 3 theta squared d theta. So this is really 1 half r squared. And the reason it's 1 half r squared is instead of having a whole circle which would be pi r squared for 2 pi radians, I only have d theta of the circle. So I want d theta over 2 pi is the fraction. Instead of 2 pi radians, I want d theta of that. So I want d theta over 2 pi pi r squared. That's where this quantity comes from. And the pi is cancelled. So I can do this integral if you want. But it will take an extra 5 minutes. It is squared out. You can integrate by parts. And then you use the cosine squared formula, which is the half angle formula that's turned into 1 half, you know. Let me not do that because that will take another 5 minutes that I don't want to do. Is it theta over 2? Inside? I don't understand your question. If I put the 2 on the inside, it's still the same. The biggest loop is the last loop because this theta makes the loop bigger and bigger as theta grows. Over 1 over theta, then of course the first one will be the biggest. Yeah. You said theta over 2, so that is just half. What? Was there a question related to this? No. Okay. So let's move along. More questions related to polar? Okay. Questions related to other stuff? Yeah. Okay. So the general idea in work problems. So work is force times the distance that you apply. Right? So this is the piece of physics that we need in order to do this, and then once we have this we can turn this into math instead of worrying about specific physics. So if the question is work just pump water out of the tank or whatever it is do you want to do a work and pump water out of the tank or give you the general sentence? Huh? Just do one? So let's say that we have I don't know a conical tank we want a cone tank we want a different shape I don't care what shape it is that's fine. And let's say that it is 10 meters high 2? 2. Okay. So it's 2 meters radius at the top. And the tank is so we want it to be full, half full I don't care half full. Okay. So the tank is half full of water do we have a spout on the top which will just come right off the top? I'm not even saying there will be one of these problems on the test because we need to know how to do these problems maybe there's one of these problems on the test so asking will there be a spout on the test is a stupid question. Yeah. Do you want a spout? Okay. How long should the spout be? Give me a number. 3 3, okay. So there's a spout here 3 meters long and then the water comes out. Okay. So what we need to do we need to calculate the amount of work which means we need to calculate the force that we need to lift a little bit of this water all the way to the top and it's distance. So we're going to set this up let's let h be the height from the ground or wherever it is that the water is sitting and then what we need to do if there were just somehow just floating in the air a little tiny slab of water oh I guess I also need some things I need some more constants some more constants I need um I don't remember the density of water I guess it's a thousand kilograms per meter maybe it's not water maybe it's molasses so it's 2,000 kilograms per meter per cubic meter is the density of molasses okay so we have some it's not water now it's molasses or maybe it's okay so it's a big party and this is Guinness staff so it has a density of 2,000 kilograms per meter and you have a bunch of very thirsty people down here and the constant the gravitational constant is 9.8 meters per second square and everything is in meters or kilograms so we don't have to worry about adjusting units okay so now we want to pump this out so what we need to know is how much work does it take to take a little bit of water of some little height how much does it take to get it out so if we have a little bit of Guinness at height h we need to know several things we need to know the mass at height h and then the distance so it's the mass times the gravitational constant this is the force needed the distance that we need to apply that force which is from here to here so that would be 13 minus h because to the very top here is height 13 okay then this will be so this is the mass at h and it has to travel 13 minus h if you prefer you could use let's let another letter t for from top be this distance then this will be t and this distance would be 13 minus t so you have two things that add to 13 and you can choose which one to be from the top and the bottom so let's let this be h and then this here is 13 okay would you prefer it started from the top I don't care so we have to know the mass at a given height h and then the distance it has to travel here is 13 minus h and you could set it up using the mass at 13 minus h and the distance is t but 13 minus t is the same okay and so now we need to figure out the mass at a given height well the mass at a given height is going to be the volume at the given height times the thousand so this is the volume at a given height h times the density 2000 times g which is 9.8 and the distance we already have is 13 minus h and now we need to turn this into figuring out what the volume is at the height h so again I'm just going to draw the mass of this I'm going to put the central line in so this is 2 and this distance is 10 and we have a little triangle here of height h and its radius is r and we have a similar triangle 2 is to 10 as r is to h and I want to solve for r in terms of h because the volume here is pi r squared dh so I need to know r in terms of h so that means that r is h over 5 in this case so that means that the volume at height h is pi h over 5 dh because it's a little cylindrical slab of radius h over 5 and so now the integral to do the work is pretty easy we integrate this quantity pi h squared over 25 times 2000 times 9.8 times 13 minus h dh so that's the integral we do and what does h vary from well it goes from the very bottom of the tank where the volume is 0 to half way up which is 5 meters so that's the integral we have to do which is easel notice that you could also if you prefer we could go from what is 13 minus 5 8 from 8 to 13 pi 13 minus t square over 25 2000 9.8 t so if you prefer to measure from the top you were the one who got it right someone over here suggested measuring from the top you were the one who got it same this is just make the substitution do you want me to do this integral I mean let's just multiply it out plug it so is this alright and all of the work problems are of this nature you figure out what force you apply and a given thing and for how long you need to apply it and you integrate force times distance the brain problem the force is proportional it's kx and you figure out what the force is and how long you need to apply it if it's lifting a chain the force is proportional to the length of chain you have to let me hold that one for a minute yeah let's change this problem where the cross section is a square so instead of being a cylindrical tank it's now a pyramidal tank what will change all that will change is this little calculation right right here instead of being h squared over 25 the side length of the square will not be h over 5 it will be 2h over 5 and so this will be 4h squared over 25 everything else is the same because the side length is a square I mean the cross section is a square it's now a pyramidal tank so this cross section is a square and we need to figure out its side so the thing is exactly the same except this volume becomes a slightly different formula yeah it's the same the cross section is now so if I have a spherical tank we need to figure out what is the cross section there so now the thing that will differ is the relationship between r and the height because this is like a volume of a surface of revolution so we have the property that well I don't know who's r so let's say a spherical tank of radius what radius would you like 3 so we have a spherical tank of radius 3 so at a given height here well we find the distance from the center let's call that I don't know c which is halfway up minus h and this gives us a triangle and so c squared plus r squared sorry c squared plus r squared equals 3 squared and so r is 9 minus c squared square root and so the volume at well c is well if I measure from the center so let's instead of having the h here let's vary the ground so that the center is halfway up so c is the distance from ground level then this height distance seen from the center is pi 9 minus c squared it's the same idea always figure out what the cross section is and whether the cross section is a circle or a square or a triangle or the head of a cat is and you just calculate the area so I am assuming that you guys can calculate areas of simple geometric shapes no more complicated than a triangle so no octagons although you should be able to figure out an object rather than doing turning this into a class on work I want to give other things a chance because we have 20 minutes left and 20 topics still to review so let me comment that there is that's a reasonable thing so let me comment that the other class is also reviewing so is your free at 5.20 no 5.20? and want even more review you can go sit in on the other lecture okay error on sums of series so we have two types of series where we can calculate error readily we have alternating series well I guess we also have geometric we can do that so we have an alternating series so in an alternating series which looks like minus 1 to the n times sum vn where v is positive and I don't care where it starts then if we add up so if we add up just the first m terms this will be off from that by the last by the first term we did not use so this is the remainder formula it says if I have an alternating series and I stop after 10 terms then I'm off by no more than the absolute value of the 11th term so that's one that we have another one is using the integral test so if if my terms a n are positive then the sum equals 0 to infinity of a n minus if I stop at the nth term then I'm off by no more than the integral of the tail let's call this f of x where here n and you can also do a similar thing with the geometric series just by summing the tail because the geometric series will not sum the tail so the index where I didn't use I'm going to put this up in just a second I need to do an example which would you like me to do an example of one where I use the integral or one where I use alternating series well let's do it all let's compress four weeks of class into four minutes it's like that one minute Shakespeare thing uh ok so let's start with the alternating series no you want to start with one of the integral I heard no the nth term 8 let's make it n do what I need to ensure that the sum plus or minus 1 over 1 1, 2, 3, 4 ok so this is trying a question that would be asked so what do I do I just take the next term and I solve for when it's less than 1 over plus or minus I take 1 over n plus 1 to the fourth oh shoot plus 8 I want that to be less than 1 over 1, 1, 2, 3, 4 so I solve for n so I cross multiply should be less than n plus 1 to the fourth plus 8 so now I subtract some big number 9, 9, 9, 2 fourth root so how big is that this is less than 8 so the fourth root of 1 and 4 zeros is 10 if I subtract 8 it gives me 9 so that means n plus 1 is bigger than 10 so I take 10 terms so that means that I would well I add 10 terms now for an alternating series it's actually pretty easy small you can just start adding until you get to something that's smaller than your error and you stop it is this clear anybody confused by this so if we change the problem just a little bit in a different series but the same question so if I have so let's make it not alternating and I'll use just n to the fourth so same question I'm not alternating so I can't use the fact that it will be the 10th term I have to do it slightly differently I have to integrate so this one's alternating this one's not so here I need to find some number m so that people from n plus 1 to infinity of 1 over x to the fourth dx is less than 10 times well now my number is going to be a little nastier so I do that integral ok so it depends on what you want m to be is m the first term I don't use or the last term I do use if m is the last term I use then I want n plus 1 I don't have to go over that so we do this integral we get so that's x to the minus 4 so we get x to the minus 3 times 3 evaluated from m to infinity the limit is n goes to infinity of this is 0 so the infinity term is gone so we get 1 over 3m cubed we want that to be less than 1 over 1000 probably should have done 8000 so that I could take the cube root so 10,000 should be less than 3m cubed so I don't know 10,000 over 3 so I want m bigger than that so if this was so that's 10 cube root of 10 thirds cube root of 10 thirds so that's about cube root of 3 which is like 2 I mean by 10 thirds is 3 and change it's cube root is slightly more than 1 which is like 1.1 or something like that this is around the limit so I would take 11 so it's similar but not exactly the same so obviously on an exam I would choose the numbers so this arithmetic works a little easier so that means that I would add so I would have to take 1 plus 1 over 2 to the 4 plus 1 over 3 to the 4 plus up to 1 over 10 okay other questions we still have 8 minutes left isn't that what I just did error or remainder the same thing so it's remainder and error is the difference between the sum and the remainder I mean the sum and what you get so it's the same question exactly so and I will not use either of the words remainder or error I will say how many terms do I need so that the sum is within blah blah blah yeah okay so let's rephrase the question correct to 3 decimal places means that your answer the first 3 places here are right and then this one you don't care but you care a little bit that you need so that when you round it it doesn't affect that so the words correct to 3 places are the same thing as saying to within 0.123 and 5 different ways of saying the same thing is more than correct this is more than correct to 4 places it's not quite correct to 5 places because the 5th place is within 1 digit but it could be off by 1 digit in the 5th place okay other questions we can do in 6 minutes where you're all experts and you're all going to care yeah probably I can do it no sum of 1 over k factorial and you're supposed to say whether it converges or not so I don't see what the question is is it say find the sum or just 1 over k factorial 1 over 2k so what do you think well you have the answers so does this converge or diverge so it diverges so we can just look at it and say it diverges because this converges and this diverges that is not good enough for you we can put it over a common denominator and use one of the standard tests probably the ratio test so if we put it over a common denominator so this diverges so let's do it this way first it diverges because this is a number and this diverges because it's a p series with p equal 1 or a harmonic series but if you don't like that you can do it other ways do it another way yeah so you can compare this to the series minus 1 over 2k so this part I can compare for this part but I can also put them over a common denominator use the ratio test I can't use the integral test but the ratio will be equal all of those will work so let me just say in a few minutes then in terms of tests that you want to use for checking convergence if all of the terms are positive then you have a choice it's not always obvious which one is the right choice but our choices are see if it's geometric see if use some kind of comparison test so when I say use a comparison test you throw away things until it looks kind of like a series you know and then you try comparing them you might use the ratio test so the drawback I guess here is it an alternating series so if it's not alternating or geometric which you can do immediately then you might try comparison by throwing away terms until it looks like something you know and then comparing the ratio test the drawback of this is maybe it won't look like something you know maybe you can't compare that the ratio test, the drawback maybe the ratio is 1 in which case it doesn't give you any information you can try the integral test the drawback of the integral test is maybe the integral is too hard or do stuff so by do stuff I mean maybe you can manipulate it into a telescoping series maybe you can do some algebra to make it look like something or just try calculating the partial sums and see what happens other questions so just to remind you of the topics that you need to know you need to know how to do volumes both with a known cross section or volumes of revolution volumes of revolution you need to know how to find the arc length you need to know how to calculate work you need to know how to manipulate polar equations understand the graphs find the area you don't need to do tangent lines you don't need to do arc length and polar you need to know how given the sequence the size of the convergence that is just take the limit see if it's 0 sequence is a list of numbers given a series which is a sum of numbers apply the various tests to the size of the convergence for certain series like a geometric series you might need to sum them up like a geometric series and find its sum oh yeah average value I forgot average value average value is fair game for the test I just forgot to mention is power series on the test so only determining interval of convergence not manipulating one power series would be enough so what we did on Friday not necessarily only what we did the first 10 minutes of convergence so interval of convergence was on the test and in fact I can tell you it is on the test what about work? I said they were coming out either coming out water or spring or something work like is fair game for the test let's have a seat I don't have that it's a new rig it's really fun