 Do you have any questions? Let's go. Can you feel it? It's a goal of 5 to the negative 1 X. Yeah, the arc sign. The arc sign. Right. I was thinking, I want to know what sign X is. Also, who makes that a goal? It depends on whether you do this. No, they're not the same thing at all. If X is 0, the value of arc sign X is 0. If X is 0, one of the signs X is not 1, who's not the same age? If you make the same use, do you think they're the same thing? We need to go away and take that to a 22. That's what you want, right? No. No, that's fine. It's a negative 1. So, do you want to know what sign X is? I do. I mean, either it is this or it means this. Which one is fine? I mean, that's all good. I mean, when you say it's 9, then maybe it's 1. I don't know what sign X is going to be. Okay. So, we need this. So, what would you do? Well, look. Anybody have a clue? Yeah. Yeah, you would do parts. But what you're thinking is parts. Again, no clues. You're in trouble with it. So, we don't have a lot of choice. This has to be one part, and this has to be another part, because there's nothing else to do. So, we take equal r times dv equal dx. So, v, and the derivative of r times is 1 over the square root of 1 on this x square. It's du dx. That becomes easy. Because this is... So, remember integration by parts. That's the integral of u dv minus integral of v for u. And here we have all this in our case. So, we're good. So, that means that this is r times x, the integral of v du. So, that's x dx over the square root of 1 minus x squared. Now, to do this, we make a substitution. We substitute u for 1 minus x squared. There's du, well, almost. So, now I make a substitution. So, I need to divide by minus 2, u over the square root of u. And so, that's u to the minus 1 half. So, its integral is 2u to the plus 1 half. This is 2u to the 1 half. But I'm taking half of it so that cancels. So, this is x or something. Is there 2? Is that right up here? So, I think that's right. You can check it. If I take the derivative of the x bar sine x, that can use the product rule. So, I put the r sine derivative of this. This is some stuff to the 1 half. So, the derivative of this is stuff to the minus 1 half times the derivative of minus x squared. Yeah, 1 half is stuff to the minus 1 half. So, that's the r sine. So, it's good. Why are people looking something like this? You want to get to this? You're the one that asks. So, you're happy to go. Okay. So, like this. Okay. So, I'm planning a derivative of this. Not on 68. It's this. And so, I get an r sine from the product rule. But then I also get x times the derivative of r sine. And then the derivative of this is the minus x derivative. Right? Because I get a 2. Okay. There's a 2 here and a 2 there. You don't see that or do you see that? And so, this is 0. So, this stuff is gone. So, then you're happy. Do you have a question or are you... Okay. Your turn. Yeah, did I just do that one with the answer? I don't know. I think I did do that. So, this one kind of sucks in that you have to make a substitution and then you have to be clever and then you have to do some stuff and then you do this, but okay. So, first, there's not a whole lot you can do straight off. It should be obvious that, I mean, you can't do parts really because, I mean, I guess you can take the derivative of this and pick up an x. It's not going to be good. So, I will try... So, du is 1 over x dx. And now you might say, well, now what? I don't have 1 over x dx. But then you'd be clever. x du is dx. And if u is log, then x is e to the u. And what I write to this as e to the u du is dx. And so that transforms this into something that we can do because now this becomes... So sine of log of x becomes the integral of the sine of u and dx becomes e to the u du. So this is one that maybe looks a little familiar. It asks me if we need to memorize the identity that is this. This is one where you integrate by parts twice and you get to say you think back. Which I can do what you want, turn on if you don't. So here we integrate by parts twice and get back to what we had, plus some junk, and it's 1 half the sine minus the cosine or the cosine minus the sine or something like that. If anyone wants me to do that, I will do it. If you want me to do that. So now we can parts two times. And unfortunately, I'm going to call this instead of u I'm going to call it a because I want to use u for parts. So the substitution I made was with a. So I'm going to do parts now and I can choose either part to be u and the other part to be v. So I'm going to choose u to be the sine and dv to be e to be a. And so that means v is e to the a and dv. And so now this becomes, it becomes uv. So e to the a sine a, the integral of v to u. So since it's a minus a minus it becomes plus e to the a, cosine a, and then I do parts again. Yeah. Okay, so this is minus and it's uv minus v because I know what's the name. It's still minus. Okay, thank you. It's minus next time. If I make the same mistake twice it'll cancel out. Okay. All right, so it really is minus. I don't know why. Okay, so this time we do the same trick. We take u to be the cosine and dv to be e to the a dA. And so du is now the minus guy. This equals, well I have that e to the a sine a from before and this time I pick up another e to the a cosine a, but it's negative because I'm subtracting the a right at this place. And then I have an integral left over which is minus the integral is minus a minus so it becomes a plus of e to the a sine a. And so I have equals this stuff plus the same thing. So I can solve for the thing. So that tells me that two times the integral of e to the a sine a dA equals e to the a sine a minus e to the a cosine a. So that means e to the a sine a is one half of this. Sine a is one half. Put the e to the a outside. E to the a sine a minus cosine a plus a constant. And a is to log. So this is, well e to the log is just x. So yeah, would a problem of this be a part two problem? This would for sure be a part two problem. Part one problem. Part two problem. I mean integration by parts problems could be a part one problem. We have to be just as straightforward. Here's one part, there's the other part, go. So the arcs for line one was an easy part two problem. It's borderline hard for part one. So any questions on this? So I don't want to do all 86 problems on web time. But yeah, a sequence or a series? You have one in line. Long like natural law, you're not going to have rats, right? It doesn't matter. One in line. OK. So really this is asking this angle of continuity because it's a sequence just a list of numbers. So does the limit of the angle of continuity along two grand over n exist? And if so, what is it? So if you just plug in infinity over infinity, so this should be, oh, right, low p-tals rule. But maybe not. Is it not a right low p-tals rule to you? OK. Oh, right, low p-tals rule. So if you plug in, this is a uniform infinity over infinity so we can use low p-tals rule. Instead, we take the derivative of the top and the derivative of the bottom and the limit you need to say. And we're taking the derivative of respect to n. So the derivative of n is 1. The derivative of the log of 2n is 1 over 2n times the derivative of 2n, which is 2. So this is the limit as n goes to infinity of 1 over n. So the sequence converges and it converges to 0. Do you have one in mind or do you need me to make one up? Then I'll probably blow it. I prefer you to have one in mind. So what I'm assuming you mean by this is a problem where I would say, how many terms of the Taylor series do we need in order to ensure that if we take k terms of this sum, we want to be within, I don't know, one hundredth of the answer. Yeah, sure. Whatever that is. Yeah, yeah. So sure. So how about I'm going to approximate, I can write it right away. Yeah, I don't care. No? No. If you can have one in mind, if you don't have one in mind, I'll make it up and it might suck. Okay. Okay. Oh, so this is even easier than I was thinking about. Good. So we want the Taylor-Cohler polynomial of degree 2 here to be even easier. Okay. So. Yeah, just say how much it is. Okay, so even easier. So first, part a, find the second Taylor polynomial in there a equals a, and part b, how good is it for x between it's easier than the one I was thinking of, so good, I like that. So, to find the second Taylor polynomial, this means we're going to find a degree 2 polynomial that is clearly the first three terms of the Taylor series. So this is going to be, well, so f of x is x to the 1 third. We want it there a equals 8. So f of 8 is the cube root of 8, which is 2. And now we need the first derivative. So our first term is the Taylor polynomial. You just write the answer over here as we go. The first term of the Taylor polynomial is 2. And then we need the derivatives. The next term would be the derivative times the x times 8. So the derivative here is 1 third x to the minus 2 thirds. So f times 8 is 1 third times 8 to the minus 2 thirds. So that's 2 to the minus 2. So that's a 4. So the next term here in the Taylor series is going to be f minus 8. So let me remind you, t2 of x is f of a plus f prime of a x minus a of a over 2 factorial of x minus a. So the next term here is just going to be the derivative times x minus a. So this will be 1, 12, 8, 8, 8. And then it's the derivative of this, which is minus 2 ninths x to the minus 5 thirds. When I plug in 8, it's minus 2 ninths. And then I have to raise 2 to the sixth power, which is 32. So 1 over 32. Which is somewhat 1 over 32. I don't know what 9 times 16 is. It's 9 times 8 to 72, 145 times. So I get 144, but I have to divide by 2. So it's actually 288. If I had asked for t3, you'd take the derivative again, plug in, and then you would have whatever number you get here, divide it by 6, multiply it by x minus 8 cubed. Then you just keep going until time's done. So now the second part says how good is this for x between 9, between 7 and 9? So Taylor inequality says we look at some error less than or equal to m. So m is the maximum of the next derivative over 3 factorial in this case, pi x minus 8 to the 3. So m here is really the mass of the fourth derivative, of the third derivative of, let's just call this, where the maximum is over the range in question. So we have to look at the third derivative. I guess I'll go and erase this guy now. So we look at the third derivative, and we want to know when it's the biggest. So the third derivative is 27s. x to the minus 8 thirds is smaller for bigger x and bigger for smaller x. So we want to evaluate it at x equals 7. So this is biggest t7. So the biggest, and here's this, f triple prime of x is less than the next derivative. Is less than 10, 27 times 7, let's put it on the bottom, times 7 to the 8 thirds, which is some number. And so that means my error is going to be 10, can you see this? Or you don't care. Can you see where I'm right in here? Okay, well I can write it somewhere else. You can. Okay. Or you can just go back and see. 10 times 7 times 7 to the 8 thirds, which is just some horrible number, times 6 from the 3 factorials, times 1 to the 8. So this is my error. Pretty small. I don't know how small it is because I don't have a calculator. But it's like 100 or maybe 1,000 or something. You asked the question, you asked me the question you wanted to ask. Anybody who's confused about this? You're confused. No, it can't be anything. Yeah, but it's the answer to that. Because you don't have a negative error, right? The amount you're off is not a negative number. On one side you'll be negative, on the other side you'll be positive. So this is over the whole range. So your mistake is the absolute value. Yeah. Well, this is really one unit away from 8. So it's minus 1 and plus 1. So I'm really thinking of this as 8 minus 1 and 8 plus 1. And so I'm saying, so if this were a 12, then I would have to use 4. I mean I would have to use not 12, but I would have to use, where did I get 12? I would have to use 4 instead of 7, because it's not centered. I don't expect you to know, I mean I might ask you to derive it. So I don't want you to memorize any reduction formulas. You have to know them and produce them. You have to ask one that is one that you know. It's okay. It would not be out of bounds to ask you to do something where you have to integrate by part 27 times and so therefore you would want to derive your own reduction formula and then just use it. But I don't want you to memorize a reduction formula. This is only the off chance that I might ask you to, I don't know, do the integral of x to the 12 e to the x. Probably I will be asking you to do the integral of x to the 12 e to the x. But if I did, you could just figure out what using integration by parts x to the n e to the x gives you and then plug in n equals 12. Does that help? Do you understand the question? So much together. So, if I... Okay, so here's a question that I'm not going to ask because I'm telling you how to do it right now. Suppose the question were find the integral of x to the 12 e to the x, e to the x, okay? Or 216. I don't care. How would you do this? You would integrate by parts 12 times. But the pattern is pretty obvious. Once you do it once or twice, you see what's going on. So, if you integrate so, you integrate by parts 1 to get an x to the 11, you do it again, you get an x to the 10, blah, blah, blah. So, you might as well figure out what the formula is once and for all of n plus n. This would for sure be a part 2 question. And it would be a hard part 2 question. And I might even say, okay, n, figure e to the first derived reduction. So, okay, let's figure out x to the n e to the x first. So, x to the n e to the x is, well, we take u equal, so du is nx to the n minus 1, and dv is d to the x, so dv is e to the x. And so, this becomes x to the n e to the x minus the integral of n x to the n minus 1 integral of n x to the n minus 1 e to the x dx. So, now we're formulaing our hands, and we just use it for n equal 12. So, I mean, this is going to get long. So, let me just start using it, and then I will get tired and stop. Which I get to do, you don't. So, that says that x to the 12 e to the x dx is x to the 12 e to the x minus integral of 12 x to the 11 e to the x. Well, what is it when we use n equals 11? We get x to the 11 e to the x minus 11 x to the 10 e to the x. So, and then there's a minus, oops, sorry, there's an integral here. I lost the place. Yes, so minus the integral of the 10 e to the x. Well, what's the integral of 11 x to the 10 e to the x? It is, so this part expands to be 10 e to the 11 something. Yeah, and you just keep going. Let me stop because I'm tired. You just keep going and you'll get x to the 12 e to the x minus 12 x to the 11 e to the x plus 12 times 11 x to the 10 e to the x minus 12 times 11 times 10 x to the 9 e to the x plus da, da, da, da. So you just pick up another factor and the power goes down by one and the sum is all right. So it's long because it's 12, but I do not, I certainly do not recommend that you memorize this formula because this formula is only useful in the context of this problem. But you should know how to derive such formulas, which just pre-cabin is available. We don't. That's why I have the back. I think I can do that. I think I just did that one on Piazza with a step. But that's what you step, right? This is a big part of one question. But everyone do this. Then why would you ask? You can't do it. So we know that, so you supposedly know this fact. So we want to make this look like that. So divide by 3. So when you divide by 3, well, we almost have it looking like that. Except I really want to think of this 3 as the square root of 3 squared. That makes the substitution u equals that of square root of 3 to u vx. And so this becomes the arc tan of u. And I pick up the square root of 3 and put an arc on question. Yeah, somebody in the back, wow. Can you do one where you have to do a solve or a constant to make the solve creative? Ah, okay. Do you have one in one? Yeah. Okay. From one? Okay, better. Otherwise it'll, then work. I mean, now it's easier. Okay. So we just can't use the number. I don't know what it is. And we use the ratio test, for example. The ratio test might be our friend here. I think so. So let's try the ratio test. So that means I want to look at the limit. I think so. Now I take n plus 1 factorial over plus k factorial, right? That's the next term. I divide it by tan factorial squared. And divide it by n factorial squared. Okay. I think that's something over there. So then I'm going to go up there. So this n plus 1 factorial over that n factorial cancels leaving in n plus 1, plus there's two of them, so it's squared. So that's gone. And this k n factorial over that k n plus k leaves me down to k n plus 1. So this is k. I should go ahead and do this. And we want this limit to be less than 1. So this gives me a k. I kill off one of those. And we left with an n plus 1 on the top. And a k here. And then we go to crap. And is there a specific number or a rank of k's? Yeah. Oh, it's a positive integer. Yeah. Okay. I need to make sure that this is all bigger than, so I need that this product here is bigger than n plus 1. So how many terms do I have there? Well, it is bigger. Did I do something wrong? So how many terms are here? They're all there. k minus 1 times. This looks, I mean, did I do this one wrong? Does anyone need some algebra error to see if it works for any? Okay, so let's, let me just check for k equals 1. So if k equals 1, I get n plus 1. That's 1. And that's 1. But if k equals 2, equals 2, then this is an n plus 1. And this is 2n plus 2 times 2n plus 1. And we're done. And the limit of that, for sure, is 0. This works for any k bigger than 1. Is there a specific k or does it say 4-1k? It says it is. So it doesn't work for k as 1, but any k bigger than 1 improves. Okay, I thought there was a specific value of k greater than 1. So is this, did I lose you in this? Okay, so let me explain again what I did. Right at the k. Here we go. Now, expand all of this garbage and cancel everything you can. So here, well, n plus 1 factorial over n factorial is n plus 1. There's a square here and a square here, and we're done with the n plus 1 square on the top. You can do this one again a little more slowly. If you're clear why this is k n plus k, okay. So k n plus k factorial is k n plus k times k n plus k minus 1. k n times k n minus 2. Duh-duh-duh-duh-duh-duh-duh. And eventually, we'll get to k n plus k minus k. So we get on the bottom, k n plus k, and then the next one is k n times 1, and we keep going until we get down to k n plus 1. And then the next one is k n here, but this is a factorial, and I had a k n factorial here for that campus. I want to know what's the limit of this as n equals to infinity. Well, this one I can factor a k out. So this is really k times n plus 1. So I can cancel, I can cancel one of these squares. I can cancel one of these squares with that. Now the question is, for what k's will this limit when n gets big be less than 1? Because I'm doing a ratio test. I want to know, for what k's will this be less than 1? Well, here, I have an n plus 1 on the top, and I need something smaller than n plus 1 on the bottom to know that it goes to infinity. Look at the degree of this. As long as I have at least two terms here, I have, so this is bigger than an n plus 1 on the top. Bigger than an n plus 1 on the top, on the bottom I have k, and then here I have something like n plus 1, and I have lots of things with n's. So this will be n to the k minus 1. So now n is going to infinity. Well, as long as the degree here is bigger than the degree here, it goes to zero. So this goes to zero, but k is bigger than 1. I think this is just a k. Maybe it's k is bigger than 2, no. How many do I have? It's got to be 1, yeah. It's just a k. This is definitely a part 2 question. This is a hard question. So let me just emphasize again the structure of the exam. Part 1 exists of very straightforward questions that anybody who deserves to pass the class should be able to do basically all of it, making it be 80% so that you can make a mistake or two. But basically, part 2, part 1 is essentially equivalent to the placement exam that would allow you to place out of this class. To place out of this class, you have to get 80% on the placement exam. So if you get 80% on part 1, then I will pass you. Part 2 is harder stuff. Part 2 is like the midterms. So some of the part 2 questions will be easy and some of the part 2 questions will be hard. This will be a hard part 2 question. So hard questions I don't expect anybody to get. I don't expect everybody to pass this again. Yeah? The parts are graded separately and then I put them together. So part 1 just says, see or better, yes or no. And then part 2 says, so really it's part 1 and then your grade on the final is part 1 plus 2. There's a bar you must clear and then there's how high you need to jump over the bar. We're not this much shocked, so we need to be done. Yeah? I'm not going to show you examples. I don't know if you can try it. So a logistic equation type of question, you would have something like this equation like y is equal to some number 0.03. There's a differential equation. The kind of question that could be asked on this and then maybe I can use some initial thing as well. So one thing I didn't say is 0 except I'm writing answers when they should be lost. So several different types of questions could be asked about this. One, just give me a formula. So that means you separate variables and you integrate or maybe you memorize the formula or whatever. Another type of question I could ask is maybe instead of giving you the 0.03 and the 100 I give you constants a and b and I give you some data and I ask you to find the constants a and b. Or I give you an initial condition. y of 0 is 10. What happens? y of 0 is 110. What happens? Those kinds of questions. You want me to do any of those? Or you don't want to do all of them? Do that all. So I didn't give you 0.03 and this is usually called a and I didn't give you this. Then I could give you I mean these are all so I could give you for example that y of 0 is 10 and y maybe I could give you y prime of 0 instead. y prime of 0 could be a formula. Tell me the population and your question with unknown constants and blah blah blah. This is one where you can find the formula rather than analyze the face and analyze the pictures. So what do you need? What do you need to do? You need a formula. So let's solve this technical differential equation. So we can solve this differential equation. Maybe memorize the formula already. To memorize the formula you don't need to solve it. Yes? On the test let's say let's ask us to solve the equation and then say I memorize this so that means I'm not sure So I'm not going to ask a question that tells you how to do the problem. I would say this is the question. So you want to skip the next step that I'm going to do and write it down. That's fine. I don't care. But suppose I changed this and put a square here. Now you're screwed. You didn't memorize that formula. So if this is the problem and you memorize the formula then you can say this is the logistic equation. I know the solution to the logistic equation looks like this. But write I know the solution to the logistic equation is this. Don't say my friend has this formula on his page so I wrote it down. So just in case you didn't memorize the formula let's solve the equation. Even though you've done it 100 times already y over let's leave it here. There. By over y times 1 minus y over m equals k separable. So I separated variables. You know actually let's find k first. k is obviously a half. Is that clear? Let's see. I guess it's not a half. But it's okay. It's obvious to me. So I'll do this side. This is a partial fraction. Do this and we have to do partial fractions. I'm going to rewrite this just to make my life a little easier. So I need to do the integral of this and the integral of this. So I need to know how to do this integral and this is a partial fraction. So I need to know for what values of a and b I have this and in fact I'm going to write this just that I don't have fractions of a fraction. Is that okay? So now I need to find a and b so that's true. So that means times m minus y times m is 1 by cross multiply. And so if y equals m then b is 1 over x. Oh yeah, thank you. Did I get that right? Is this right? Why do I have top and bottom? Okay, so if y equals equals 0 first if y equals 0 then we see that a m equals 1 and if y equals m then we have b m squared and b is 1 over m squared. Something's wrong here. What did I get wrong? One of the signs should be next to it. This is 1 over m this is 1 over y plus 1 over m minus y is 1 minus y plus y. Oh it's right, okay. So then this becomes the integral of the log plus minus y. Why? No it cancels. I think I have a negative sign missing. So this is 1 over m times the log of y plus and that equals kt squared plus 1 over m kt. So this is the log for y minus m over there m kt plus some constant to differentiate both sides an easier way to solve this isn't there? We just want to define the formula anyway. So I have y over y minus m ds e to the mk to solve for y and you have to do some dancing around to solve for y across multiply and burger burger burger and you could get something that I already forgot. y equals I don't know this is right Okay so now we know y of 0 is a half y prime 0 is a half and y 0 is 10 so y of 0 is 10 so y of 0 is 10 so y of 0 is 10 and I guess I got to solve for y first we have to chase them all away so solving for y yeah algebra algebra is 1 so you memorize the formula solving the formula solving for y that sounds right I guess I need another information I'm sorry I have three constants so I need a few pieces of information so I can't even do the problems I need to know I need to know like what y is y anyway do you see how I'm doing the problem now? yeah I'm going to find a okay I use the fact here that y of 0 is 10 so I look here when y is 0 I have 0 when t is 0 I have a equals 10 over 10 minus 10 I don't know what's that yeah well that's where I screwed up but I need to give you another piece of information in three pieces of information you need to be a problem and I'm going to give you two you need to be a problem I'd rather give you y1 I would give you the ability to find it which I didn't do before or I would give you k or I would give you some stuff so let me point out that both exponential growth and logistic growth and all of these things are just special examples of several differential equations the main goal of this class is now to teach you the two models two biological population models that we have but to teach you to deal with several differential equations when they arise so I am not wedded to the logistic particularly or to the exponential particularly what's important is the technique of being able to solve it and being able to find the constant no questions in it it's okay you have a question can you tell me how does this solve the problem so you cross multiply band it out and you bring the AT to the plane and over here factor that out and I got tired so I stopped I mean it's really just the algebra but this seems to be most calculus students biggest problems are actually pre-calculus calculus is easy yeah do you have one of mine tell me again do you need a circle I don't understand this question okay that makes sense you have the cross sections for squares and the base itself find the volume to some surface being a circle or I just couldn't read her mind this is the real problem okay the solution is close to it okay so let me just do it and I'll probably say exactly what I know is the solution for the cross sections and this is perpendicular to the line you have to say which ways the cross sections are going because you get a different thing let's say perpendicular to I think the line this is something let me try and draw a three dimensional picture so x squared plus y squared equals four is a circle of radius two in the xy plane this is an x y and coming up out of the board are big squares this is a big square this is a smaller square this is a smaller square this is a smaller square that's a point square so this thing comes up like this if I try and draw it from the side that's looking at this line so maybe I'll draw it the other way if I look from here that's my eye can't tell you that this is my eye from there then we see a thing the base is a circle and there's a big square cross section out of the back and then this sort of shrinks down to a point like that and the cross sections are squares with a little like that I don't know if you can visualize it it doesn't really matter too much these squares come up out of the board this way so to find its volume you'll imagine that I'm just going to take out the saw and I'm going to cut this up into a bunch of little slices of like cheese slices cheese cheese cheese cheese cheese and then I just want to add up all of their parts so the volume of the whole thing is the integral of the volume of the slice or the area of the slice the, well which way am I cutting it I'm cutting it up this way so my slices are debauched I need to know what is why I range from slicing well here the biggest slice y is 0 here the tiniest slice y is 2 so this goes from 0 to 2 of the area of the slice now if I take a particular slice how about this one at some given height y when I look at it it's a square so this slice here looks like that and it's got a little thickness of dy and it's a square what is the distance from here to here well this is if I tell you why I know that x squared plus y squared is 4 and so I want to figure out if I tell you what the y value is what the x is and then you're wrong so x squared 4 times y squared that means that x is the plus or minus so here is the plus square root and here is the minus square root plus the distance from here to here 2 times the square root so this is 2 2 times the square root of 4 minus y squared and so is this it's a square root so that means that I integrate if y goes from 0 to 2 of 2 times this 2 times the square root of 4 minus y squared square d y there's a minus sign here so when I square a square root the square root goes away so this is the integral from 0 to 2 of 4 times 4 minus y squared d y and so that is 4 y minus 1 third y cubed d x from 0 to 2 so this is hello I don't know 32 minus 4 third of 8 something wrong here well 32 minus 32 over 3 which is 60 through 1 because it's a square root if it was a triangle then I would multiply the days times the height of the life I choose if it was a circle I would multiply our squares if it was a sneaky mouse I would have to work a lot harder I would have to tell you what shape of triangle is it an equilateral triangle is it a really tall triangle I would have to tell you some relationship between the height and the base because all we know is the base so for example I can say it was an equilateral triangle or if it's an equilateral triangle of course I know where to write if it was an equilateral triangle for example if I know the base here and this is also the base that I would have to use but this is 60 degrees so this is a 1, 2, square root 3 30, 60, 90 triangle and so this would be b over 2 so this would be 8 over equilateral root 3 and so that what I mean is b square the life I choose 2 root 3 oh no yeah 2 root 3 b over 2 square so I would integrate b square over 2 2 over 2 but you understand that and if it was a semi-circle then it would be something else it would be high high square over 2 so all of these problems are all the same once you figure out what the slice is the hard thing is to remember that you have to integrate these parts okay are there questions? if you have no questions then we're done are we done? oh we have a question now yeah well okay so I don't have credit or credit formula in my brain so let's just make a picture okay so well okay so there's many variations on these credit or credit problems that could be happening that's why there's many variations on the homework one thing could be here's an equation tell me what happens another thing could be here's an equation does this describe a predation or a cooperative or whatever another one could be here's a face portrait tell me what the graphs look like or here's some graphs tell me what the face portrait looks like I was I mean or a problem here's an equation tell me everything so you have a preference okay let me make that up so here's a graph so this is time and this is the predator so the predator is I don't know leopard population the leopard population over time well initially they start here it's a 10 they grow it's the graph of gazelles here's the population of gazelles and given time in gazelles there's 100 gazelles to start and they're doing just fine they start to fade off and then they grow and I need to make these matchups somehow here well okay I'm going to do something like this too and the bumps match up in a nice way so what does this look like in the face portrait so I want to draw it in I'll give a map I want to draw a graph in the first gazelle lab so when we start here we have 10 leopards and 100 gazelles so we have 10 here and as time goes on the leopard population increases up to about 20 so the leopard population will grow up to about 20 and the gazelle population grows a little bit and then it starts to fall so if we start here at 10-100 both populations grow a little bit and then the gazelles start to die off so that means that the leopards continue to grow but the gazelle population drops until the leopards population reaches some maximum of about 20 and the gazelles are continuing to drop off and so they drop off and now we might be so the leopards reach their maximum of about 20 and the gazelles drop off so now which is here and then the leopards start to then they both start to fail and they go down not as much and I get something like that the leopards grow and the gazelles fall off and the leopards start to fall off and the gazelles start to come back and they make this oscillation or something like that so this will become a spiral that limits on some equilibrium because if you look at these two graphs eventually they start to stop wobbling and they're wobbling some more if on the other hand so let's turn it around suppose that instead of this picture we saw this so now I'm going to give you the picture and go together so if I give you a picture like this then this is saying if I look at the gazelles population starting I don't know they grow to some maximum and then they decrease back to the original amount if I make the picture of gazelles versus time I don't know 100 and this is 200 they go from 100 to 200 so they grow and then they fail and they oscillate like a sine curve and the same thing happens to the leopards but with two different maps and they're shifted by a chord period but again the leopards is also going to look like a sine curve but shifted by a chord period and starting here the leopards shrink a little then they come back a lot and they shrink that same amount and come back that same amount is that the question that you asked that you're on so I could ask you if you need to grasp like this I could ask you to draw this graph I could give you several graphs and take a pic I could ask you to answer a question like what expected to happen to the leopard and gazelle population after 100 years something like that or I could give you an issue like this and say what will happen after a long time to the population or I mean those kinds of things probably I won't ask you to draw the pictures because this health grade so more likely I would give you a bunch of possible pictures and ask you to choose one or choose one and interpret it or something like that no problem you're talking about too yeah has there ever been a case where in a base graph where the direction of the base will blend I could yeah sure it's just that these guys would these guys would shrink as these guys grow sure I can make them do whatever you want I can't make them cross but I can make them write my name on them I don't usually cross the t's so you just switch the axis and you alter the t's so that's another kind of question so you asked I don't remember who asked who asked so I have some equation so let's say I have an equation so let's say I have r prime is 3r I mean so the numbers the numbers are going to be yeah it was a logistic plus an interaction term so this was something like 3r this is the prey so let's make this be the predator let's just make it easy minus r plus 2rw and the prey w prime is oh, except r's oh well so the prey is now weird and the predator it doesn't w times 1 minus 1 minus w over 10 minus so actually the prey is now the predator so the prey are along that and the r's are rabbit decos r is the predator here well, yes, r is the predator here and w is the prey because the interaction here is favorable for the r's and un-favorable for the w's the predator and prey are swapped here so the rabbit decos are along that so now we want to find the equilibrium well, the thing to notice is that let's say in this one there's an r in each term so we probably want to factor so I get r prime r prime minus 1 plus 2w so this would be 0 exactly when r is 0 or w is a half so now I know when one of them is 0 and I'm looking for equilibrium between both of them are 0 so I also have to figure out when w is 0 so as I notice that there's a w term well, I can factor this let's just plug it so if r is 0 then I have this so if r is 0 and w prime is w times 1 minus 7 is equal to 10 so this is 0 w is 0 or w is 10 so that means that I get equilibrium and I get one here and one here at 10 and then the other case is when w is a half and we'll also make r prime 0 so if w is a half times 1 minus a half over 10 minus and I want to know when is that 0 so that's 0 and what that means is that w is 0 and w is 0 and w is 0 and I want that to be 0 so that means half r equals to that so r is 1 minus 1 over 20 which is a strange number but it's 19 it's 1920s so this must be whatever it is so I get one here at 1920s so here okay, what was this number 0 is 10 so it's way back here so I get a half 19 so anyway, there's 3 of them how does the face look how does the face look well, let's first notice here so r is 0 p grows and wait then r just shrinks so here, that comes down so that means in this region it's going to go this way because I have this is still there this is going to go this way what am I doing wrong here something's wrong so what we do is we divide this up into regions and look when the slopes are positive and negative and so on so if if if w is small r is small and w grows needs to be less than 10 and w grows and r decreases so in this region it goes like that now when when that w is always negative so if r is small I'm missing something here I think it spirals but something's wrong with this so this will so let's see if w is large yes if w is large if w is larger than a half then notice that this is positive so this goes up on this side so the large goes up over here and the w's go up over here and down over here so that means that we're going to get something like this and we'll get a spiral out of it I guess things give him a minute could you have a private scene next to you now I don't know could you put someone else with you but he's gone so I got confused okay, other questions yeah so could you look over here okay you have one in mind so do you want area or what do you want area so let's say so this one is extreme for the month you got what you want there you go so what does this mean well if I didn't draw the picture for you then you have to understand you should probably figure out what's going on so when you think about the pictures here r is the sign to theta so when theta is 0 r is 0 as theta increases r originally increases but then it hits its maximum when theta is pi over 4 and then decreases back down when theta is pi over 2 and then decreases back down in fact this graph would look like a four-leaf clover and let's make it a four-leaf just to emphasize so that means it's going to do this and then it comes down here and do the digits so I get a bunch of it to get these so that's something I've got to want just one of them so I need to know what value of theta I care about since I've said one loop well at least 0 at theta equals 0 when it comes back to 0 when theta is pi over pi over 4 I'm sorry because the sign pi is 0 so really I want to look from theta equals 0 to theta equals pi over 4 I want to integrate from theta equals 0 to theta equals pi over 4 and when I do my integration my integrals are going to be my things that I'm adding up are little guys like this little sectors of a circle that's the area of the little sector of angle d theta well it's going to be the radius square one half of the radius square there's no pi because the pi's cancel out because I have that means this integral is going to integrate from 0 to pi over 4 one half function so now I've turned that into an integral so now I have to do this integral an even power of sine which means in order to do this integral I have to turn it into an odd power of sine and cosine so I use the fact that sine square let's call it x the integral of sine is 2x square x is 4 theta so this becomes one half and then I get another half the cosine of 8 theta d theta and 0 and so when I do that integral I get one quarter the integral of 1 is theta and the integral of cosine negative cosine of 8 theta over 8 so this half is just so there's 2 halves and they come from 2 different places this one comes from the fact that the area of the circle is pi one square and I want some little fraction of this I want one theta of d theta of this square all the way around there's 2 pi I just want a little d theta of it so that's going to be 2 comes from here right so I have pi, r, square and I want each thing that holds 2 pi and so the pi is canceling and you can be in r squared over d theta because they don't want a whole sandwich, no they won't so this half is the 2 and 2 pi and this half is the 1 half so it's a quarter so the quarter of theta minus the sine of 8 theta over 8 and now I'm going to get it from 0 pi over 4 so this is pi over 16 minus 0 plus 0 which is pi over 2 of course to make the problem more complicated there could be 2 guys that you want to figure out where they cross or stop I could have another curve here find the area outside and I'm going to find out where the 2 curves cross and then I integrate from where one goes to the other goes and then I'm going to take the area the outside where it's the area of the inside there's lots of variations essentially polar area once you set it up is not too bad but you have to remember that it's 1 half r squared or figure out that it's 1 half r squared and you have to be able to figure out where things cross even if it's not where you figure out where the regions you're talking about yeah well so the definition of a sequence getting bounded means it never gets bigger than someone so if the limit if it has a limit if the limit is less than something then it's bound if it doesn't have a limit because it never gets bigger than something the idea is you want to find some number that's always less than and if that number exists then it's bound usually this will be something where the power on the bottom is bigger than the power on the top or equal to the power on the tops or you have a sine or a cosine laying around or e to the minus s I can't just say I can't do it because the method is going to vary by the function the method is find the bound and if there is no bound then I need the bound but that's not a very satisfactory answer if you what are we going to do in series 4 are we going to repeat that part so this is essentially geometric series in stocks so let's say I have I don't know 2 3 2 3 2 3 2 3 so what does this mean this means this is the sum if I notice it repeats every 2 so it's 23 over 100 23 over 1000 23 over let me write it this way 23 over 1000 0 to 0 23 over 1000 0 to 0 huh this is 23000 yes what am I saying thank you anyway 100 square so this is 23 over 100 square 23 over 100 2 to 0 so this is the geometric series 23 over 100 times the sum from n to 0 to infinity 1 to 1 1 over 100 to the end okay well you know this sum this is geometric series so this sum is 1 minus 1 over let me just write it up here so this is 23 over 100 times 1 minus 1 over 1 minus 1 over 100 because it's a geometric series so this is 23 over 100 times 100 over 99 this is 23 over 99 so it's just really the geometric series in disgust and the disguises in verbal it's like you know state of paradoxes it's as bad as part of Kansas so here are the questions we have the room for 10 more minutes no 445 we have the room for 40 more minutes and we have the room for 10 more minutes and that's it okay so if you have no more questions then I don't want to ask you yeah do that do you solve the geometric series and do you have any speculations of 1 minus x squared 1 minus x squared so this is I mean I can't right and if you notice that's not how it's worded it's that which of these is the solution and then instead for that solution you found what is it given in initial conditions that's because we didn't cover how to solve these kinds of different equations they cover that in the three of them in the help of the school we don't need to know how to solve an equation like this but if I give you 3 possible solutions or 10 possible solutions you should know how to check it so do you want to give me the 3 choices I don't understand yes there's a way but I don't want to cover it so there's been great difference there's something that you multiply both sides by and then you integrate that yes there's a way but I'm not going to show you and apparently I forgot to write the second part of the equation the solution of the second part of the equation I would put this in the list do you have a question yeah I want to be on a practice exam for the last question do you want to conduct it in the sand okay read it to me wait wait wait slow down wait wait and I have a bucket so I have to lift so if I have to lift the bucket 40 feet and since everything is in patterns our answer will be in the foot paths I mean it's bizarre but this weight is actually a force I'm not a mass so 40 pounds so we don't have genes so that means so our work is the input of the force times the distance that's applied the bucket is here at height h we need to figure out the force that we need to apply for it over the distance that we need to apply okay so that means we need to figure out the weight of the bucket at height h so since it so h the number is the number of feet from the bottom so the bucket has to travel from the bottom to the top they're going to integrate it into the weight find how far it has to travel so the weight here if it starts at 30 pounds and it loses a quarter of a pound before it then the weight is 30 largest h over 4 and how far does it go I don't know do I want something to go h it's just the force if it's not a factor of h nor is there a factor of 2 minus h so that should be it am I screwing this up am I missing an h so it has to travel the distance of h oh yeah yeah okay it has to travel the distance of h there we go and then is that it do you want me to do it so the question is whether we're talking about a force or a mass if you have a mass which is in kilograms this is the physics method if you have a mass then the term mass the force that is involved is the force times the acceleration of the ground I mean force equals mass times acceleration so if the units are in kilograms then that means kilograms are not a unit of force they're a unit of mass we have to multiply kilograms times the acceleration of the gravity to turn this into a force kilograms per second square or a kilogram meter per second square but weight pounds is a force so here's how you know that you don't need G the work is in units called pounds if I integrate pounds guess what I get what's pounds so I don't need anything to adjust the units a Newton is in kilograms per second square so that means I need to multiply millimeters per second square to convert kilograms into kilograms per second square so joules is kilograms we test we don't need a G because this is not a huge supply so I think it's a problem on the other hand the pressure going at large these problems so we have to fight over them every time yeah I don't think I do multiply by H so let's think about the amount of work I need to do over the little interval the amount of effort to raise this thing from here to here is it's weight times D H I just do that all the way up I don't think I do multiply by H did I multiply by H in the solution yeah I was going to say that yeah I don't think I do so the other reason I multiply by H is because everybody kept yelling multiply by H I don't think I do no I don't so if instead of looking at the bottom out I was looking at water then I would have to multiply I was pumping out the water so suppose I have water here so suppose let's change the problem where I don't have to multiply by H or 40 minus H instead of looking at the bucket out I have a well-insult of some substance that gets lighter and lighter as I get to the top and it gets lighter in this way it gets a quarter pound lighter for each foot so at the bottom it's sludge and at the top it's water and it's a nice mixed mixture and I'm going to pump it out and then I'm going to have a sludge here at the bottom and pump it out then I have to do a weight times the distance it has to go so then I'm going to multiply by H and then a little sludge up here and it has to go all the way up so then I'm going to multiply by well actually 40 minus H but so then I'm going to have a factor here of 40 minus H because the pounds would be the weight of the distance it has to travel but here for this bucket when it weighs 30 minus H over 4 how far does it have to travel just the H and then it weighs something else so the amount of work to raise the bucket from here to here is this amount these go all the way to the top so what we're doing is we're adding up how important is it to get from here to here and the amount of work to get from here to here is this much if on the other hand we change the problem we just try and do the two problems that are almost the same but different thick and this is like if I have some slab here basically it weighs very much H over 4 pump it all the way to the top that's what I have to do so I have to take a little slab of 30 minus H over 4 VH pounds because that's its volume and I can get it from here to here how far does it have to go from here to here it weighs the same all the way so when I get it from here to here I have to travel 40 minus H so the work to do this little slab is 40 minus H and it's 30 minus H VH and I have to do it for all the little slabs so I have to do it for all the slabs between 0 and 4 so if I'm pumping stuff out then I have to multiply I have to figure out the weight and how far I have to move that weight here I just have to figure out how far do I need to extend to move that weight just a little bit but since it's a different problem it has a different set but if you yell loudly enough you'll probably make me change it and then I'll get it off are there questions? yeah yeah I gave it to me because I don't know where I was I just tell me just read it yeah I don't know but we want to know for what X does this converge okay so figure out for what X does this converge we just say X is some number and we look at the ratio test and we take the limit just n goes to infinity of the next term so the current term we want that to be less than 1 to make it converge so this is the limit this n goes to infinity of the next term so that's n plus 1 over e to the n plus 1 times 3 x minus 3 to the n plus 1 divided by times 3 that's the limit I need to do and so stuff cancels this doesn't cancel the n plus 1 is e and x minus 3 to the n plus 1 over x minus 3 to the n so I want to take that limit as n goes to infinity this part goes to 1 as n goes to infinity and plus 1 over n basically 1 this part has nothing to do with n so it just falls out now we want to ask the question for what x or e is equal to 1 in other words for what x does the ratio test give us something that converges so now we just need to solve this inequality so that's going to multiply through by e so I have x minus 3 is less than e I'm going to get rid of the absolute value by saying x minus 3 is between e and minus e and then I'm going to add 3 to both sides x is between 3 plus e and 3 minus that means x well x is between 3 plus e and 3 minus e now remember the ratio test tells us what is less than 1 tells us what is bigger than 1 and when it equals 1 it doesn't tell us anything so we have to worry about what if x equals 3 plus e and what if x equals 3 minus e so if x equals 3 plus e my series becomes 3 plus e minus 3 is e so that means the series becomes 3 minus e x is 3 minus e I have 3 minus e minus 3 so that's minus e to the n my series becomes the sum minus 1 because 3 minus e to the n is minus e to the n and over e to the n gives me a minus 1 to the n so again this starts first so in fact it doesn't converge at either end so my interval of convergence is x between 3 minus e to the 3 plus e not even e to the n sure but I think you made a mistake when you read the geometric series so the r-value I thought your r-value was wrong which one do we want 0.232323 again we add that one because the summation was 1 plus 1 over 100 plus 1 over 10,000 times 23 over 100 yeah I just thought you'd get 1 over 10 didn't I? I used 1 over 100 and I wrote that you might be interested to play that I don't believe you because I thought it was 1 over 100 so the answer when I wrote on the board I think I can still see it it's 23 over 99 so it's 23 over 99 and r-value is 1 over 100 so you just didn't write to 0 everybody knows everything? welcome to Dunmer, I'm happy to be done no more questions? I have a question are there going to be more easy questions? depends on what you mean so part 1 is all over there we should take you around a half hour I think there's a chance but they're all easy part 2 more nothing is true there's three levels of questions almost trivial everything is almost true if you know how to do a ratio test you should be able to do this one if you can improve your knowledge you should be able to do that one they're all very easy part 2 is nothing is look at it right now and answer but some of them are pretty easy some of them are not easy please press and confirm and it is okay