Surface Area, Related Rates and Differential Equations

So hopefully he will be one of us now in about two months. And today I begin with the usual. I like Ivy's homework very much, and I haven't had a chance to look at Ivy's and Elaine's homework. My apology, there's no excuse. I've been busy, so I have to ask you. And do you have any question from homework? But I don't think so. What about Elaine and Ivy?I don't think I did. And then, are you? No. Alright, Elaine. I'm falling behind your homework now because you lead me to a beautifully written. This is actually a merit, and which is a file actually. Is that Google Doc? Elaine, I'm asking you about the format you're submitting your homework.That was a lesson recording. I sent. Oh, I thought that that's how you do homework. That's why because I couldn't sign in. I have trouble signing in from the computer I often use now. That's why I couldn't. I said, "But where is your homework?" Oh, I haven't turned it in yet. I'll turn it in once the lesson's over. Ah, usually that shouldn't happen. No wonder I thought I'd been behind your homework. Okay.But Andy, what about you? I haven't turned it in either. Yeah. Hey, yeah, yeah. Well, I was just checking with Chad. I saw you handed in while I was working. I think there was a period of time you were all good because I was keeping my wits on you. Just because I don't mention it anymore, and you don't hand in homework, the loss is all yours. Okay, please do hand it in. Alright, well, if you guys really have no issue with whatsoever.Integrating for the volume now. We are actually starting a new topic now. I'm gonna actually do this a little out of order because usually after doing this year, we're gonna actually look at and Riemann sum and numerical integration and approximations. But because that's a specific kind of an, I would say algorithmic machinery, which has really has no conceptual nuance behind it. I'm gonna save that for the baritone right before the test.I'm going to go on to a topic where it takes time to sink in. I want you to wrap your mind around this topic better than just doing the remainder sum, which is numerical calculations. So knowing we're putting an end to integrating to find the volume now, and in fact, I still had some other material planned out. For example, finding the surface area by integration. However, they never show up on on the AP anymore. Meaning those problems are considered too difficult. They're going to flunk the entire.So I'm not going to cover them, this year. I because I really just said that I love those kind of gymnastics, but they're not necessary for your test. Now I'm going to actually open a new chapter, which is all about, and solving differential equations. We're going to do a second round here. So first round, nail the AP essentials, differential equations. Well, first of all, what do we even mean by differential?Equations, I do equation with differentials. Give you a few example. I can say when when you did related rates, you could really have a volume of a sphere would equal to a four pi r cubed over three. I didn't even prove to you. Oh, we did prove that lately. We proved that. But then what you started with was a differential equation. But we actually write out a shell of this spherical. Oh, I didn't prove to you that the surface area. We proved it by using a disk method. 对对对，对对对。嗯，所以，就 D V， equals, um, twelve, twelve pi r squared over three.Brilliant. And Eddie, you did well. Except for, is that a two or three that power here? Two, very good. And the way Eddie is writing the V and R to make them hard to recognize, so you could actually write it as a big R just to make it more obvious. Very good. Eddie, you are getting it right too. We're getting actually the four pi R squared dr. And had I begun with this equation itself without giving you the formula for the volume of a sphere, this is a differential equation because it contains differentials.I could also write the differential equation in the form of a derivative instead of a differential, but you know derivatives really are ratio of differential. So what you can do is to divide that over. I like like I did that. He wrote it as a derivative. He's saying the v as a derivative as a change of the r, not rate of change of r, really just equal to the four power squared. Both are called differential equations. As long as you can see, there are some derivative.But if there's some differential, it's called differential equation. But what's the meaning of the differential equation? Using edit way of writing it now, and the minute you're writing is a derivative, it means a rate of change, something regarding something else. Along that line, could somebody give it a physical geometric meaning? What this differential equation actually tells you. I need to fetch my charger. My computer is zero percent. I'll be back in forty-five seconds.``````Thank you, thank you. Well, I think four pi r squared is like the surface area of the sphere. I think it like like when. The radius increases by like a tiny amount, it also increases the like the the surface area by a tiny amount. So you have. Wait a second, hold on, and you're saying when the radius increases, and it would increase associated with it. I'm gonna call the surface area the S now, which is also a function of r, as opposed to the volume. These are all the characteristics of a sphere. These are meaningful variables.相对于sphere now, and I'm saying as the r increases, there's gonna be a d s, the surface area will increase a bit. Continue. And then that form like a kind of like a like a layer, so like this is the inside layer. This is like the. It forms a shell, right? Yeah, like like in that picture. Beautiful. Which means it's like an infant testemalis.Correct.We say the DV here should equal to the shell volume. Whether the thickness of the layer of the DR. And the shell volume here should equal to the surface area multiplied by the thickness. So therefore, and I think Eddie, what you meant is that a it's not we're really not focusing on the increment of the surface area. It's a surface area itself multiplied by the DR that should give you that increment of the d volume.Right.我叫The Apple Peel Theorem, and it's a party trick. So I am going to actually prove to you, use some other method now, and indeed, the surface area is four pi squared. Does anybody know that formula? Of the sphere, yes. Yeah, good. Do you know how to prove it? No. Well, you have to tensor. You have to prove it to us. Yeah, but.No.We can have surface area. We still chop it up into infinitesimals, and maybe, ah, we we event earlier we divided it up into disks. We're going to do likewise. So basically, you're cutting it up into one small slice at a time, which brings in its corresponding part of the surface area, which would be actually this whole band here on the surface. Ah, why do I refer to it as a party trick? Because on the party here. You'll be given the apple, which is approximated as a perfect sphere, although of course it's not. But you know, we're not getting down to infinitesimal little deviations. So imagine an apple is a sphere, and we're gonna cut it with the same thickness. Notice I didn't say same volume or the same arc length, which is the same thickness. And meaning, we're gonna cut it like this. And they don't have to be infinitesimal. Just.Strictly speaking, the same thickness, and I made sure they're white. Eh? Oh, Eddie, you're just turning on your recording. Yeah, I forgot to turn it on. Oh, do turn it on. Well, then you forgot to record your brilliant performance. You've been really doing well since the this session. Okay, so we have these lines.Well, I'm not going to bother drawing. It's too difficult for me to maintain. They're all equal distance now, and fundamentally, you're cutting them into same thickness delta h. Now, why don't I just call that thickness the little h? All the same height, which actually means, yeah, it's the same height h, h, h, h, etc., etc. And you know, apple peel has magic healing powers. It's loaded with vitamin C.Fibers, da da da, it's magic, it would make you live forever. So, which piece you shall select with the most amount of apple peel? I remember something about like they're all the same. Yes, that's exactly right. That's the apple peel principle, and they're all the same. In fact, although it's a trade-off at the equator, you know here, and in fact, that arc length is a lot smaller than the arc length toward the pole.But this is filling up the entire dome. There, the area must be rotated around that bigger equator, which has a much bigger radius. So eventually, when it boils down to which one has a bigger surface area, we shall have to investigate it very carefully, just quantitatively. And for that purpose here, I'm going to actually prove. And first of all, during the process of calculating the total surface area, we would have already proven and, in fact, each one has the equal amount of appeal.At this moment now, for clarity of illustration, and I need to draw infinitesimal layer, but think of it as infinitesimal. Okay, it's not really infinitesimal, but think about it as infinitesimal. And we're going to call that the dy here. In the y direction, and I'm looking at this piece, and if that's the piece here, this is center, and why are you looking at? We are only looking at this part of the area. I know this is the unexpandable. You can't really.We expand it out, but we're going to do our best. Try to expand it out. Whatever this rotational area now, which is really like that, okay? It's made up of it's the arc length, which I call a ds, sweeping out the total circumference. So I'll say this da here. That da really just equal to the drclines, unknown times when you expand it out, it's becoming something like this now. What is the total length of this?And that is actually two pi of the little r. The little r is right here, correct? I'm using this little r to generate the circumference. Do you guys follow me so far? Yeah, then we just need to integrate. We just need to integrate, except for we're integrating over the dy now, and this stacks up in the direction of y. What none of the variables we're coding here is. Exactly why? So we have to find a lot of related rates. It's not hard to find that r now in terms of the y by Pythagorean theorem, and just mean the r squared minus y squared. Correct. By looking at this little triangle here. Yeah. Okay, we also need to relate r now, and I can relate.Let me read them. But in fact, it's easier if I introduce another angle as a parameter. And we notice here: if I call that angle theta, and this is the angle formed between the r and the your big r now, then the same angle. This is a piece of geometry. Okay, we we need to realize this is a tangent line, which is also the instantaneous direction of the arc length now. By being perpendicular to the radius, it's also forming the same theta.With the horizontal. Can we make sense of this geometric fact? Yeah. Good. But I can also, since I'm introducing the theta now, can I also write the R as a big R times a sign theta.嗯，yes。 And what about the s now? S in the direction of this theta, it's aligned with actually the horizontal, but the projection of this s now onto the vertical, it's always a d.My final goal is to write everything in terms of y variable. So can I say my ds now multiplied by the sine theta would actually give me the dy. Two projections. Does that make sense?Yes.Yes, now, and the dy is determined by the angle in between, which is also the theta. I know this is so crammed here, but if we're talking about the ds here now, that angle would be very big because, in fact, if you project it onto the dy, and that's your angle theta, right? Don't you still need to multiply the hypotenuse, which is the ds now, by the sine theta? Oh, why did I write.That as the theta, I meant to write as the y. That's the dy. Uh, and I still kind of don't get like the projection part. Let me actually draw another picture. Here, you you got a circle, and we're choosing a a location to start it at, where within the angle of a theta. Now, instantaneously, the DS part is going to be the segment of the arc lying on the circle. Now.And the DS is parallel or it's in the very direction of the tangent line, because just a straightened part. The tangent line is the linear approximation of the circle locally. So first of all, do we understand the angle between the DS and the horizontal? This angle is theta. That.That's the horizontal. Why? Because I V you think about these two, and this is perpendicular. That's perpendicular to the horizontal, and the other line, parallel for the two sides of the angle. Now that's perpendicular to this. So the two arms of the angle are respectively perpendicular to the two arms of another angle, and these two angles are the same. They're both theta.嗯。嗯，我 still kind of don't get why they. Yeah, Ivy, you have a trunk.You have an angle, and if you turn both sides ninety degrees here, I'm gonna say this is a BAC now. I'm gonna turn the AC vertical, and AB also, really just a perpendicular to. Basically, you rotate in the whole angle with a ninety degrees, so that each arm turns to be perpendicular to the original. I'll call that the B A A star, B star, and C star, and that's the same angle.Simango theta. Oh, wait. Maybe it's a little easier if I actually draw that angle on the other side. Yeah, that's the same angle. As if I'm going to call that the original y-axis, and that's the a-axis, and basically you rotated it so that the y would become this now.And this will become an A star, which is marked by the fact that the Y star is perpendicular to the Y because that's the horizontal, and the A star is perpendicular to the A because A is a radial direction, A star is a tangential direction. Do you see it now? Wait, so like the, are they like, um, is the theta is like vertical angles. Ah, yeah, these two theta, right there, they are vertical angles. Oh, okay. And we have proven that theta two is is equal to theta, therefore the theta one is also equal to.In fact, you don't even need to. We just need to know the angle formed between the tangent line, hence between the direction of the DS now, the local arc, and the horizontal direction. It's the same as this angle theta. Beautiful. I'm glad that you really nailed that and made sure that you fully understand what's going on, and then do. Are we okay by saying I V that when you project DS?Onto the vertical, you're multiplying by the sine theta to get the dy. Because you got the ds over here, this is my ds, and the angle with the horizontal is theta. But you want to project it onto this central diameter now. But this is.The same as that one, that's the dy. Yeah. Don't they follow that relationship? Yes. Beautiful, and uh, thank you Elaine and Eddie for waiting. You're kind to each other, so this little r I can replace it by the big r sign data, but the sign data could be re bundled up with the dy's to give us.The D Y isn't it? Yeah. So the two unknowns between the little r and the D S both are changing magically by cancellation of the angle. That just give you the integral, which is the two pi r of the D Y. But two pi r is a bunch of numbers. You can move them directly outside of the integral. The integration of Y goes from a event. Could you give me the lower and upper boundary of this in order to achieve the total surface area of the.Entire sphere. A center of the sphere is the origin, right? Yes. Oh, so it would be negative r and r. Brilliant. Beautiful, and you can see. Ah, Yilin, why didn't you finish the whole integral for us? Oh, would it be two pi r squared over two? Hey, you're forgetting which is.It's a variable. Now, y is a variable. R is a constant. You're integrating the location of the slices layer by layer to end up with actually total surface area of an entire sphere. So I want you to give me what is antiderivative first. Before you plug in the boundary.It is two pi r dy, the integral. Yeah, it is. And I'm asking for what is the antiderivative before we calculate the final number. Two pi r y. Yes, beautiful. And here we plug in the true boundary upper boundary minus lower boundary, so the y would yield two r altogether, another four power squared. So I think the theorem was confirmed. That's the surface area, but of course we infinitely prefer taking the difference of the volume. Interpret that way, the surface area area is really just a rate of change of volume.Regarding radius, isn't it? And later, when you're done with your with your test, now I'm gonna actually well next year, as we're doing physics together, we still need to add a lot of content to the calculus. Now I'll show you how to integrate for fun the so-called volume and the surface area of a four-dimensional sphere. And we can't visualize it. How do we even cut it? Right? What is a way to do it?And then you can see we are again going to find out the surface area of the forty sphere by finding the volume first and then take the derivative regarding r using Edison's idea directly we're getting the surface area pretty much for free. We're good. Yeah, but today it's all about all about differential equations. I've given you this example of the differential equations. Now we rotate it two ways. You can look at the raw differential or.You can already change it into a derivative, but how do we next time? If I just give you a derivative here, how do we recover what is the original function behind it? You might be thinking something akin to taking the integral would work, and absolutely you're right. And let's see how that plays. So detaching ourselves from the geometric setup, I'm just going to give you a differential equation, and let you use our common sense and eyeball techniques here to try to find out what is the.The equation or the solution to it, meaning recovering what that function is, where the derivative follow a certain equation now. For example, if I tell you that the derivative equal to the seven f itself, f is a function of x now. You can write is the y prime is equal to seven y. Could somebody guess what what is that function y equal to f of x? What function when you take the derivative, you're getting pretty much just yourself.Therefore, it would be scaled up by seven times. E to the seven x. What a brilliant idea! You got there, then right away. Well, let me ask you: What if on the f prime equals to the seven x times f of x? In this case, we not only get an actual profession, but also has an actual power of the x in the front.Use your eyeball power. e to the seven x squared. Almost right, Elaine. That's very very good, except for not.All together right here. So we're gonna actually follow Elaine. Why don't you share your motivation? Why you hypothesize something like e to the seven x squared? Ah, well, we know that derivative stays the same. Ah, so it's like e to the x because the derivative of e to the x stays the same, but there's like an extra constant in front because of the chain rule. It will be multiplied by something. Yeah, beautiful. She is actually using what she has already has already worked before. Exponential. Function has this very nice feature. No matter how many derivatives you are taking, we're always getting back to ourselves. They're always getting back to themselves. But in addition, we have this, and Elaine very brilliantly recognized: "Oh, it could come out of the chain rule. It could be the result of the inner function. So instead of putting x here, we're putting the x squared on the exponent." Eddie, could you perfect it? It's not right yet.Excellent. Adi, do you realize what she did? Divide it by two. Ah, why do we need to divide by two?X equals the X squared over two, meaning, and so that only if we actually have that one, when you take the derivative, that's going to give you X. Otherwise, we would end up doubling. So fine-tuning the coefficient comes after. I do highly encourage you to do what Elaine did here on the spot, meaning recognizing the overall shape of the function, and then you're going to tighten up a little detail.That's beautifully done. Well, I bet you guys know how to do upturns power now, right? If I start from here, what is my antiderivative? Really, I'm finding the antider. No, it's not the antiderivative. It's antiderivative of inner function. Would really just be e to the seven over the n plus first, and x n plus first power. Correct.Give me a confirmation. That's very powerful. Now let's try something in the close proximity. What if I actually look at the derivative equal to the cosine x and times f.E sine. Hmm, beautiful. And Eddie recognizes, look, still look at the cosine as actually coming out of the chain rule. It looks like we're done. I mean conceptually, whenever you look at the f derivative equal to the derivative of sub multiplied by something, that's something. When you find the anti derivative, throw it on top of the exponential of the e. Now we're done. And what is a.Proper language EnglishProper language. Language of solving a differential equation. We call that separable. Let me actually talk you through the formal steps. You guys got so good at eyeballing the results now. If you really want to solve it as an equation, we want to spell out what is the meaning of the f derivative, and equal to the df over the dx. Correct. Yeah. And we call that separable, and in fact on the AP there are only two.Kinds of differential equations you are ever required to solve, either separable or, it's homogeneous in a very simple way, and I can give you the definition. But once, well nowadays, the other slightly subtler kind has already vanished from the AP. I mean, the AP has been dum dum dum dum. We just don't see any interesting problems anymore. So this is basically you're stuck in trade. If you know how to solve this type, you're you're fine. And we call it separable, in what sense. In the sense that you can push all the terms here, be it a differential or a function, everything having to do with the f to one side of the equation, clear it, while the other side has only to do with x. Can I do that here? Yes, I can actually do the f divided by the df really just equal to one over the cosine x and dx. Oh, wait a second, that this must be multiplied onto the other side. Sorry. Hey.Would actually equal to. Let me actually. That's a poor way to to start. You always keep the differential on the top. What I did wasn't wrong, okay? It was right, except for that's not how how you want to write it. You rather want to keep the differential on the top. It's this way, and that's the cosine x dx. What did we accomplish? You end up actually two questions finding antiderivatives.On both sides here, that really looked like that begging for an integral. It is a differential form of a function of x, and that is a differential form of a function of f. Now, if they are equal to each other, their integrals necessarily would equal to each other, barring a constant on the outside. But if you regard that as a indefinite integral, and that just means finding the antiderivative, that means we have gotten rid of the differentials by finding the antiderivative separately for the f function.And for the x function. Do you follow me? Let's find the antiderivative on both sides. There's something tricky going on. Later, I'll show you.Real left side, can we use the natural log? And Eddie, only you I've mentioned this now to be rigorous. It's the anti-derivative of the f necessarily the natural log. Because remember, the minute you putting it inside, I'm talking to the whole class now. I only had a chance to mention.mention it to Ivy. Oh, sorry to Ivy, and the anti derivative one over the f is natural log of the f, all right. But the minute you're putting it inside the natural log, are we already assuming f must be a negative? A strictly positive absolute value. Yes, did I mention that already? No, no. So where did you actually learn that? Ah, I was just like, ah, using ah, like so.GoogleNow, and nothing is telling me that f must be positive. So, in order to hide ourselves here, we could add the absolute value. This is an even function, therefore the derivatives it's odd. And indeed, and this one over the f has potential to be either even or odd. So that's odd function. Very well done, Andy. What about the other side? Well, surely we do know an average sine x, and you can add up your constant c now. Correct.Absolute value would equal to the e to the whole thing, which actually I'm going to reabsorb the constant here as the object constant e c, and I'm going to let it into let it name it as actually some kind of a parameter a. It's just a number, and it's a e to the sine x. Makes sense. Same conclusion. Except for there's a few things going on now, they're not exactly the same. Well.First of all, earlier we eyeballed the answer without the A. Can the A be anything? Can this A be a little different from what Eddie eyeballed for us? He was he was giving us the answer very fast, which is simply e to the sine x. That's brilliantly done, but is that the only answer?中文嗯。I'm not going to answer it for you. You have to carefully double check. And by the way, would this be different from the answer we had before? In fact, this leads to the generic answer where the f would equal to positive or negative of the a e i. Correct, meaning your coefficient here could either be positive or negative.Is that justified? Eddie, what do you say?```嗯，Can we like try plugging in concrete numbers? You don't mean plug into the x or y. You mean you mean plug into the a, right? Yeah. Yeah, wonderful. Yeah, I think, I think, it can only be positive. The a can only be positive. Yeah, but could the solution to the f? Because eventually, when you put the absolute value over here, and for the absolute value of the f to be positive, that's by definition. I agree with you. That's true because after all, on the left-hand side, it's absolute value. But then finally, the solution for f though, could that be both positive and negative coefficient.In front of the e to the sine x. Um, I mean the only way for the f to be negative is is just like the a is negative. Like as a negative coefficient. But of course, a can't be negative because a comes from that exponential. But then no, it can't ever be negative. But Eddie, you yourself gave us the answer you're really here is not the f or not.Solving for a, we're solving for the absolute value of that. A can only be positive by a green. If it's like a complex number, maybe. Oh, that's not a complex number. We're by the magnitude here. But although we our kids here know how to take the natural log of a complex number, that's not the high school or the AP curriculum. They would never ask you to take the natural log of one plus i. People don't understand what that means on the.AP test. But idea, I like the nimbleness of your mind. I'm not saying you shouldn't know that we sure that you know because we're thinking just better on most of the high school requirement. But come back here, and this is if all these are real functions, they're not complex. By the way, let's go up with your AP calculus. You're always dealing with real functions.People don't even know that the e i x are equal to the cosine x and the plus i sine x. You know, we know that, and you you you you can feel free to use it on the AP if it's needed. I don't think it's ever needed. Elaine, what's your opinion on the choice of the coefficient here? I think Ivy is giving such a wonderful idea. She's saying, "Why don't I plug it back in."And test out whether it fits into the original differential equation. It's the only fitness test, right? That's only the only thing we care about. Yeah. Now I will go ahead and do it for us. If we go from here and we just take the derivative, what are we getting? We're getting plus minus a e sine.So does that recursively equal to the original f scaled by the cosine x?Yeah, mini. All these parts now really just at the term by term. They're gonna be recovered all over here. That's actually replaced by the f again. So this is rigorously true, regardless the coefficient is positive or negative. So I'm actually contrasting the two ways of solving it. You could eyeball answer, but that answer is not. Exhaustively all possible answers. Now this is why we do need to learn this technique of really separating them to two sides now and just go through every single step rigorously. And this is not even one of the tricky cases. Okay, later you shall deal with a lot trickier. But can we compare them and master the advantages of each method now? Both are fine as long as next time when you try to come up with the answer, remember the answer to this kind of a differential equation.Can be scaled by anything, because if you scale it by a constant, I mean a number now, then both side of the equation will be scaled accordingly. Not ruining your answer. I mean, whatever this new answer still fit into the same equation. Are you with me? And that coefficient, because of the absolute value, could come out. The final coefficient in front of the exponential now could come out as both positive and negative.Let's look at another example, but in this case here, it's not going to be just as so simple as f derivative equal to some kind of a function of a x times f anymore because it looks like their solutions are all exponentials. But let's actually do something entirely different. Why do I give you the y minus the y prime x squared plus a three equals to.Executed. My question is: Is it separable at all?EDIT. It is separable.I don't think it's separable. Correct, and how do we try to solve it in connection to what we have done earlier? I did. Do you confirm? Is it separable? I don't think it is either. Which part is winning the game? Just tell me one term. If you erase it, it will become 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minus a dy over the dx now x squared plus the three, but the point is when we multiply all the sides by the dx, and what you're getting is actually this plus the dx now. And at this moment, you could actually push all the y to one side, x to the other. How do we do that? Well, first of all, we're going to actually divide up, and first everything by the y, or rather, let me just solve it in this form now. The my dy over the the times x squared. Actually equal to the three plus y, this many of the dx, and I can actually divide that over. And divide that over. And so we end up with a separable differential equation. Do we know how to solve it? We do. By the way, we may even want to do so. We're still facing the task of solving this differential equation. Any suggestion? Maybe we could actually solve the part that is separable and then focus on guessing.The part that's not separable, meaning eventually you can make a guess now, find the religious eyeball a solution which is going to produce that executable. Think very hard, invent a way to solve it.嗯，所以，我们去试一试，一试这个。Antiderivative`````````I don't expect you to have the answer. Final answer, but I really do want to hear some suggestions. Which way we're going. Could we try to find the antiderivative?Antiderivative of something, and then well, I'll show you what it is, and then plus a y over the x squared. You want to take the antiderivative of that one, but it has the unknown function y in it. It's not only a function of x. Remember, taking the antiderivative worked before only because the variables are separable. This is premised upon we could separate the two sides of the equation as pure and simple a differential equation concerning one.Variable, and then you could actually perform your antiderivative. Is the left side just natural log of y plus three or three plus y? Oh, Eddie, you're actually doing for this here, meaning it's a simplified version, pretending that we got rid of that square. Yeah, we can do that. It is the natural log of the absolute value of three plus y. That's fine. And the right hand side. What about.Negative one over x. I totally agree, and don't forget to add in the object constant c here. At this moment, we're fine. Yeah, I meant taking the antiderivative of that equation. Of which equation? The one you circled in red. Ah, meaning you're suggesting the same. Yeah. Beautiful, it's a good start, but we do know it's not the answer though, because we can't just neglect it.And that term here, right? So how do we bring that in? This is what I mean by maybe you can eyeball a few actual terms here just to make it work. If you make a wild guess now, and it's what kind of category of the functions when you take the derivative would form this kind of relation. In fact, at this moment here, when you plug this in, the left-hand side is already zero.It's guaranteed to cancel out now, so it looks like we just need to come up with some other combination to actually give me the x cubed. Well, if you want to take the derivative and do some kind of a combination with the original function itself, and then still end up with a simple power, what would be your first guess in terms of just the major family of the functions that could possibly serve as a function yYou repeat the question. I'm saying, can you make a wild guess? Just what form of the function y when you do this kind of combination, you're going to end up with a power. It's cute. For example, is this a trig function? Not very likely. Because when you take the trig, the result is another trig. When you multiply that.That's not going to cancel out with the previous trigger, and then that doesn't seem to work. Is that exponential? No, because no matter how hard you're taking the derivative, exponential function, there's unconscionably a exponential term there. You can see, yeah, there will there might be some mutual cancellation between these two, but they won't completely cancel out. Give you a leftover of the SQ the minus three. Right. So basically, this is what you're getting. What we want to do is do identify what would be the plausible candidate for this y now, so that when you do this kind of combination between the y and the derivative, you're gonna end up the polynomial on the right. I was thinking of polynomial. What a brilliant idea! What's the degree of your polynomial?I didn't. I'm sorry, Elaine. Come on. A highest power would be like x squared or something. Wow, why did you suggest x squared? Because when you take the derivative of that, it is the power is one, so when we multiply with x squared, the highest power would be cubed. Great. So Elaine is recognizing between these two. If we guess why it's a polynomial or not, and then between these two, this is the one with the higher power, because when you take the derivative, you drop the.Power by one, but you multiply back by the x squared, bumping it up into the original power plus one. For that to be matching with the highest power three, we need the y to be a quadratic. Go ahead, find it out. If you don't know what it is, just set it y equal to the ax squared plus bx plus c. You know Elaine, this is a really I think you have fully arrived at being a true student now, even before you're given a set of formal tools to solve the equation, and you can.Go your way around toward it. Actually, just know what to guess, how to guess, and do it very smartly. I'm very happy. Anyway, go ahead and find out what exactly polynomial fits. There will be one.```嗯，嗯。```You're seeing some contradiction. Tell me. Can we locate such a three coefficients a, b, c to make it work? I think I'm having like some contradictions between a. B good. What's the a you are solving? I got negative one half. Positive or negative? Negative. Yes, beautiful. I am reaching the same conclusion: a equals negative one half. And then what's your b? Well, her if a minus b equals zero, then b has to be negative one.Well then, the x term would be zero. Brilliant. That's right. So in fact, we don't have enough coefficients to accommodate because on the left, you're only free to set three coefficients, but we need to fit into four coefficients on the right. So unless we're just lucky, we still didn't get a solution. But Elaine, your guess was actually wonderful. That's exactly how we're supposed to think in calculus. Well then, you know how to solve this now. I'll.You first, it's beyond the AP curriculum requirement, meaning you're not required to solve it. Maybe many, many years before you were, but no longer, okay. But I will show you how to solve it here because it really illustrates how to solve in general this kind of a combination between the y prime. So can I have your attention, please? Alright, I'm going to actually the idea is to bundle up this part or every single part with the y into a complete.Derivative, as well as come out of something which is a combination between x and y now, so that I can bundle this into a whole function taking the derivative. And if I do that here, I'm gonna actually first clean up the y, and because first of all, I think of the product rule. Taking the derivative of product rule would immediately give you two terms. So my tentative hypothesis: Eddie, I really want your attention. Can I see your video, please? Thank you.What did I just say? What are we gonna do to the left hand side in order to make it solvable? Separate out the derivative. What? Separate out the derivative. Separate out. No, what do you mean by? What do you mean by separating out? Were you listening to me? Yeah.And Elaine, what did I say? What did I propose to do? You make the left-hand side in a complete derivative. Beautiful. We're trying to actually make it into a one single function, the complete derivative, so that you can find the antiderivative. And this, whatever inside, would be a mixture between x and y, because you begin with two terms here. Our I would say, in the natural way to think, would be consider the two terms as a result of the product rule.But if you do so, you think this y times another f, this is another y now. Then I'll call that another function g. G is also a function of x. Oh, I I haven't found out what it is now. If I take that derivative here, it's going to come out as y derivative times g, and then plus y times the g derivative. And in order for that to fit in here, so looks like we need to choose a g equal to x squared. But you can try it out. If I do so, the other term doesn't.So what could we do? We can still actually scale it by something though. So we want to hypothesize, and what do we do here? Just so that when do when we're doing that here, the ratio between the two covariances between the y and the y, uh, y derivative term here, the ratio between that should fit into what we have here. That's going to give you additional differential equation regarding what g we're picking, and what is that criteria for g. In order for this to fit into that, barring some kind of a scaling factor, scaling function, what can we say about g?```Eventually, they should almost equal to each other. They can't strictly equal to each other, but they could be scaled version of each other by some function of x now, which I can push to the right hand side and join the part where taking antiderivative. The whole point is, I don't have a want to have a leftover wise. We need to actually deal with the inseparableness between.By creating another function and make it separable. What we can do is let's actually try to fit the requirement now, and the ratio between the g and the g prime. I'm writing the g prime over the.For the reason, which will be obvious soon, would equal to the same ratio between the coefficient in front of the y and y prime. It would equal to the one over the negative x squared. And this is a separable differential equation concerning x and y in itself. I'm gonna leave it as homework. Solve that differential equation. This is well within the range we've been doing today. Because if you expand it out, oh, doesn't that look familiar? You can do it by.Just eyeballing the answer. Keep in mind there's that arbitrary coefficient this time because you're getting solvier, or you can solve it using the formal steps required by the AP. Meaning separate out the g and x. Now remember the g itself is a function of x. Ah, once you solve the g, you can actually plug it back in and rewrite it in in that form now. Then you actually have a new differential equation between this function and the other side of the equation. That's not what we're having here because at the actual.Scaling coefficient. The scaling function is a function of x. This is the very first time I am leaving you a serious homework, which is not only calculational steps or recap of what we have learned before, but rather it has some genuine integration to do. Not integration as a technique of integration, but rather synthesizing what we've been talking about. Take it seriously, because you are moving up to a different study method now.But did I lay out the road plan clearly? Yeah. In fact, this G has a name. It's called the integrating factor. I believe it's out of the AP curriculum now, but nevertheless, in practical application with calculus, this is just so popular that you cannot do without it. It's homework clear. Yeah. Alright, do keep up a good work, and thank you for coming. Dating my time. Yup, I'll.

Surface Area, Related Rates and Differential Equations

Surface Area, Related Rates and Differential Equations

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