Series, Inverse Derivatives, Polar Curves and Differential Equations

Perdence you, where the having to format your files, that's not the priority. So I would say, use all the time just do problem solving, and when you do meet here, bring whatever the the issues that emerge out of your own problem solving. For example, analogy point is not solid, or a kind of problem is taking too long according to the standard solution, and you wonder whether it's a better way better often. I would say ninety percent of the time, there definitely would be a better way to to solve it.And I will be able to feel that. Okay. Oh, actually, now why don't you share the screen? Ah, can you conveniently share the screen? If you could, then find out the twenty twelve. There's a public, or you could you don't even need to name the year now. I just do public sample test of a twenty twelve, or rather just AP calc BC. Multiple choice.Well, we had to complete the test, including not only the multiple choice, and I think they only post one on record. This one's a fall twenty fourteen. Oh, that's good. That's good. Okay. Ah, yeah, it was ah May eleventh. Thank you. Thank you.Alright, like I said today, I'll do all the talking. It is only sample exam questions, or is that a complete exam? Please scroll down. Let me actually see how many problem they give us. Wow, that's not. Ah, we're going to do these, but on top of it, try to find the.Just keep the official test APBC calculus. Meanwhile, let me actually do the search myself. Hmm. May I stop sharing and find? Yeah, of course. Alright. Find another thing. Hmm.Yeah, it is actually twenty twelve. You can type in twenty twelve. I think that's the only complete set they post. Twenty twelve. Yeah. Public practice next time. Okay. They give.View the entire booklet. A lot of that just waste. Okay, I'll share it. Yeah, please.Very good. I would go through the following process: I will first name the knowledge point, and then the major procedure, and then I'll start solving it. And I do it in the hopefully in a simple way. And those are not necessarily the textbook version of the. Solution.As exercise, because I don't believe you you would have any trouble. It's just a straightforward taking the derivative. However, okay, let's go from number one, please. Okay, so for this one, using the chain rule, what are you getting? It's a sine of the x squared and then times the cosine x, and definitely that's the one using chain rule. Number two, I'm gonna as I read here, I.We'll mark the the keywords. So we're saying by the ah, so you know the x and y both return the parametrized function, and we're finding basically we need the dx and we need the x dot equal to the y dot equal to zero. The two components of the velocity both equal to zero. So this one requires that the three t squared minus six t would equal to zero. That gives you two solution t equal to two and t equal to zero. But at the same time, the other one, the y.Dot equal to the twelve minus sixty equal to zero, which yields only one solution t equal to two. Therefore, the t equal to two is the only answer. However, we're not answering the question in terms of the t, so we have to plug it back in. If the t is equal to two, then the x would equal to a negative four and the y would equal to twelve. So a would be the correct answer. Alright. Makes sense. Award.So we're given a graph now, and we want to do an integral. This is interpretation of the integral as the signed area covered under the curve. So we're integrating from zero all the way to the four, and we're saying the positive area is six units, the negative area is negative six, so they totally cancel to give you a zero. Makes sense. Okay, I want to. Uh huh.So we want to find the length of the curve, and you're given explicit function. And we do know the length of the curve is the integral of infinitesimal length of the curve, which is going to be by Pythagorean sum. That just give me the dx squared plus a dy squared. So when you pull out to the dx now, what you, hey, come on, I come to right here, and the remaining part is just one plus the y derivative squared. This is from the infinitesimal analysis of the of the arc length.Next time, in case you forget the formula, I'll just look at the infinitesimals chunk of the DS now. Decompose it in Cartesianly into the dx and dy, and then. Yeah, you're right. Because the derivative, yeah, you got it. Well, what if, in case they give you a weird function which is hard to take the derivative? For example, they give you y equals to arctangent. Oh, that's also weird, but let's suppose, okay, they give you arctangent of that.Now instead of taking the derivative of the y as a function of x, now you could equally pull out the y, and then find the y boundary from the y one to the y two, and inside this one plus the derivative of x with regard to y, x actually equals to the tangent of the y. You could also do one plus a second y squared. These are both the same. You could take it as a y as a function of x, or x as a function of y. Notice the symmetry of the beginning formula. They don't really differentiate. It actually puts the x and y on equal footing. Okay. Makes sense, ah, onward. You know, I have the feeling the problem would be easy for a while. I have, I actually want to do it backward so that in class here I would hit more difficult problems. But let me finish this one. We're given a power expansion. Immediately, you see it's a geometric sequence.So I can directly write it. They share the common ratio of the ninety x over four, and they properly start from zero. So the entire sum is simply going to be one plus x over the four, and then you can just plug in, and the answer is four sevenths. Right. Yeah. Well, in case you don't remember the formula for that, how would you? What would you do? You write out a few terms here. This is going to be one, and then plus whatever that are, and plus r squared plus.RQ the infinite item. You just notice this sequence has a self-generating pattern. So if I want to find an x, we might as well just multiply by the r now, and almost you're getting itself. So we do know minus i just use one, meaning I'm showing you a way to derive your geometric sequence formula. So be smart about the pattern. We're good now. Okay, uh, when we get to the the end of the multiple choices, and I'm going to do it backward.Because the beginning section usually they're agonizingly simple. Very good. Okay, so we're looking for a limit, and apparently this is zero over zero case. Otherwise, it's not even worth doing. And the regular students will do Laplace's rule. In this case, it's not hard because you're taking a test. I want you to really just go for the shortcut of doing Laplace, which means once you secure zero over zero, take the derivative.On the top, that force into the fundamental theorem of calculus, the derivative of the upper boundary just give you the g of x. Now, on the on the bottom, you got the derivative of x, and this is evaluated at one. And this is no longer zero over zero. You can just plug in what you're getting is six over three, and then the answer will be two. Makes sense, ah. However, let me actually show you. I think I'm even faster, but I want to think intuitively. I don't want to go for a rule, meaning.On the numerator, I just I know this is zero over zero, so I want to do infinitesimal analysis. X is extremely close to one. X is approaching one, so I write the one x as a one plus a delta. So since it's a sliver of the area integrating from the one to one plus a delta, I just equal to the base times the height. Right? Yes. Yeah, because that's the function value on the bottom now, and this is a taking the function nearby.That's just equal to the slope on the multiplied by. That's the rise I want to do slope times a run. The run is a delta. Oh, so I don't mean the derivative at six. I mean derivative at one. But this is saying the g of the one plus a delta minus the g of the one. That's the rise. It's a difference in the y value. Really just equal to the derivative. We're doing the linear approximation. Derivative function like a slope. The multiplied.By a difference of the x value. So you can see, why when you cancel the delta, it's the same conclusion. It's the six over the three. Okay, makes sense. Okay, however, on the AP, if you see a limit of an indeterminate form zero over zero, let me tell you, right away, that's the quickest way to solve it. Twenty-seven. Okay, we actually want to find out. They both converge. You could eyeball it. This is a proper geometric sequence, so we need the common ratio to be less than one. So that just means the p must be less than two, right? The p would be from negative two to positive two. However, that's a p test, so we need the two p here to be greater than one. Remember that's a p test now. We need basically there's a power function to decrease fast enough. So that requires the p strictly. It's not the magnitude, but the p strictly.Okay, it needs to be greater than one half. So this is the answer. All right. Makes sense, ah, and remember both are open. You can't include either because when you include either, and this is adding a bunch of ones. That's definitely divergent. Okay. Okay, ah, the surface line tangent to the polar curve. There are several ways to go about.And remember, if you don't remember any formula, I hope you don't. You go from the raw derivation. I'll show you two ways because I'm because I'm good at it. But I would say probably easier way to go for the Cartesian. If you want to find the dy over dx now, and keep in mind your x is the arc cosine data, your y is the arc sine data. So your dy, remember when you differentiate, you have to do the product rule. The dy is sine data dr.Plus R cosine data D theta, right? Yes. And your dx now is the again product rule. It's the cosine data of the dr, and then plus R negative sine data of the D theta. Now we need to actually plug in the dr over the D theta. For that purpose, here I'm going to divide both the top and bottom by D theta. So if that's the case, what you're getting is I'm copying that term first: R cosine data, and then plus the sine data.D R D theta, but the D R D theta is really the derivative of this function that is equal to twice of the cosine theta, right? Yes. So I'm multiplying twice of the cosine theta. On the denominator, likewise, we're getting a negative sine theta, and add on to the twice of the cosine theta squared. I'm again plugging in the D R over the D theta, D R over the D theta by twice of the cosine theta.Now finally, I'm going to plug in a number. If the theta equals zero, then the sine terms are gone. We're left with only the arcsine divided by twice the cosine theta squared the cosine of one, and then the R here. We have to plug it in when the theta equals zero. Are actually equal to one, so the answer is one over two. Do you fully follow me? Yes. Okay, very good. And I said earlier, this is not usually my style of doing a problem because.Because I want to stay very close to the graphical meaning, so you think about the grid lines in the polar coordinate, and you think, you think about the local one. First, I want to find out where where is the location. I'm actually looking at theta equals zero. I plug it in the r equals one. So in fact, we're looking at this location where the theta equals zero, r equals one. And now, gauge from the tangent line. Let me actually change your color. The meaning of the dr over the d theta, it's actually saying, if I have a.Small increment in the theta direction, which is the rd theta. Remember that's not just the d theta, the increment in that direction, the rd theta, but r equals one, so that just going to be the d theta. And then I have rise, that's the dr. I did the dr over the d theta, but I need to scale it by the r now, and that's going to be this rise over run. That's going to be this slope. That tells you in the radial direction how much you increase, and in the theta direction how much you run. That gives you that slope, slope of that angle, gauged from the tangent line, and that when I plug it in, I'm we're actually getting a two. Right. Yeah. Okay. Well, then fortunately, you could eyeball locally. This tangent line is vertical. So, in fact, what I do know, whatever this tangent line is forming, the if I put that the theta, then the tangent of that theta is equal to two of such angle. But we're looking for the slope, which is.The tangent of this angle—that's the complementary angle. So it's a reciprocal thereof. So it's one half. Alright. Okay. So the point is, you should understand that the derivative in the polar coordinates, the d theta over the d r, the d r over the d theta, also has an angle. It has a two it has a geometric meaning. Generally, if my curve is not so fortuitously located on the.But rather, it's right here. And then, what is the meaning of this r d r over the d theta when you scale it down by r? It actually means locally, according to the radial direction, and what is the angle here? It's the curve, the rise over run. It's the d r as opposed to the run, which is r d theta. And that's the relation to z center, right? Yeah, that's right. That's right. So this is the angle.Form the between the local tangent line and the radial direction. And first, if you want to translate it into the the angle formed between the axes, so you need to actually subtract it from that or attach onto that the angle here, this angle here, which is a local radial direction. Do you follow? Okay, twenty five.Oh, this is just a so easily evaluable a chain rule because we're noticing and the inner function is x squared and that happened to be the inner function. So my hypothesis would simply the anti derivative of just this. But now I need to fine tune the coefficient. If I take the derivative, I know I not only get that, I also get negative two x. But I don't have negative two x; I only have x. So therefore, I need to do negative one half. But I just don't.Don't write it there, but rather, this is a negative one half. I mean, that's my anti derivative. Okay, crystal, and then we plug in the boundary one and infinity. Infinity side is zero, so then here is a negative. So in fact, we're just getting a negative one over a two e. But because we're subtracting the lower boundary, so that becomes positive. Alright. Crystal.So I would say the moral lesson is: you look for if something is more complicated than immediately eyeballable derivative, then you look for the possibility of a chain rule, meaning part of the integrand needs to come out of the derivative of the inner function. Our two twenty-four. Okay, so we're actually saying it's a differentiable function, so the we're given a equation concerning it. Now I want to find.We want to solve the f, so we need to actually get rid of these integrals. And what is the quickest way to do so? These are antiderivatives of why don't we take the derivative on both sides of the equation? So immediately this becomes the f sine x now. I mean, I skipped the same f of x now, and would equal to the derivative of this using the product rule would equal to unfixing the a the f now taking the derivative of sine to give you the f times sine x, and meanwhile fixing. The cosine minus f derivative times the cosine x, right? And then plus the derivative of that is simply going to be the four x cubed of the cosine x. So they go away. What you end up is a differential equation. Exactly, the f derivative actually equals to either. Remember, either you have the cosine x equal to zero, but it can't be perpetually zero. So we can delete it. f derivative equal to the four x cubed.So, makes sense. Yes. And therefore, the ice could be at zero fourth, but wisely they say could be. That's not the unique answer because we know the derivative, the anti derivative could be plus the arbitrary c, right? Okay, so that's actually this is really a differential equation in disguise. But what motivated us to try to get rid of the integral is that we're.Looking for the f now. Then surely we want to get rid of the integral, and the easiest way to do that is to just take the derivative. Makes sense. Yes. Okay. onward. Oh, this is a logistic function, not quite because ah none of that is directly a quadratic equation regarding the p, except for this is a quadratic, but it's not in the same direction. But any what, in any case, these are just differential equations.And we want to say linear growth in the size of the population. Linear that really just means the DP over the DT is constant, right? Linear as constants. Yeah, yeah, you got it immediately. But if I ask you which one is concave up, which one is growing faster and faster with the t, then what would you do? How could you quickly immediately eyeball a solution? Well, that really just means we're given the derivative now. We want to find out.Which one would give you a constant positive second derivative? Right. So definitely, yeah, that works. For this one, it depends on the t. It's positive or zero or negative. So whether this is a concave up or down, in fact, that's ambiguous. In fact, there's an inflection point. So this is not always a concave up function. But it gets trickier if you're actually looking at a differential equation that depends on the p, not dependent on the t. However.I'll tell you a trick. Now remember, if the P is a concave up function regarding the t now, then why don't we just look at it as a function of, for example, you're looking at a graph like this, and you think of the t as the dependent variable, the P as the independent variable. Then, if you think that way, is that concave up or concave down? You have to flip it basically across these angle by.The forty-five degree. Then let's come get down. Let's come get down. Now that actually means if you find out your differential equation is written in terms of the p, then you might as well think of it as as dp. I'm giving you the most complicated one, which is the e naught dp over the d dt over the dp is one over the one hundred p squared, right? Right. And we want to find as the p increases, does it increase or decrease? Is aCan't give up or can't give down. In fact, that depends. The p is positive or negative, so this also has an inflection point. But if you look at the d now, then the dt over the dp is going to be the one over the two hundred p as p increases, decreases. Therefore, it's can't give down. That means that's also can't give up. In fact, the answer is exponential. This is the exponential solution because the derivative really equal to a scaled version of the function itself. So indeed, just keep in mind, if the function is a concave up, then its inverse function, it's actually concave down, because graphically, when you do the reflection across the angle bisector, that actually changes the concavity. Precisely. Yes. Okay, our. So what what must be true? We actually say it converges at.Actually, go five now. When you plug in the five, you're getting two to the n power. Well, this is not helpful for convergence, and instead, it's definitely dependent on the convergence of an. So this is telling us: if you do the an plus one over the an times the two, this magnitude must be less than one. That's the condition for convergence, right? Right. Okay, but when we plug in each one of the numbers here.And what you can see, which one is going to actually make that magnitude greater than one. So when you plug in a zero, this is a three now, which is bigger than a two. So we actually don't know. That might be divergent because we don't know whether the three times the an plus one over the an is still less than one. Fundamentally, we're comparing. Let me actually give you the intuition first. The convergence, a guarantees that the radius from the center three.It's a tool. So this one simply tells you the radius of convergence of the greater angle of the tool. Meaning, when we have the distance of the tool now, then it is convergent. So you're comparing which one of these is closer to the three. And you can see the zero has a distance of the three, which is greater than the two. So we don't know anything about it. The one is magical about five, but it doesn't. Oh, I mean, I should say a few words.Why we cannot actually draw the conclusion: it must be convergent at one or divergent one. Neither is right, meaning we cannot know that for sure. But how come though they share the same radius? The five minus three equals two, one minus three is also equal to two. So we're going to come back to it. Two, it's a smaller radius, so it's guaranteed convergence. Six, that's even bigger. Okay, so now tell me why both could be possible, and we don't know until we know more about the sequence. For B and C.We can't say either must be true. The magnitude could, could the limit of the magnitude could be one. Yeah, it could be one. That's right. Meaning it's the same radius, but if I know it's convergent at five, why don't we know anything about the convergence at one? You are right. It could be one.It could be one. So I think you're trying to tell me two different things could happen on the boundary. This is really the three plus a two minus the two here. Two basically, these are on the boundary of the negative one to five, right? Right. We just don't know which is closed, which is open. And the reason that both could happen, either could happen, is.Let me give you an example. I'll say my an happen to equal to a ninety one to the n power of the n times a two to the n power. Then can you confirm at i equal to five, it's really convergent. It's negative one to the end over n. And it's convergent by what? By what test? By the alternating series. Very good. So not because of the magnitude, it's convergent. It's not absolutely convergent, but it's just the fortuitously convergent by the alternating series. But what's going to happen at x equal one?When you plug in that, you do one. What are we getting? One over n. Indeed, and that's divergent. Meaning it's at the one. Apparently, this is alternating now, but the whole sequence becomes positives. We lose the alternating.Therefore, we lose the convergence. So remember, pass down the two boundaries here. That's the only location where the convergences of divergences could differ, although they share the same up. Oh, can you mark the ones that you definitely want to take a second look before the test? Ah, twenty-two definitely mark it. I think trace back a little. There was one too because it's a little irregular. Oh, that integral function.When you actually look at that, they give you that integral, they want you to look for what is Alpha X Market, not because of the problem itself is so difficult, it's not, but usually they're just not so they're not so cliched in terms of the problem's type. So I would rather that you take another look in the future. Alright, ah, may I scroll down to find it? I, yeah, please. This one.Yes. Hmm. Alright, the others are trivial. Alright, we actually say, ah, we're looking for given this is a horizontal asymptote, and we're looking for which could be the case. Well, to have a horizontal asymptote, the degree of the top must be smaller than or equal to the degree of the bottom, and but it must be equal to, otherwise, the horizontal asymptote would be zero. So.So this will do, and does it qualify? Yeah, because that ratio. If you just look at the leading terms here, that ratio is indeed five, so that's correct. Oh, that's negative five when you look at the leading coefficient. So that's wrong, and this is approaching zero, and that's divergent, and this is approaching zero. So that's the correct answer. Alright, make sense, ah. Yeah, this is behavior of the rational function, asymptotic behavior of the rational function, and you need to pick the predominant term.We only care about the leading terms and nothing else. Twenty two. Oh, I mean twenty. Thank you. Oh, this is partial fraction. You guys are very good at this. We break it down into x plus the one and x plus the two. We decide what's on the top by plugging in the ninety one to get what is number here, and which actually going to give us the three. And then knowing that the three, this is just a two.This sign? Do you plug in the other one? No, no, no, no. We plug we plug the negative one into the remaining part of the denominator. I did. I plug the negative one into whatever that's left, which is including the numerator and the exponential. That is what I did. And we plug in the negative two into the remaining part of the numerator and the negative denominator. Right. Right. So, and we're just looking for the natural log of the x plus one, and q because. It's the coefficient, and then multiplied by the x plus a two squared, right? It's the whole magnitude. And when we plug in the two boundary from zero to one, now at zero, and we're just ending up the minus the log of a four, right? At one, we're actually looking at, uh, eight times the three nine, so we're looking at the natural log of the seventy two minus the natural of four, but according to the log.Get identity. So this is what you end up with, which is the natural log of the twenty-three. Ha, why do we not have that answer? Natural. Natural twenty-three. Oh shoot! Shoot! Sorry, I misdivided the eighteen. Oh, I did seventy-two divided by four to get twenty-three. I don't know what happened to me. Seventy-two divided by four.Sure. It gives you idea. So that's the correct answer. Okay. Do you really follow all the details? Ah yes. Okay. Good. Ah onward. So we're given the function on the group of have property the tangent to has slope for one half. So we're just taking the derivative and let's be smart. And first, I want to rewrite it's the one on the minus a two of the.x plus two, instead of doing the original, we have to use a quotient rule. This only takes the power rule. So the derivative just equals to a two over x plus two squared. Because that's just going to be a negative power, right? Right. So we want that to be one half, meaning we want the denominator to be four, which means x either equal to zero or x equal to a negative four. Right. These are x value.So I can plug in the y value, and the both would fit. So in fact, this is the correct answer. Makes sense, ah. Okay. So we're looking for knowing. Oh, that's the derivative. So we're doing the integral by looking at the area, and because we know the f of two equal to one. Ah, that's the beginning point, and now we want to find out the ninety-five.So I integrate backward. I actually know the f of two, and minus the f of negative five by using the idea of the integral gives you the difference in between. We'll give you the area in between, and this is a function increment here. And if you look at from here to here, what is the total signed area? Now this one has the area of three units. That's positive three, but this is negative of a two pi because the total circle is four pi. So we're doing the three minus.It's a two pi, correct. Hmm. So, f two minus f ninety five, is actually equal to the integral from the ninety five to two of the f prime of x, which we worked out. Crystal.And then you could isolate it just equal to the one minus three plus a two pi, which would be this. Okay. Nice answer. Just be clear headed about from where to where. What is upper boundary? What's a lower boundary? Seventeen. Oh, we are very familiar with this. A this. Function now, and by the way, you could probably just recognize what it is. It's the sine. Except by the way, you're not required to know it. It doesn't really matter. But I hope you could recognize going from our familiar because the minute you see in the alternating and only the odd factorial, definitely you're thinking about the sine. The sine x is x minus and then plus x fifth over the five factorial. Da da da. And this is really just the sine x over the x. Yeah. Right, so that's it.Right. Okay. However, just in case, okay, you actually don't remember, and what is the expansion of the sine or cosine. Well, then this is where you really just can take advantage of the fact is multiple choice, and you can do the match. Well, apparently it's not cosine, apparently it's not sine, and then the rest of it. If you want to do a comparison, what you can do is to take its derivative, because if it is any.One of this, at least you remember what's the expansion. Or, I think what I want to say is, at least you remember something. You can just plug in the expansion of the e, and you can see it doesn't fit. The point is, in case you don't remember all the formulas, yeah, do smart elimination. Okay. Uh, onward. Alright, we're given a differential equation, and unfortunately, it's not even separable. But we're not supposed to solve.We're using the Euler method with the two steps of equal length. Remember, this is a linear approximation, and we set a table. We set our x, and that's the y. That's the y derivative. We start at x equal to one, and we want to look at two. So with the two steps here, we want to look at one, and one point five, and two, right? Yes. And we're given f one into the three. We plug into that slope field. We can work out the local derivative, which is negative two. And this is a slope we're using.To do a next step of the estimate, so the y here would be the rise over run. Well, basically, it's the a the function value here is base of the three, and plus the rise. But the rise, it's actually the run, which is a point five times the slope, which is negative two. I'm using this value here as a slope to multiply the delta x. And add onto the original function value. Alright, so the.Formula would be the y at x plus the delta x, really just equal to. This is a McLaurin expansion, where the only two terms, y at x, and then plus the y derivative at x, and multiplied by the delta x. That's called linear approximation. So when you calculate it, what you're getting is a two, right? Yes. And you plug them again back into the differential equation. What you're getting is a negative zero point five, right.And then you do it again. It should equal to a two, and then add onto another zero point five times negative zero point five, so it's two minus one fourth. So the answer will be seven fourth. Chrisa. Yes. Question: Is it the overestimate or under? Meaning, you need to look at the concavity. You need.Need to understand that that's the derivative increase or decrease with your x. Yeah, the it increases. Ah, meaning yes, it I mean it might be decreasing, but it's getting less negative now. Yeah, you're right. So this increases as x increases. Therefore, is that over or under estimate? Hmm, under estimate. Brilliant. That means your function can't give up. That means the tangent line stays under the curve. Well then, that's very good. onward. So looking at the graph, which of following is true? Oh, let's just reading the graph now. We're given the f, we're required to find something of the age. Well, to begin with, because the h derivative is the f, right? Right. So we know that the h derivative is six is zero. So we know for sure that's zero. Now what about the.Function value at age means it means the integral, and remember we're integrating from the zero to that number now, and however from zero to here it's negative area, so we actually know that the integral, which means the area, is negative, so we actually know that h is less than zero. But what about the second derivative? The second derivative is the first derivative of this function at six. That's increasing, so in fact, the second derivative.It's a first derivative at six, that's greater than zero, right? Right, yeah. So A is correct. Okay. Ah, could you also order them at two? At this place, so which one would be also in the ascending order?So, uh. H of six is less than. And you mean H of two. Or H of two, yeah. It's less than H derivative two. You're very smart. You have to explain that to me. It's not obvious. And then, what about second uh H double derivative.Ah, what do we know about the H double derivative at two? It's zero. Yeah, and what is the derivative? A negative four. Uh huh, but how do you know that area is smaller yet than negative four? Because this is the area, right? Right, which is indeed also negative.You get me the the answer so fast that you were right. I thought either you were being really brilliant or you just were lucky enough to bump into the right answer. Let's, left, Reiman can. Yes, oh, that's really brilliant. That's correct. In fact, because of the sidelines, is a tool now, so that area is indeed less year to that ninety four. Left Reiman sum is smaller in magnitude, but it's actually greater in real value. Brilliant.Good, our. So, the logistic differential equation. Oh, you can't tell which one is logistic. What is the definition of logistic? If you don't remember, I'm going to remind you: it's a fish pond, it's the growth rate. Experimental. No, no, no, it's a not exactly exponential.The growth rate follows when there is no fish. The growth rate must be zero. When there is a carrying capacity of the fish, the growth rate must be zero. And it's the ecology says ecological system where the growth rate reaches a peak with a medium population, and it's quadratic. That marks, by the way, this is a population. The growth rate depend on population, it doesn't depend on time. Which is marked by e. Remember logistic. Definite equation marked by the rate of increase. It's the upset down parabola regarding the function population itself, not regarding the independent variable t. So that marks a logistic function. Remember to solve it. You have to do partial fraction, and the the solution is not exponential. The solution looks like this. It's giving something. Come on, sorry. There's a carrying capacity, and then divide by one plus.So minus some kind of alpha, and this is the e of the. Well, basically, there is a negative coefficient of the KT. So when the T is approaching infinity, the pound is approaching the carrying capacity. When the T is approaching negative infinity, and this one, the it's actually approaching zero. So in fact, the solution is an S shape that would actually look like that.Remember, you're not required to memorize all of these, although it's better. You're better off if you do remember them. Otherwise, it takes quite a few minutes to resolve the logistic function. I would say from now on, you start building your personalized cheat sheet. You want to bring these templates of the problem-solving types—the most important ones—and just derive them once and write your conclusions.There, sort of, but before the test, you just commit them to memory. Okay. Next one, please. Again, it's a proper geometric sequence, and if you want to look at the ratio between the adjacent two terms, it just can be the x over the fourth squared over a three, right? That is the r and.We need that to be smaller than one, so we actually need the x minus four to be smaller than the root three. So the radius of convergence now. There's no need to solve for x. This is the radius of convergence. It's just root three. The four is the center of the convergence interval. Interval so that means your x would range from the through uh four minus root three to the four plus root three. All right.Both are open. Both are open. Although on the positive boundary, it is alternating. Although on the negative boundary, it is alternating. But there is no remaining presence of n to make the magnitude approach zero. So definitely, both sides are divergent. Okay, number twelve. Oh, we're actually looking at the must be true, and we want to find a differential equation satisfying it. And but these are.About the local minimum or maxima. What does that mean? Now, surely, if you're looking for the optimization, you're setting the derivative equal to zero, or on the boundary or undefined. But this is everywhere well defined. So if you actually set that to be equal to zero now, what you're getting is the y would equal to a negative x squared. This is a critical point, and where is that even satisfied? Well, in fact, these are the locations where this.This is even satisfied, correct? Yes. Now we need to know whether it's a local min or local max. Well, let's look at the derivative. When we plug in a negative one, negative one, now we're actually getting zero, but a little before it, and if we actually looking at the x is a little less than the negative one, basically we're looking at here now, then this magnitude would be a little bigger, but the key is I actually do not know the why, so it is actually hard to tell whether. This is going from the negative to positive. So how would you do it? We actually want to graph the slope field. The quickest way: if you're given a slope field and you want to do optimization, instead of ah calculating, doing it analytically, and instead the quickest way, and we don't need to take the second derivative, it should actually graph the slope field. And if you do so, you're gonna see at the negative one, negative one, or getting zero here. But if you move a little to the.For example, looking at negative two, and if I still plug in a negative one now, if I plug in a negative two, negative one, and what you're getting is a much bigger positive slope. It would actually go like that. But if you fix the negative one, and you're going to actually make your y into a negative two, and then what you have is a negative slope. And then when I move to the right here, if I give the x equal to zero, and this is going to be actually a negative one. Ah, yes.That doesn't make sense. Right. So it looks like if you just look at what's nearby, and then it looks like this is a local maximum because apparently these curves will actually go like that. Oh, we don't know for sure, but I would say this is a crickets' judgment, so it will be a local maximum. Okay. However, however, and I want to look at what's going to be the zero. You can see somehow these numbers are shifted.Right, and if the y basically, and these are where the zero points are. All of these are zero, so therefore it doesn't look like an inflection point. It does mean on the left here your slopes are positive, on the right the slopes are negative. So looks like this. This is indeed a local maximum. That's enough for taking the AP test, but if you actually, I would really solve that differential equation. But that's only because we're good at solving.Differentiate equations. I'll quickly show you what is the solution. It's the y prime minus the y would equal to x squared, right? Right. Okay, in order to bundle this up into a complete derivative, what I can do is to multiply this by the e of the negative x, and we're rather taking the derivative of this guy. If I do so, what we're getting is, come on, hey, what, what we're getting is the e of.Of the negative x, and then you have the y derivative minus the y. Right. Right. Just apply the product rule, and that would actually equal to x squared times e to the negative x. Makes sense. Yeah. So now I want to remove that derivative. So I want to take the anti derivative.This is e to the negative x y would actually equal to. I guess and double check. In order to get this as the derivative, I need this term here. Then using the product rule, if I fix the polynomial part, I just take the derivative e to the negative x. I will definitely get that term, but I also get something else. I also get negative two x of the e to the negative x, correct? Yeah. That's also the derivative, which I need to.Cancel. I need to put another term to cancel that out. So I need to subtract by the x of the twice of the e to the negative x, because I can take the derivative by fixing the polynomial. Just take that term, which create a positive term, so that cancels out. But it also creates additional term. It also creates the negative two e to the negative two x. So I also need to add onto twice of the e. Then, oh no, subtract. Because you can double check when you take that derivative. You end up adding twice of the e to the negative x, which cancels out. So give it a shot, and I want to rewrite it. The e to the negative x inside you got a negative x squared. In fact, why don't I plus the negative here? I'll now plus a two x and plus a two. You can take the derivative of this.Which would exactly equal to the left-hand side. I mean, would equal to this. This is a that's the derivative. So this is actually your e to the negative x y. Of course, there's an arbitrary constant there. So eliminating this now, your y would equal to x squared plus a two x plus a two, and add on to oh minus. Sorry, this is negative there.Ah, would equal to the c times e to the x, and the minus x squared plus a two x plus a two. Now I use the initial condition, plug in the ninety one ninety one. If I plug the ninety one into the x here, what I'm getting is one, and the c of the e ninety one minus the one would equal to a ninety one, so that must be zero. So forty.And then in the C at zero. Now I double check. When I plug in one, do I get ninety five? Yes, we do. So this is a complete solution. It's just a ninety five equation. It is pretty close to what I actually earlier I have. All the times the the graph looks like this, right? Yeah, and it does. So therefore, that is a local maximum. Sorry, local maximum. Right. Yet one more way, and because this is pretty complicated, I don't want you to do that in on the test.I only want you to, if you're doing problems by yourself for your own conceptual improvement, and this is the definitive and totally rigorous way to to know exactly what the function is, you just solve that differential equation. But in the test year, we actually did it enough to actually choose the correct answer. But I also want you to do a double check. What about the second derivative? If you just take the second derivative from the slope, a slope field, what you get is a two x and then plus y derivative. But you have to.So plug back in the y derivative, which is x squared plus the y, right? When you plug in the negative one, negative one, is it positive or negative? It's negative, right? You announce actually negative. If that's negative, then we know for sure it's a local maximum. Right. Okay, so all of these are different ways you could tell. I would say that's quickly that that's pretty much the quickest way. Anyway.Our word to number eleven. So we're actually looking at the first. This is a symmetrical line; it's at two. So first, we're going to actually graph that one, but then you have to flip it over. Remember, having the magnitude guarantees that the whole domain here are all the reals, and the whole graph is symmetrical about x equals two, right? Okay, or looking for continuous, not differentiable. Yeah, immediately.Yeah, you can tell it's a sharp turning point. But in case you don't graph the accurate graph right away, and then how do we test out whether it's differentiable? Then you take the derivative. You have to take its derivative, and which actually gives you the square root of the x minus a two, and then followed by there is going to be a one half. Right here, I'm neglecting the absolute value, but we do. No, when the axis approaching two, that's approaching infinity, so it's not differentiable. It's also leading to the same conclusion. Uh, number ten. Oh, you can do that. That's that's too easy. Uh, number nine. Okay, which is series convergence? Well, we take the ratio test or the p test. Use your intuition. The last one is.The power, there is no geometric sequence. So roughly, the power is equal to one over the n, right? Oh, sorry, I don't mean that. I mean one over the n squared because the first power on the top, third power on the bottom. So by the p test, necessarily this one converges. Right for this one, although you do have a big power, but this factorial wins out everything. When you take the ratio, in fact, we end up with n over it's n plus one over the n of the one hundred power. Eventually, that's approaching.呃，one here，so basically it's divergent for sure. The each term, it's actually divergent. It's not even approaching zero. For that one though, the factorial on the bottom wins. So when you take the ratio, you're getting the eight over the n, right? And then definitely the. Convergent interval is all real. That's right. Yeah, and they all and they're all convergent. So just remember, factorials are very powerful.They're going to win out to pretty much everything else. Okay, now let's go to number eight. Oh, finally, we're getting a remainder sum. We're looking the approximation in the tangent at t equal to fifteen. We're doing the remainder sum now. We're given one the time equal to four. So this is a the f of t now. Oh, they didn't define that function. We're looking for what is actually the number of liters of oil.然后求得f，我们 looking for the fifteen should equal to the f four， which is given at fifty， and then plus the integral， which is the whole function. And if we do go for the Riemann sum， did they tell us exactly or use the right Riemann sum? That really just means we're doing the interval length of the three multiplied by the six point two， that's the first right end of the interval， and then plus the five multiplied by five point nine.And then plus. Do you need a four times six point five? No, because that's a right Riemann sum. We're told to do right Riemann sum. So the function value on the left boundary doesn't really feature. Remember the first interval is between four and seven, but we're supposed to use a function value at seven. Ah, okay. So we skip the left end here, and then finally plus the three on times of five point six. Okay. And you add.嗯，onto the fifty。没想象。Yeah. What if I'm asking you, judging from just this information, can we know our integral is greater than or less than or somehow can I know the comparison between the red room and some we're getting and our integral? No. No good. Well, even if this monotonic is increasing, like so. In fact, this is monotonic decreasing, and you didn't buy the temptation. You didn't say.The right Riemann sum is less than the integral because for the decreasing function, the right Riemann sum is truly smaller than the integral. When you do the right Riemann sum, but there's no guarantee that whatever the representative value at four points really tell you anything in between. Your function could be really wickedly oscillating with only these points given to you. So just remember, unless they tell you something behavior in between, for example, the second derivative is all. It's greater than zero or anything like that. Knowing the particular values on the boundary doesn't yield any information about the values in between. You can't judge anything about the concavity about the left region and some of the right region, some which is bigger. In fact, because these intervals are not even evenly spread out. Okay, good, huh? You just keep in mind: don't buy those temptations. I have to go, but we're going to actually.We do this for a week or two. At the same time, I would say you focus on developing a infantry, fill in some knowledge points. Meaning, if you don't remember something, ask your ChatGPT or WeChat me, and I'll find some time to to help you catch up. And this is really the last month now. Your labor pays off. I know you're very smart. So could you promise to concentrate on calculus for an hour a day? That's a huge commitment. I know your other subjects are also.Getting into an intense final period—it's not easy. I do not assume that you don't have much of anything else to do. Nevertheless, please concentrate because it pays off. Alright. Alright, wish you the best luck and happy problem solving. Take care. Goodbye. Bye.

Series, Inverse Derivatives, Polar Curves and Differential Equations

Series, Inverse Derivatives, Polar Curves and Differential Equations

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