Well, we are at the last day, day 59. I thought it would be somewhere close to 60, so didn't quite make it to 60. I think the task today is to kind of go through what you did on that first, and then after that task is accomplished. Then we'll go back and look at things that have made it to tests that you might see again on the final exam, and what have we covered since test four that might also have a chance of making it to the final exam. So I think we have addressed some of that already. What have we done since test four? We've done a little bit more with the binomial series, and hopefully I've connected those well enough to you that really the binomial series and the Taylor polynomial, yeah, you guys help yourselves there because I don't want to take any of them with me. They're really one in the same. The binomial series kind of was rooted or founded using the Taylor series, so when they talk about a Taylor expansion and you want to use a binomial, that's fair game. They're all the same. They're in the same category, but we have done more with the binomial series expansion than since test four, so that's potential final exam material. The applications I think we decided last class are probably too lengthy. They're good. They're valuable. I'm glad we got that far in the course, and we looked at them, connected some things maybe in physics that you have used before that kind of wondered why we had two different formulas and how we could get away with that. But those problems are too lengthy for a final exam. And we have, I think, did we finish the most recent test as far as going back through everything on test four? Do we get all the way through that one? Okay. Yeah, we didn't address the bonus. Actually the bonus, I mean it's appropriate, but probably would have been more appropriate for test three, because that was kind of the topic there in test three, but it is a couple of infinite geometric series. The ratio is 4.4, which absolute value of that is less than one, so it does converge. But just a simple one. The first one, it drops five feet, and then after that, the amount that it comes back up, it also drops back down. And theoretically this process of responding to a height of 40% of what it was prior to that, in theory that ball never does actually sit on the ground, right? Doesn't it kind of continue to respond to 40% of what it was before? We know eventually that it does just sit there on the ground, so it is a mathematical model. It does a great job, but the reality of it is not completely modeled by the infinite geometric series that we're trying to use. So I think I drew a diagram of that, and then the one five foot trip down is kind of off by itself, unless you want to add it to all the other trips down, and all the trips up, the first trip up is 40% of five feet. So I just split it apart into two different infinite geometric series, and then added them together. Did anybody get full credit on the bonus? Yeah, I think you did. What if you got the right answers, but you didn't do the work that you did? I don't know. I'd have to see something. What I don't want to do is reward, oh, I think it's 19 and two thirds. I think I'm going to write that down. And then you happen to get the right answer. But if you've got anything else, all right, so I think we're done with test four. So let's just kind of continue our way back. So let's look at test three. I am hoping that as we, more importantly, that as you look back at this stuff that you'll say, oh, yeah, we've been doing that a lot even since then. For example, on test three, question five, the first one, five A, you've got an alternating series. We've dealt a lot with alternating series. So we could probably, if all we need to do is determine convergence, use the alternating series test, but the additional directions, if it converges, find the value to which it converges, that's not going to get us that. So we've got to come up with something else. It is an infinite geometric series. Because it alternates, the ratio is negative, right? And it's a matter of figuring out what it is that we multiply through by in that particular series. Well, a dead giveaway, if the first term is one, isn't the next term the ratio. That's the only way you can get from one to something else is to multiply by that something else. So the ratio is negative 0.4. I think if I remember right, the most common error on this problem was that people had the ratio at 0.4. The ratio in this case is negative 0.4. And the first term is one. So the sum of an infinite geometric series that does, in fact, converge is what? What over what? A over 1 minus R. A over 1 minus R. So in this case, 1 over 1 minus negative 0.4. What does 5B look like to you? It's harmonic. There's a 5. You can pull the 5 out in front. It's not going to save it. If it's diverging, it's not going to cause it to diverge. If it's converging, it's just five times whatever we get with that what's left in the argument. So that is divergent. And that's enough of a reason that it's harmonic. We've used that pretty much consistently. What do you think of as a possible method when you look at 5C? And I don't really care right now if you're looking at your answers, just kind of looking at these problems as if you were looking at them for the first time. Partial fractions, decomposition. And when we did that on these kind of series, what type of series did we, I think, just about every time come up with telescoping. Now could you determine convergence without using that? Yes. You could kind of multiply the denominator out, 5 over n cubed plus some other, or n squared plus some other stuff. You could use a p-series, right, comparison test. You could determine convergence, but the best way to determine the value to which it converges is to decompose it into partial fractions and write out several terms of the telescoping series. See what remains, see what drops out. 5D, that may be the easiest problem on any test this semester. Some of you rejected that gift, that was a gift. What is the deal with 5D that makes it less than a minute to do that problem? It's the limit. The limit is one big. If the limit of the nth term is not zero, we know automatically it diverges, which is the converse, I mean the contrapositive of the theorem that's in the book which says if it converges, then the limit approaches zero. Take the contrapositive of that. If the limit is not approaching zero, then it does not converge. So I think they call it a test for divergence in the book. So 1n over 3n, the limit is one-third, which is not zero, therefore it diverges. That's the end of that problem. Integral test, if it doesn't say use the integral test, one of the ways you might want to look at it is this argument, if I had x's instead of n's, would that be integrable? If it is, then you could use the integral test. So you have an n squared or an x squared in the numerator, n cubed or x cubed in the denominator, looks like the numerator could potentially be the derivative of the denominator, so that'd be a natural log format. I think this ended up diverging, right? Is that correct? Because we had a natural log of some number and the number itself was getting larger and larger, therefore the result was not converging. What is your line of thought in 5f? When you look at that, what do you think of? Okay, comparison, maybe. I would recommend when you first look at that the plus two is not going to be very much of an issue, right, for large values of n. So it's really kind of 1 over 3n in a lot of ways, and what is 1 over 3n? That's harmonic. So we're thinking that this diverges. If we're going to use the comparison test, what would this argument have to be in relationship to just 1 over n or 1 over 3n in order for it to also diverge? It'd have to be greater, right? It's got a larger denominator, which makes the value of this argument smaller, so even though it kind of falls out of the comparison test, we still have a general idea that it is divergent. So the next kind of more powerful test that will give us a conclusion since the comparison test fails is the limit comparison test. And it doesn't matter which one you put in the denominator, which one you put in the numerator, if you get an answer and one of them diverges, then they both diverge. If you get some finite answer and one of them converges, the other one also converges. No, in other words, if you get, let's say, five-thirds and the one that we knew about was, in fact, convergent, well, either the one in question is either five-thirds times the one that was convergent or, if we put them in the other order, three-fifths. Either way, it's also convergent. It's either five-thirds as large or three-fifths as large, but in either case, it's still convergent. Isn't the answer that it converges to? No. That's correct. So we get a decision, but we don't get a value to which it converges, because this one does not converge, right? No, it does. So if it's... Can you do that one? Oh, we've got the... Yeah, you've got the solution. Yeah, okay. Go ahead. If you get an answer, then it's convergent. If the one that we're comparing it to in the limit comparison test is also convergent and you get two-thirds or 11-seventeenths or whatever you get, the one in question also converges. So it's not really... So you get the one in question to be related to the one that you know about by some numerical scaling factor. So if the one diverges that we know about, and we find that this one is two-thirds of that or three-halves of that, either way, it doesn't change the fact that it's still going to diverge. How would you never get a number? How would that... If it does not exist. Or really, I think we've got some potential problems if you get zero, because if you think about that zero times a divergent series, is that going to diverge or converge? I don't think we really know a whole lot there. But two-thirds, five-seventeenths, those are all fair game. Does not exist and zero are potential problem quotients. Obviously, the bonus there is a third order. You're not responsible to solve third order differential equations, but some of the same techniques you used in second order, which is the earlier part of the test. You should be prepared to, as you look back, do all three cases of second order differential equations. The three cases are where the characteristic equation had two distinct real roots. That could be with a homogeneous or non-homogeneous. Be prepared to do that. Find the two R values, decide on the nature of the solution. If it has a real root, but it's a double root, what does the solution look like there? And the other one is what? Two distinct real roots, a real root that is a double root, imaginary or complex root, and that's when we enter into the trigonometric type solution, the oscillatory solution. First one is homogeneous. Second one is also homogeneous with some boundary or initial conditions. And the second one, or the third one, is non-homogeneous, kind of left me there for a second. I haven't looked at it for a few days. The right side may not be, so you don't just look at this for review. The right side could be a polynomial, which this is. What else could appear on the right side? Some trigonometric, and really not any trigonometric, but sines, or cosines, or sines plus cosines is kind of the subset we looked at in here. And what else? Exponentials. Now if it happens to be a constant, a constant is really a polynomial, so that fits into that category. So you would need to find the homogeneous solution, the particular solution, the final solution is the sum of those two. And then we had a spring problem, an application problem, remember in a spring problem, Hooke's Law, the force required to stretch or compress a spring is directly proportional to the amount you have stretched it or compressed it. Any issues, questions about any topics that you see on test three? So that's going to be a pretty good review for you. With the exception of number three, you're going to have to look elsewhere to get an exponential equation on the right side and or trigonometric sine or cosine on the right side. Questions, issues, test three, content. Alright, test two. So I came home last night and my wife had the TV on, and I don't know what time it was, probably about 5.30, and this was on.  She goes, how do you get so excited about math? She goes, I have no idea what you're talking about, and you just really seem to be all into it and stuff. I said, well, no. You made the decision to kind of get married 27 years ago, so you know what you were getting into, some weird math person. So I do, I like it, I like this stuff, and obviously she doesn't. Really getting to be a marital problem. And she's from Carolina, so I don't know, it compounds the situation. Test two, orthogonal trajectories. We have a derivative, we want to, or we have an equation, I guess here, and we want to solve for the derivative. We want an orthogonal trajectory, orthogonal means perpendicular, we want to flip it and negate it, and the one thing that happens in this problem is that if you have a k in your derivative, dy over dx, you don't want that k to appear in the flipped and negated version, you want to be able to get rid of that k. What is k in relationship to the original x and y quantities from your original equation? So this is an example where you had a k in the derivative. You can flip it and negate it, that's fine, but don't try to anti-differentiate it in order to solve for the orthogonal trajectory, you need to get rid of the k. So k in this problem was replaced by y over e to the x, and the nice thing about that is that it drastically simplified the thing that you had to anti-differentiate. So instead of being this ugly thing with e to the negative x's in it and k's in it, it ends up being 1 over y, that's a lot easier to anti-differentiate than had you not eliminated k. Remember, when you do the anti-differentiation of both sides, kind of it's a separable differential equation format, is that you include a plus c on one side or the other. At that time in the solution, don't think you can recoup that at the end of the problem, just put a plus c or k on your last step because you're going to do things beyond adding the c or the k. In this case, doubling both sides and taking the square root of both sides. So that c or k could be under a radical, could be in the numerator, in the denominator depending on what algebra solved for y from the point in time where you anti-differentiated the cope. After you took the negative reciprocal, what did you do in the next step? That's a separable differential equation. So multiply both sides by dx and multiply both sides by y because you want to get the y's and the dy's on one side and the x's and the dx's on the other. And that would have to be a separable differential equation for us at that point in time. I don't think we had any other exact techniques that we did at this particular stage in a problem. Is that okay? Second one, center of mass. I see center of mass as a potential exam question. I mean there is some memorization there. If you had it memorized for test two, you might want to kind of refresh it. Rememorize it if that's the case. I think that's something you might see in another math or math related class, center of mass. Force due to liquid pressure or hydrostatic pressure, number three, that's potential. Memorization there, just some words, the density of the stuff, doing the pressing, the depth of each horizontal strip in terms of variables that we set up on our own axis system and then the area of each horizontal strip. And then the accumulation of all of the hydrostatic pressures on each of the strips of each of the kind of described areas, we don't have to worry about that, that is added up through the definite integral. Also be prepared on the exam for directions that say set up only. It is, I don't think time will be a factor, you have three hours to do the final exam. But it is a time saver and if I've found out if you know how to integrate something like this one in another problem, I don't need to necessarily find that out in this problem as well. For differential equation, but this is Euler's method, I think that's a method that we'll have some carryover beyond this class. Any iterative approximating technique like this? I know you'll revisit that, certainly if you go to math 341, which Sarah was talking about before class, she was looking forward to all the other math classes that she would be going to beyond this one, so you'll see that in 341, this Euler's method. Just remember that as you compute kind of the intermediate x and y values along the way, that you're using the prior information. If you're trying to find x3, y3, you're using x2, y2, so that's any time you need to sub in an x value, using all the previous information, none of the newly found information along the way. This is one, and I kind of like to ask the question this way, obviously I ask it on the test this way, so it might appear on the final exam this way. Here you've done this approximation in part a, using Euler's method, go ahead and solve this exactly using separable differential equation techniques, see if by chance our approximation is in fact related to the exact value, so it kind of serves as a little bit of an internal check, sometimes the incremental approach to Euler's method, you start to deviate from the exact curve, and it's kind of hard to get the answer back close again, but at least it kind of tells you if it's reasonable, so I like that method of asking that question and kind of get two birds with one stone there too. And everybody's favorite, the tank problem, so I have never given a Math 241 exam without a tank problem on it, and here's a hint, I'm not about to start now, I'm not ready to turn over a new leaf, so tank problems will be on the final exam, tank problem will be on the final exam, this isn't a literal tank, but the room here with a certain volume to the room is our tank for this problem, just remember to pay attention to the stuff that in the problem that we're trying to analyze, in this case it's carbon dioxide, so we need to pay attention to how much carbon dioxide's coming into the room or tank, and also how much is leaving the room, so it's the rate at which the stuff that we're paying attention to is coming in, minus the rate at which the stuff that we're paying attention to is going out. Rate in, minus rate out, certainly you want to pay attention to the concentration of the stuff coming in, as well as the rate of flow of the stuff coming in, same thing on the stuff going out, why is this a problem in this particular part of our course? Because once you get it set up, if you look at problem five, you should get a separable differential equation, right? And usually the, if you establish your variable, like in this problem, C is the amount of CO2 in the room at any time t, that will come into play somehow in your differential equation. Notice the concentration of the rate at which carbon dioxide is going out in the solution to this problem, how much is in the room, C, out of the total volume of the room, 16,000 cubic feet, wasn't it, yes, so that gives us a concentration. And these have some little tricky parts to them because you're going to have to exponentiate both sides and then put in the initial conditions in the problem. And we also had a problem six, you had your choice, Newton's law of cooling, and a logistic population problem, I guess in a sense each of them made it to the test, some of you chose one, some of you chose the other. So they have a chance, so if you did the first number six, you want to either make sure you study this or look elsewhere, back at your notes, back in the book for the logistic growth problem. Anything, any formulas that were given to you on the tests throughout the semester will also be given to you on the final exam. And in like fashion, anything that was not given to you that was expected for you to know on the tests will be expected of you on the final exam as well. You know, I always have to do this. So if we had a three by five card or a sheet of paper on the final exam, how many of you would like to have that? Just a little kind of formula sheet, three by five card piece of paper, okay, we're not going to do that. I was just kind of curious to see. But that's nice that you would like that. I appreciate knowing your opinion, but we're not going to do that. Sorry, you knew that was coming in here. You knew that was coming. It just kind of adds fuel to that fire that he's just a mean spirited person, sorry. So that's how I kind of raised my children. Honey, you want to go get some ice cream? You do? We're not going. What are you talking about? We're not going to get ice cream. I didn't do that, really. I didn't do that. We have to do that once each course, so we've got it on tape now. You know, I happen to watch the ones that happened in last semester. That you... It sounds worse on tape. Pretty horrible. It's so bad. That you were laughing at me? Yeah. Okay. Do we have any snorting? We don't have any snorters. All right. Test two. Anybody? Any issues? Any...that's kind of loaded. Test two is loaded with stuff. Applications. Yeah. Yes? Are you going to give us any evidence of which type of...all of the problems that we're going to have? You know, this lindrical oil tank, one like the pressure one, that's compared to the work problem, because there are like six different types of those and it's pretty cumbersome to... Yeah. It'll kind of be a work or hydrostatic pressure. Work is you're integrating force times distance, right? So it could be one of those where you're kind of...you've got this tank full of water and you slice it and then you figure out how much work is required to move. One of the slices to the top of the tank, one of those, force times distance. Or this one, density of the stuff, doing the pressing, take a look at horizontal strips parallel to the surface of the liquid, what is their depth and then what is the area of one of those strips, invariable form. I mean you're not...you don't want a number. So I'd probably come from one of those two. Anything else from test two? All right, test one, a lot of this was probably stuff that you had or at least potentially had, maybe kind of in a hurry at the end of math 141, trig integrals, powers of sine and cosine, decomposing into partial fractions, which we also then use later in this course with series. Trapezoidal rule or Simpson's rule, I think that was technically new. I don't think that was covered in 141. If I remember right, 141 ends at 5.8 in the text and trapezoidal and Simpson's rule are 5.9, so that's kind of new material. Table of integrals, subset of the table of integrals is possible. Match up the integrand after you've kind of made it match up exactly by correcting for coefficients that aren't there, that you need, kind of a comparison of what you have after you doctor it up to the table of integrals. That shouldn't be a difficult problem, but you may have to kind of doctor up the integrand to make a match. Improper integral, we had a variety of those. You may want to look at the one that kind of made it to the test and then look back in your notes or at the book at some other types of improper integrals, some other things that could happen that would make an integral improper. Area, I guess bounded between two curves is what it looks like to me. Number six, and if you took that area and revolved it around the x-axis or the y-axis, what would be the volume of that region? Both of those are, I think, important models in Calc 2 because they are accumulation models, that whatever it is you're talking about, you're talking about an area of a region, you slice it up into skinny little rectangles, you describe the area of one of those skinny little rectangles, let the calculus do the work by adding an infinite number of those skinny little rectangles together. Same thing with volume, you slice it up, perpendicular to the axis of revolution, they get excited about areas, I wish I could get them excited about volumes. Slice it up, perpendicular to the axis of revolution, describe the volumes, remember they might be solid discs, they might be washers. I think probably I'll steer clear of cylindrical shells. So solid disc or washer, when you slice it up, that's what each disc is going to be. And length of a curve, there's technically I guess three ways there for length of a curve starts out being 1 plus first derivative squared, right, is one of them, if you have y in terms of x and that works, it might also be 1 plus derivative of x with respect to y, you can integrate that with respect to y from a y value to another and I think this is parametric in it on the test, what's it look like, right, so it kind of depends on what you're handed in the problem, all of them are kind of rooted in the same thing, this is kind of a description of an individual hypotenuse and then we use the accumulation model to add up all these hypotenuses, if that's a word, from starting point to ending point. Any questions, issues with test one, that's a lot of stuff. And I honestly, I think final exams are helpful, I know you don't think that in the place where you are, try to convince yourself that they are because they kind of force you to look at all this stuff again, to hopefully see that this course, even though it is somewhat disconnected, it is all connected and some of it's connected that the first time we had a concept, let's say partial fractions, we didn't know how we were going to see that again in a sequence and series problem, but we did see it again, it's the same technique, it just helps us solve a completely different kind of problem. So the course is connected and I hope that you find that to be true as you prepare for the final exam. Now the date for this group of students for the final exam is Friday, May 1st from 8 to 11 and that is in this room. And I know that a couple of you have issues getting here at 10.15, make sure you set a backup alarm because 8 is very different from 10.15. Questions? I bet you, you folks, I sent a couple emails out to you, they were kind of harsh initially that if you weren't going to come to remember those emails, because I had some people in the 141 class that just kind of disappeared and it just looks weird when you look at the tapes and this person's never here and she's never here and that's why I was trying to make sure, plus I knew there were some people that wanted in this class that it was full by the time their registration window came open. So hopefully it didn't turn out quite as badly as you might have thought initially. I think that you have been a very pleasant group of students. I don't know if the camera affects that, it kind of affects me. I'd probably be a little bit looser, maybe a little bit less guarded in some things I say if the cameras weren't running, but hopefully it was a good course for you and you can finish strong with a strong final exam grade. If you are on the border, and that is not just for students in this class, but all my classes, somebody is on the fence between a B and a C and they really do the final exam right. I mean they get a 92 on the final exam and they're on the border between a B and a C. What is their final grade? They get a B. I mean they prove that they deserve that if they're, but you've got to be on the border. I'm not going to take somebody with a D that gets an A on the exam and say, well they get an A. I mean that's not representative of the body of work so to speak, but if you are on the fence, the final exam grade could in essence be more to you in terms of waiting than just the waiting that it's given as far as the percentage. So finish strong and I will see you on the first.


I want to thank Tim for the super sticker. By the way, if you purchased a super sticker super chat, I'm so grateful. Thank you so much. Okay, Elaine says, question, do you have to be exclusive to build a relationship? Do you have to be exclusive? Okay, so folks, everybody has a different opinion. I know a lot of dating and relationship coaches suggest dating multiple people, but I want you to think about dating multiple people. Okay, if you're dating three or four guys, and the guy is dating three or four women. First off, when do you have enough time to date multiple people? Now, I want to differentiate between meeting people and dating people. Meeting people and dating people. So let's differentiate this. First off, listen, if you meet someone and you never see him again, that's just a meeting. If you meet someone on a Monday, never see him again. You meet someone new on a Wednesday and you never see him again. You meet someone new on a Thursday or Friday and never see him again. That's not dating three people. That just happened to be meeting three people in a given week. Now I want you to change this scenario. You're actively dating three people at the same time. How much, so in the other person, let's just say the guy isn't dating multiple people. Let's just go with that for a second. All right, he calls you up and says, hey, what are you doing Thursday night? You know, one of this guy says, and you go, well, I've got plans somewhere else.  Okay, and then, you know, that guy, okay, let's say the one guy calls you up on Friday night, what are you doing? You've got something else going on because you're dating multiple people. Well, you might be missing out on a great guy who goes, look, you don't seem to have time to be in a relationship with me. So that's one of the challenges. Now I know a lot of coaches will tell women to date multiple people so they don't get attached to the wrong guy. Let me think about this. If you have a propensity to get attached to the wrong guy, maybe you might want to heal the wounds that cause you to get attached to the wrong guy. For example, a lot of women get attached to the bad boy and the player. Let me repeat that. They get attached to the bad boy or player. There's predominantly two reasons why this happens. And if you're not familiar with the following two books, attached by Amira Levine and Rachel Heller and Getting the Love You Want by Harbell Hendricks and Helen Hunt, okay? Why you want to read these books is first, love attachment is when you get biologically attached to another human being based on three attachment styles, anxious, avoidant, or secure, okay? So we can actually be attached to the wrong person because of a childhood wound that happened in our lives that causes us to get attached to the wrong people. In addition, Getting the Love You Want talks about the amago, the amago. So what the amago is all about is we oftentimes choose partners that are very similar to one or both of our parents and in their personalities. And usually the unhealthy aspects are one or both of our parents. So here's the thing. You're dating multiple people. You're not really paying attention. You get attached to the wrong person. You let go of a great person because you're dating multiple people and he loses interest in you. And then where are you left? One or two years down the road, heartbroken. I'm a big proponent of dating one person at a time once you've agreed to explore a relationship together. And again, most men who are genuinely serious about seeking a life mate, a partner in their life actually make time to date only one person at a time. So I think as a common courtesy you want to also date one person at a time. That's just my belief system. Everybody has a different opinion around this. You have to ultimately do what's right for you. I do want to caution everybody on this because of the internet, because of swipe dating. We are in a unique period in our lives where we have this thing called perceived choice, perceived choice. In other words, because we have this belief that there's so many choices because we can swipe we have actually now been desensitized by human beings. And if you're not familiar with the TED Talk the paradox of choice, I'm gonna say it slowly the paradox of choice, the paradox of choice. It's a TED Talk. I highly recommend checking it out. That 20 minute TED Talk will give you some, has nothing to do with dating but has everything to do with the problem we have of dating, it's not dating at all but the fact that this belief that we have all these choices. And let me just tell you this folks you may miss out on a great guy because you're dating multiple people and you get hooked with the wrong person and you might let go of a great person because you're spreading yourself too thin and I'm not a big proponent of that. Look it, it only takes three weeks to decide if you like a person. If you see each other once or twice a week over three or four week period of time you'll ultimately know whether or not you wanna explore a relationship with them. And if you followed my principles in this conversation about being radically honest in the get go, you might avoid the need to date multiple people to protect yourself.  I'm a big proponent of having an open heart knowing yourself, having that self discipline knowing that you won't get attached to the wrong person because you've done the inner work to get there. And if you need some help doing the inner work I highly recommend doing the Hoffman process. The Hoffman process. This is a great book to really heal the childhood wounds and traumas that cause us to choose the wrong people in our lives. This is a 50 hours of worth of work and I think every one of you is worth it. So that's my thoughts on dating multiple people at a time. When you come from a healthy place within you hold space for the right person to come in your life and then you only need to date that one person and stating multiple people to protect your heart. Hope that helps. Thank you so much for that question. All right.


What's going on everyone Sam here with late sports cards ripping one box of 21 22 panini immaculate Basketball hobby for Michael B. So good luck here Michael. Let's see it Frosty. Come on, baby. Come on Out of Bowman University first. That's nice dude. That's solid I know he's got some autos in Bowman you and then in leaf as well here in Fox to 49 base card We've got a Fred Van Vleet 34 of 99 That's going to the well. It's for the Raptors Very nice Three colored patch. We've got 10 of 30 Jeremy Grant. That's gonna be game used as well Game used for the pissed in Jeremy Grant There you go frosty 30 or that is 29 of 49 campaign modern marks For these sons cam pain We've got a 12 of 12 that's Joe Harris two colored patch auto game used for the nets Nice one there Joe Harris and very nice one here 15 of 25 three colored rpa quentin grimes For the next who's actually turned out to be a very solid player former top recruit To Kansas transferred to houston first rounder for the Knicks in 21 and it's carved out like uh Very good role in new york. So congrats on that. They're uh frosty Um, that is gonna be empty empty. Thanks again for the support buddy get them right out to you man


Hello there! Continuing our series of videos about how Moon Cheezer was developed, today we'll talk about how to manage the achievement system. So to begin with, I had to take what could be done in my game, in other words, the mechanics. After that, I stated some cool events which could happen in my game, such as gather a certain amount of cheeses, witness a supernova, like the collision of two stars, play the game on a full moon day, etc. The easier way I found to keep out this data, the description, the condition, the fulfillment of the achievement, was to write everything into a .json file. So I created this file which contains the dictionary key for each achievement, and inside these keys, which are dictionaries too, I created keys regarding the total amount that must be accomplished, the description of the achievement, the current amount already accomplished, and the name of the achievement. That's all for the data. Now let's move on to Godot Engine and the logic behind the achievements. To begin with, let's talk about the structure of the scene. I created a canvas layer as the root node of the scene. This way I could easily set this UI above the game player layer, making sure that players would always know when they fulfilled an achievement. Then I started to build the actual UI, a simple panel with two labels, one for the actual warning and the other for the message. In this case, a achievement unlocked followed by the name of the achievement. After that I added the nodes responsible for the animation, a twin to interpolate the position of the panel, and a timer, which will be used to tell how long the panel would be displayed on the screen before it did disappear. Now to the actual logic. First of all, it needs to know the path to the file that contains the data for the achievement. For that, I simply exported a string with the file flag to the editor. After that, I created the respective variables regarding the dictionary, which would handle the data and gather the needed nodes for the animation and visual feedback. To read the data file, we need to do six steps. Check if the file exists in the user's disk. If it does, we will open it and parse it to a variable as a text. After that, parse this variable as a JSON file to the dictionary we've created. And if the file doesn't exist, we will need to load the template data file, parse it as a text, and parse the text as a JSON to the dictionary. And write the file to the player's disk. To write the file is simpler. Check if the file already exists. This also creates the file in the disk if it doesn't exist. Open the file with the right flag, start the dictionary as a string, and close the file. The user directory is a relative path that points to a local data on the user's disk. And it's always accessible independent of how the user's operational system handles this data. So it's the best path to store this kind of content. I also had to modify the data we needed to keep track of the players for filming. To do that, I created a single method which received two arguments. One for which achievement shall be modified and the other to develop which shall be used to modify it. If the value argument is zero, I presumed it should be set to zero. Then I just needed to access the dictionary using the achievement argument as the key and access the accomplished key to modify its value. But to add, I first had to check if the achievement is already accomplished. So I just checked it if the current accomplished is less than the total. If it does, it will add the value argument to it, and after that check if it is accomplished now. If it does, it will set the name label nodes text to be the name of the achievement and display it on the screen. To display the visual feedback to the player, I just had to treat the panel as a pop-up window. So it would start hidden and would be displayed only when I told it to. So I simply created two methods for that. The first one would display it and animate its position with an overshoot interpolation on the end. It would also tell the timer to start counting. When the timer finishes the countdown, it will tell the panel to disappear by counting the other method responsible for hiding it. This method does the opposite of the first one. It will interpolate the position in the opposite direction with an overshoot effect at the beginning of the interpolation. Then we wait for the twin node to sign out that it ended the interpolation, and after that we will tell the panel to hide. Finally, I just needed to make this scene an autoload single-tone, so it won't be clean when transitioning between scenes, and every node will be able to access it globally. Also, I had to add the JSON file to the filter of non-resources file on the resources tab of the project settings. Otherwise, it wouldn't be exported with the game executable and everything would go wrong. And that's the result I got after all this process. That's it. I hope you're enjoying this kind of content. Leave a comment below if you have any doubts and suggestions for the next topics for the upcoming videos. That's all for now guys. Keep developing and until the next time.


Hello all welcome to the YouTube live session on problem practice session This would be a problem practice session on important topics with respect to JEE main and JEE advance So request the people who have just joined in to type in your name in the live chat box so that I know who all are attending this session Good morning to all of you have joined the session So please type in your names so that I know who all are attending Hello, Aditi, Sai, Kushal, Ramcharan guys. How are you doing? Sure, without much I do we'll start with the very first problem of the day It's based on Koenig sections Koenig sections ellipse chapter So the question reads like this There is an ellipse with an eccentricity of half and one focus is at half comma one It's one directrix is the common tangent to the circle x square plus y square equal to 1 and X square minus y square is equal to 1 clearly. This is the rectangular hyperbola. This is going to be a rectangular hyperbola Then they're asking you the equation of the ellipse in the standard form Then they're asking you the equation of the ellipse in the standard form So again, the norm would be I would wait at least for five people to answer before I start discussing it Because I strongly believe that it's a problem solving session and everybody should put in heart and soul to solve this And try to be the first one to solve it. Good morning everybody who has joined the session So anybody who has managed to solve this I would request you to type in your response in the chat box as to which option is correct A, B, C or D The first one solving this would get a shout out for me so for the purpose of Realizing the situation geometrically I have drawn the graph for you so that you can refer to this So guys as you can clearly realize over here that the circle mentioned over here is the auxiliary Circle for this rectangular hyperbola, right? And that's why as you can see in the diagram I have drawn it in such a way that the circle is ending up touching the the vertices Right the circle is ending up touching the vertices. Let me call it as a and a dash Okay, so if you have to have a common tangent, it can be only at these two points Right So it's pretty obvious that the common tangents could be either of these two lines The common tangents could be either of these two lines and these lines are clearly x equal to One and x equal to minus one So now you have to figure out which one is nearer to this focus So which is the directrix that you're going to use we all know that when we know the focus and the equation of the directrix So let's say the focus is that alpha comma beta then we all know that the equation of the ellipse is given as e square Lx plus my plus n whole square by L square plus m square, right? This is the equation of the ellipse that we know. This is the most general form where your focus is that alpha comma beta and your directrix is And your directrix is Lx plus my plus n equal to zero So I'm sure the point the line the directrix which is nearer to half comma one would be x equal to one, right? So this would be your required this thing. By the way, you're a is one So I'll just replace this with one and one This will be one and this will be one.  So I've got the first response that is from poor Right poor it the directrix which is a closer to s closer to s will be x equal to one only Now everything is set. I don't see a Point not solving it right now once, you know, the Directrix is that one x equal to one and focus is that one comma. Sorry half comma one You can always write down the equation as you can always write down the equation of the ellipse as X minus half the whole square Plus y minus one the whole square is equal to e square is already given to us as half. So e square is one-fourth and X minus one square by one. Okay, so we just have to rearrange it So as to see which of the options is suiting the requirement of this So when you open this bracket, you get x square minus x plus one-fourth plus y square minus two y plus one is equal to one-fourth X square minus two x plus one So just let me simplify this further. So I'll have three x square by four and I will have minus x and plus x by two which is minus x by two and we'll have constants like One fourth and one fourth will get cancelled So we'll have y minus one the whole square is equal to zero Okay So just to further simplify I take three by four out common. So I'll have x square and When I take three by four common out from here, it becomes minus two by three X Okay, and for the completing this square over here without much ado. I'll have X minus one by three the whole square minus one by nine Plus y minus one the whole square equal to zero Which is clearly three-fourth X minus one-third the whole square Plus y minus one the whole square is equal to one by 12 is equal to one by 12 So according to the given options, I have to have a one on the other side Have to have a one on the other side. So I will divide throughout by one by 12 So if I divide throughout with one by 12 What I will see is nine x minus one-third the whole square Plus 12 y minus one the whole square is equal to one which I think Which I think is clearly your option number Option number a option number a by the way, I think there's a one missing out over it So yes, Purvik you are absolutely correct So the first one to solve it and the only one to solve it is Purvik Very good. Well done. Any question with respect to this? Please feel free to ask me right now So guys the moral of the story here is that make diagrams make diagrams without diagrams Never try to solve a coordinate geometry problem So I've been repeating this since the beginning of this course that please whenever there's a coordinate geometry problem Try to make a diagram that will always help you to find out the solution in a faster and easier way So with this note, I'll be moving on to the next problem. That's your again a question on coordinate geometry The second question so the question reads like this let there be a point P Whose coordinates are mentioned to you a seek theta comma beta and theta and there's a point Q also Where theta and Phi are complementary to each other where theta and pi are complementary to each other Okay on this hyperbola HK is the intersection point of the normals of P and Q Then find K Guys, let me tell you there is nothing important in the concept nothing rocket science in this question It's all about using your basics equation of a normal then the point of intersection of the lines and using the given condition in the problem That's it. There's nothing difficult about it and I'm expecting everybody to give me the right answer Okay, so Kushel is of the opinion. It's option D. Okay Rohan thinks it's option C Guys, they should not be anybody who is actually figuring out what is the equation of a normal at this stage of your journey So the equation of the normal at a point x1 comma y1 on a hyperbola Which is x square by a square minus y square by b square equal to one should be now clear The equation is a square x by x1 Plus b square y by y1 equal to a square plus b square Nobody should now be struggling. What is the equation of a normal? Okay, if you are means you have not Done enough practice on Koenig sections Now, let's say call this point Okay, so basically when you substitute x1 as a seek theta and y1 as b tan theta you get something like a cos theta x and b Caught Theta y is equal to a square plus b square So this would be the equation of the normal at p Similarly equation of the normal at q will be a cos phi x plus b Caught phi y is equal to a square plus b square Okay, so I is also of the opinion that it's D Purvik also says D Guys, I want the response from others. Ramcharan, Kondiniya, Atmaish, Shritvik, Naman, Sondarya, Shruti, Sanjana, Aditi So it's guys, it's very simple so the first equation that is the equation of the normal at p I would write it as x since I have to find K what I will do. I will try to eliminate x Correct, which is actually your H So let me divide throughout with a cos theta So if I divide by a cos theta, I'll get something like b by a Sin theta y is equal to a A square plus b square by a cos theta Similarly at q if you divide you get simply this So you don't have to do any hard work for the other one because it's just the replacement of theta with phi It's just the replacement of theta with phi Now just subtract these two. Let me call this as 1 and 2. So let's subtract 1 and 2 Let's subtract 1 and 2. So when you subtract 1 and 2 you get something like b by 1 by sin theta minus 1 by sin phi and On the right hand side, you'll end up getting a square plus b square by a 1 by cos theta minus 1 by cos phi There's a lot of things can be done over here So you can just get cancelled b and we can always write Sorry, this is y times so y is equal to a square plus b square by b and you will have cos phi minus cos theta by sin phi minus Sin phi minus sin theta and moreover you will have sin theta sin phi By cos theta cos phi now, of course this term would be tan theta tan phi Now guys by the way since they're complementary, this will become sin phi This will become sin theta. This is also becoming sin phi Okay, and this will become tan theta tan phi Now we know that since theta plus phi is equal to pi by 2 tan of theta plus phi will be undefined Okay, which means the expansion of this which normally has the tan theta tan phi will become Minus of 1 this should become minus of 1 Sorry, this become plus 1. Sorry. Yeah, because in the expansion you will have tan of theta plus tan of phi by 1 minus tan theta tan phi So your denominator should be equal to 0. This part should be equal to 0 So this part becomes 1 and I think this part will get cancelled off So I will get y as a square plus b square by b. Is it fine? Yes. So am I missing out anything? So, oh, I'm sorry. Yes. Yes. Yeah, this reverses Yeah, thank you for pointing that out. So this this becomes sin theta and this becomes sin phi, right? So there'll be a minus sign also left out from here. So yes, we'll get a negative sign Okay, so if this is a negative sign, of course option number D is going to be correct so I think the first one to answer this was Kushal Kushal was the first one to answer this well done Kushal very good Guys, I would rate this question as a typical JEE main question I don't think so. There was anything very difficult about this. It was all about knowing the equation of the normal and Finding out the point of intersection and this using this condition that theta plus phi is going to be 90 degree. Okay So moving on to the third question over here So this is a question which is based on functions. So there's a function from R to R and it's a periodic function says that F of t plus x is equal to 1 plus 1 minus 3 f of x plus 3 f of x whole square minus f of x whole cube So this this gap over here is just you know Unnecessary, but there was a gap in the question also. So So this gap, please ignore this gap. Okay So now t is a positive fixed positive number then the period of f of x then we have to find the period of f of x in terms of t only So I personally feel that this question is all about playing with the functional equation So on a roadmap, basically, I feel you just have to figure it out. What is what is this n over here? So when does your function become itself for a change of x by nt? So whether that n is 1 whether that n is 2 whether that n is 3 or none of these No problem. This is not the first time Aryan says B. Okay, Aryan, I have noted down your response How about others? So, Atmeshi is also getting B Sanjana, Naman, Ramcharan, Kondinia Where is Amog? Amog, Aditya, please respond Shwetha also is getting B. Okay, fine. You guys, it's very obvious that so don't consider t to be the period. Okay, see t is just a you know Alphabet which is used over here. So don't get confused. He is not a period of any function So if you see this part if you see this part of your problem, it is clearly One minus f of x whole cube, right and whole raise to the power of one third So that's going to be just one plus one minus f of x.  So that's going to be x plus t plus t plus f of x plus t is equal to 2 That means f of x plus 2t Plus f of x plus t is equal to 2 which clearly means that the position of f of x and f of x plus 2t are same because This is same as this Two is same as this. So these two have to be the same Which clearly implies f of x is equal to f of x plus 2t Clearly implying that the function is periodic with a period of 2t which is option number b is correct option number b is correct Okay, so the first one to answer this was a rn well done rn good Okay, I don't think it was a difficult problem. Well, you got carried away with the previous one Moving on to the fourth problem. The problem says GIF and fractional part are given in this function. So this is a function which is defined as to a to the power Two times gif of x plus fractional part of x minus 1 by 2 times gif of x plus fractional part of x when x is not equal to 1 And f of 0 is given to you as ln of a Okay, so we have to comment which of the following option is correct Okay, so our poor X is a C for this All right, so everybody is like okay for Vaishnavi who's saying for a everybody is replying C C C C All right, so let's have a look at it So well first we'll focus on finding the left hand limit That is a limit of this function when extends to a zero minus So a to the power two times Gif of x Plus fractional part of x minus one. I think minus one is down over here. Okay by two times fractional part of X So now when you're slightly less than zero, we all know that the gif of x will be minus of one So that will become minus two and we know that when you're slightly less than zero this fellow will actually become This fellow will actually become one minus x Right So just like you know, you can take a very simple value. Let's say X is 0.01 with a negative sign We know that fractional part of X in that case would be 0.99 Okay So it's just going to be it's going to be one plus X It's just going to be one plus X Okay minus one Similarly in the denominator will have a two into minus one and fractional part will be Will be One plus X You can look from the graph as well. So in the interval minus it was actually going to behave like this So this is a line Okay, whose slope is one so it'll be line wine equal to one plus X Okay So simplifying this I get a to the power of X minus one minus one by X minus one and as X tends to zero this is just going to become as Extends to zero minus it is just going to become a inverse minus one by minus one. That's going to be That's going to be one minus one by So this is your left hand limit Let us similarly find out the right-hand limit So for right-hand limit, I will write it as limit extending to zero plus a to the power two gif plus fractional part minus one by two gif plus fractional part so zero plus again this would become zero and Gif of X will become zero and fractional part will remain X So it'll become two into zero plus X minus one by two into zero plus X Which is actually a to the power X minus one by X which is log of X to the base log of a to the base E So what is happening is right-hand limit is matching with the value of the function at a but it is it may or may Not match with the value of the function. So these two are not matching for the same value of a Okay, so one minus a is not matching with ln of a Okay for all values of a Okay, it just means the functions limit does not exist the function limit does not exist So continuity there is no point of continuity in this case We cannot say removal discontinuity because it's the case of jump discontinuity because a values for the same a value Left-hand right-hand limit are not equal. So I cannot say this. So only option C is possible So the first one to answer this was again Purvik very good Purvik So guys any question with respect to this, please let me know this is just a basic question on continuity and differentiability Removal discontinuity is those cases where the limit exists But does that the value of the function at that point does not match with the limit? So we consider these two cases as your removal discontinuity where there is a missing point over here or where there is a Isolated point. So this is a case of removal discontinuity because you can always plug this value Always plug this value with the limit of the function at that point In case of jump or essential discount discontinuities, which we also call infinite discontinuities We cannot remove those kind of discontinuity So when this is a jump like this, we cannot do anything about it Or when these things go to infinity at this point, we cannot say we cannot remove these kind of discontinuities Exactly Sondarya when left-hand limit and right-hand limit are equal, but it is not equal to f of a those are removable discontinuities Again easy problem to score. So we'll now talk about The next problem, which is the fifth problem. So the question says g of x is defined as 1 4th f of 2x square minus 1 and half of f 1 minus x square and F dash x is an increasing function f dash x is an increasing function Then g of x is increasing in which of the following intervals then g of x is increasing in which of the following intervals Okay, again, I got a response from poorwick, which is option B Rithvik also feels the same. Okay, Krushal, Rohan Anyone from NPS Kaurangala? Atmeshe says 5d. Okay, so he differs from others. Okay. All right. So shares is supporting his Hsr fellow part. So he's also saying D Chalo, so I've got most of the responses Good to begin with So in order to know whether in which interval g is increasing the first obvious step is after differentiated, correct? And to differentiate it, I'll have to use the chain rule over here So that's going to be 4x and half We'll have f dash of 1 minus x squared times minus of 2x So 1 4th 1 4th goes off 2 and 2 goes off So g dash x will be x times I get f dash 2x square minus 1 and From here, I'll get minus f dash 1 minus x squared Now I can say that my function g of x is increasing when both of these terms are either positive or When both of these terms are both negative then only my g of x would be considered to be greater than 0 So let's take these as two situations. So we'll take case one case one where x is positive and f dash 2x square minus 1 Minus f dash 1 minus x square is also positive Remember f of x f dash x is an increasing function. So this has to be more than this That means even 2x square minus 1 should be more than 1 minus x squared That means 3x square should be greater than 2 So x square minus 2 3rd will be greater than 0 and that can only happen That can only happen when x is either less than minus of root of 2 by 3 or x is greater than root of 2 by 3 Right, so x can be less than minus of root 2 by 3 Or x could be greater than a root of 2 by 3 But this is not possible because of this restriction because they both have to be true simultaneously So from situation number one I get x should belong to a root of 2 by 3 comma infinity root of 2 by 3 comma infinity, okay Let's talk about situation number two in a similar way. So either x is negative and f dash 2x square Minus this should be negative So I don't have to you know solve again I can directly comment that x should belong to minus infinity to minus of root 2 by 3 Okay Now I have to take the cases cases means union of these two I have to take the union of these two that clearly implies option number B is correct Again the first one to answer this was Purik Bingo good guys Buck up other guys. I need to get the response from you as well So moving on to the sixth question over here. So you have been given a piecewise defined function f of x They're asking you set of real values of B such that f of x has the smallest value at x equal to 1 Again, it's a question from application of derivatives All right, so I've started getting options Ariane says B Aatmei C Kushal B Purik also B So mostly Janta is saying B or C B or C methora confusion. Hey The guys, let's try to solve this again. It's a piecewise function You know, it's a cubic equation right and cubic equation with a negative Leading coefficient. So it's graph is going to fall down Okay, this graph is going to always fall down So Let's say this is zero. This is one. This is three. So it's graph is really falling down So at zero, it'll be having some value. So it will be constantly falling Then you can also come to know from the derivative of the function So derivative of this will be minus of 3x squared from 0 to 1 which is always negative in nature Okay, and the derivative of this is always positive. So this is all is an increasing function So this is also an increasing function So guys the scene is something like this The other function is increasing this is decreasing and this is increasing Okay, now obviously if you want your Function to have the smallest value at one that means You can have these cases. Okay, so you can have these cases. It cannot go beyond this Okay, because it should have a lowest value at one so 2x minus 3 can actually You know come at these positions. In fact, it's the other way around 2x minus 3 will always be at minus 1 So this will always be at minus 1. Let's say something like this. So this should not fall below this line Right that means the limit of the function as x tends to 1 minus okay This should be Greater than equal to minus 1 right Then only the function will be set to have its minimum value at 1 Okay, so at 1 it will have a minimum value when the left hand limit of this function is Either greater or equal to the value of the function at 1 which is actually minus of 1 isn't it? So this means this means the left hand limit of this function as x tends to 1 will be minus 1 plus b cube minus b square plus b minus 1 By b square plus 3b plus 2 this should be greater than equal to minus of 1 This should be greater than Minus of 1 so let's solve this inequality.  This is always positive. This is always positive term So I just have to plot the others which is minus 2 minus 1 and 1 So it'll be positive negative positive negative. So when you want positive it has to be minus 2 to minus 1 Minus 2 minus 1 both will be exclusive Okay, union 1 to infinity 1 should be included because it can be 0 at 1 So guys clearly option number b is correct over here So the first one to answer this was Arian again Arian was the first one to answer this Okay, so guys, please be accurate. Don't be in a hurry things are easy This was an easy question. I had discussed many of such questions in the class as well So can we move on to the next one? Let's move on to the seventh question now So I think this question comes to you from integral calculus indefinite integrals Good to see many people are responding many people have responded for the previous questions So I would expect the same going forward as well Anybody any success? Okay, so at Mesh says option a okay So does Sanjana and Saimi here others please respond if you're done. Okay, so on there. You also says option a So almost everybody who has replied so far is going for option a others. What do you think Purvik Shweta? Okay, Shweta also says a just Arian has a different answer to give Yes, everybody is saying a okay guys. Let's let's look at this Now let's figure out first. What is f dash x so I'll just use this particular information I'll differentiate both the sides and For that I will use the product rule. I Will use the product rules. So it's going to be 6 f of x f dash x Okay So if I take f dash x common, I will get x minus 6 f of x is Equal to f of x in fact minus of f of x so f dash x is going to be f of x by 6 f of x minus x 6 f of x minus x Okay Now I can see this term Actually appearing over here Okay, and not only that I can also see that if I take a 2x common from these two terms I Could see 6 6 f of x minus x appearing So from the numerator what I'll do is I'll first take 2x common so that will become x minus 6 f of x plus f of x divided by 6 f of x minus x Times and we have x square minus f of x the whole square So let me take x minus 6 f of x common from both the numerator and denominator So when I take the common x minus 6 f of x from both numerator and denominator, I Would be left with this. Oh, I'm sorry. This is not the square of this term. This is the square only outside Yeah, I'm sorry.  Let me call this as t So 2x minus f dash x DX is going to be your dt That's just going to be dt by t square. That's going to be negative of 1 by t Okay, and that's going to be minus 1 by x square minus f of x is an x minus So there's a minus sign on top. Okay. I'm sorry. Yeah, there's a minus sign over here Yeah, there's a minus sign over here. Okay, so yeah, I can take a minus out common first So I'll take this minus sign common out first So yes, it's going to become this Yeah, so it's this which is going to be your answer or you can just see there's this Yes, absolutely correct guys. So option number a becomes correct in this case. So option a is correct so the first one to answer this was Arthmesh Very good Arthmesh again not a tricky question not a difficult question. It just made you think in a different way Okay, so it was just a case of a simple substitution But they complicated it by introducing a lot of functions and a function equation like this Any questions? Can we move ahead? Please feel free to stop me if you have any questions so moving on to the Eighth question for the day, which is on definite integrals So we have a function over here x by 1 plus ln x ln x all the way till infinity x belonging to the interval 1 to infinity Then find the Find the Limit of sorry integral of f of x from 1 to 2e It's a very very simple question very very simple question Okay, so shares says Option a okay Anybody else apart from shares? Yes guys anybody else So we are only interested in evaluating this function from 1 to 2e Okay, so let's say We know that ln x becomes equal to 1 exactly at x equal to e So let's see what happens when I take x between 1 and e So between 1 and e I can say that ln of x In this interval ln of x would actually be a quantity which is between 0 and 1 okay And if you see this term, it's actually ln x multiplied to itself infinitely many number of times so what will happen to Such a scenario when ln x is multiplied to itself infinitely many number of times and your ln x happens to be between 0 and 1 It's obvious that this is going to become 0 right Which means your function is going to behave as Just x when you are between 1 and e Because this part is going to become 0 So can you take a hint now and and try to complete it from here on? Uh, sure. I mean you can take a minute. No worries. So all right. Now people have started uh answering this Okay So let's see whether it's a or no Now what happens when you go beyond e when you go beyond e what happens Anyways, I'm not bothered exactly at e because that's going to create a spike and that is going to be x by 2 spike So I'm not going to be worried about it So when it is greater than e We know that ln of x is going to be greater than 1 if ln sx is going to be greater than 1 this will become infinitely big correct And if this becomes infinitely big that means the entire function will collapse That means the entire function will collapse and become 0 correct Because ln of x will be greater than 1 when x is between e to 2 e So ln x to the power n will actually go to infinity if n is going to infinity Okay So your function is going to become just 0 Now I just have to use the Definition of this function. So from 1 to e you are just integrating x And from e to 2 e you're just integrating 0 So that's as good as saying x square by 2 from e to 1 that's going to be e square by 2 minus half That's e square minus 1 by 2 and clearly option number a is correct, right So well done shares shares bhaktram was the first one to answer this and that too correctly Well done So I think guys this is very simple You just had to redefine the function and once you redefine the function according to the intervals Split the limits of integration and then just carry out the process. So I I'm sorry So next question will take up on area under curves and after this question We can expect a small break of five to ten minutes. So this is your question number nine Hope the question is legible The area bounded by the curves given by So there are three curves over here. One is this one. In fact, it's a region The overlapping of these three region is what is Required by this question Again, please do not try to solve it without sketching it I want everybody to sketch these three regions and then realize which region they are looking for Nine a okay. So athmesh says option a But not solved rigorously Purvik also says a okay anybody else So guys so far. I've just got the response from two students Okay, so arian is another one to respond as a I need at least two more responses So these two are pretty simple, right? This means x should lie between minus half to half. So And y should also lie between minus half to half So minus half to half. Let me just you know, sketch this over here. Let's say this is minus half And let's say this is plus half So x is equal to half. This is x equal to minus half Okay, and in a similar way, this is let's say y is equal to half and y is equal to minus half y is equal to half y is equal to minus half Right yeah No, I think most of the people are you know Trapped in this graph because of too many mods involved guys. Just think without mod think as if you are in the first quadrant X and y both are positive. So get rid of mod first. You cannot think with mod in place Okay So think as if there is no mod think there's no inequality also, right make your life as simple as you can Think there is no inequality also. Let's say this is equal to one. Okay. Now. How would you plot? So it's pretty obvious It's pretty obvious that when you make y the subject of the formula it becomes 2 to the power minus x and minus half Right, so just plot this graph first without any any problem first try to plot this graph Okay, so I'm giving you 30 seconds to just plot this graph not 30 second 10 seconds to plot this graph So go step by step first plot First plot y is equal to 2 to the power of x then see what will happen if you put a minus x And then see what happens when you Put a minus half So y to the power minus x is just going to be a parabola like this. Sorry. It's going to be a Exponential curve like this cutting the y at zero. So it's going to fall down like this Okay, so it's going to be an exponential function falling like this Y is equal to 2 to the power minus of x right cutting at 0 comma 1 right And at half it will be at When x is half it will sorry when x is this is x equal to half so it becomes 2 to the power of minus half Okay Now when you subtract the half from it you just have to pull this half down So this part will actually end up coming over here Okay, so this will go off This will go off and this will come down like this Remember this value will be 2 to the power minus half minus half Now when you're modding when you're saying y is less than this Or when you're saying this is greater than equal to 1 Okay This is greater than Sorry, this is less than equal to 1. I'm sorry. This is less than equal to 1 So what will happen? This inequality will also become less than equal to 1 This will also become less than equal to so less than means it is has to lie Below this graph. It has to lie in this zone Let me change the pen It will lie in this zone now when you're modding Now you're modding x and y what will happen modding x just reflects this graph about the y axis So when you mod x what happens this goes this side as well Okay, and when you mod y it just gets reflected down as well So if you would realize from the graph it would be of this nature So all you need to do is find out this area Now can you try it out? Psi is also saying a So you just have to take one of the Quadrants and find the area in that quadrant and just do four times So can I say it's four times zero to half The area under two to the power of Minus x Minus half dx So that's going to be four times zero to one Let me just write down the result over here. So two to the power minus x By l n two with a minus sign and you'll have an x by two over here Okay, so zero to half So when you put a half it becomes four times Minus two to the power of minus half by l n two Minus one by four and minus when you put a zero you get minus one by l n two And this will be zero. So let me simplify this a little bit. So it becomes four times This becomes minus one by root two ln x Sorry ln two not ln x ln two Minus one fourth plus half ln two. So it's just going to be uh One by ln two you can take common. So one minus one by root two into four minus one Okay Which is going to be four minus two root two by ln two minus one. I think None of these options are correct in this case. So option number d is correct So guys, I'm sorry. None of you are correct Okay So option number d is correct in this case If you still feel that, you know, I've done some mistake anywhere or some correction needs to be done Please highlight it right now. But according to me the answer is none of these Is that fine? Did you do you realize your mistakes? So guys, let's take a break now and uh Let's take a break and we'll resume at 11 Let's say 11 18 am All right guys, so welcome back Welcome back after the break So those who are already back can start thinking on this problem, which has been taken from our differential equations Okay Again a very very important concept for your school exams as well So please start working on this and Do let me know your response in the chat box Yes guys any response? Okay, so Purvik says option a others Atmesh, Kushal Aryan Ritvik, Ramcharan, Kondiniya Sondarya, Sanjana Gaurav, Shweta Rohan Apurva K. Sai wants to go with option c Okay, Apurva has other plans. He is saying option b Atmesh says a How about others Kushal also says a All right guys, let's try to see how we can approach this problem Again, not a variable separable neither it's homogeneous. So let's see if things work around linear differential equation form So first of all, I'll reciprocate this entire term sign of y square plus one So dx by dy is equal to a x cube y sign of y square plus xy Okay Let me bring this so y xy on the other side and we have x cube y sign of y square Now because of this problem because of this problem Because of this x x cube term over here things are not in the exact linear differential equation form So what I'll do is First I will divide throughout with x cube So let me divide throughout with x cube When I do that I get something like this Right now guys, do you identify which is this form? If I take one by x square If I take my in fact, you can take minus one by x square as let's say you Do identify that this is actually going to give you something like this correct So this term here will be half of du by dy and this term over here will be Uy is equal to y sign of y square Okay, multiply throughout with two. So when you multiply throughout with two It becomes like this. So basically if you would recall, this is actually called the Bernoulli's form Okay, this is actually the Bernoulli's form It is one of the reducible to linear differential equation form. Okay Now it becomes easy to solve this. I can always take my 2 y to be my p. So integrating factor will be e to the power integral of 2 y dy That's going to be e to the power y square So that's going to be u times e to the power integral of 2 y dy which is y square into 2 y sign y square e to the power y square dy So in this you can always put your y square as t So this will behave as your dt term. So it'll be u e to the power y square That's going to be e to the power t sign t dt right So let's follow simplify So if I'm not mistaken, it's just going to become u into e to the power y square. This is already a ready made formula So it'll become 1 by a square plus b square e to the power y square sign of t minus cos of t plus c right So let us bring this to on the other side to u and divide by e to the power y square So it's going to be sine of y square minus cos of y square Plus c times e to the power minus y square and we already know y is 1 by x square So 2 is equal to x square times x square times the entire expression which is going to be So this will this is minus 1 by x square, right? So we'll have a negative sign coming over here and this will become sine y square. This will become Minus c e to the power minus y square Yeah So I can write it this and down brackets So this will become your answer. So which option is matching with this? So 2 is equal to x square cos of y square minus sin of y square minus 2 c to the power minus y square. So it's fine even if they write c as 2c doesn't make much of a difference Okay So I think option number a is the closest Option number a is the closest to my answer So your option a will be correct in this case Option a will be correct in this case All right. So the first one to answer this was Purik again The purik was the first one to answer this Well done purik. So moving on to question number 11 for the day So this is a question which is based on vectors Let pqr be three mutually perpendicular vectors with the same magnitude and x be satisfying this equation Then x is which of the following 11 b That's what sai says Kondanya also backs him up So does koshal I need two more responses Ramcharan Sure. Sure. Take your time. Take your time You guys whenever there is a case of a vector equation I always told you try to express the unknown vectors in terms of non-coplanar vectors Right and three mutually perpendicular vectors will always be non-coplanar They will always be non-coplanar. So non-coplanar vectors are actually linearly independent of each other So whenever you get a unknown vector try to express it in terms of The non-coplanar vectors in this case. They are these mutually perpendicular vectors pqnr So I would always start my problem by assuming that Let my Unknown vector be this You can actually take it as ijk in your mind because ijk also forms a system of Non-coplanar vectors Okay, and hence they're linearly independent of each other So you can always start your assumption that your unknown vector is this And not only that we have been given that mod of p mod of q and mod of r are all the same So I may call this as you know k for the timing Now let us see what this equation has to say the equation says Now these are all symmetrical terms right so p P and q are involved over here q and r are involved over here r and p is involved over here So what I'll do is I'll just take one of the terms Let me take the very first term and try to simplify it and I can replicate the results for the others So let's say p cross x minus q Cross p So I'm applying the vector triple product over here. We all know vector triple product a cross b cross c That's going to be a dot c times b minus a dot b times c correct So I'll be applying the same formula over here as well So it'll be p dot p times that's mod p square mod p square. I can take it as k square so k square x minus q minus minus p dot x minus q times times p Okay, okay, so let me expand this up So it becomes so if I'm not wrong it just becomes p dot x p dot q will anyways be zero Okay, so this times p it will become is that fine which is actually k square x vector minus k square q vector and p dot x p dot x from here I can say will be clearly x1 into k square okay, so it'll be p into a p dot x1 p which is going to be p mod p square x1 rest all will give you zero because they're perpendicular So this will give you x1 p k square Right. This is your first term So this is your first term So similarly your second term. I can also write it as k square x minus k square r minus And I will get uh Since there's a p there'll be x2 q k square And the third term will be similarly k square x minus k square p minus x3 r k square, okay If you add the three If you add the three terms that is your first second and third term in the given expression you'll get three k square x So that's the sum of all these three terms Minus k square You'll get p plus q plus r and you get minus k square x1 p x2 q x3 r which is actually again your vector x itself correct Right, and this is given to you as a null vector So you get two k square x because three k square x is over here and minus k square x is over here So you get two k square x vector minus k square is e Is equal to zero Which clearly implies x is going to be p plus q plus r by two p plus q plus r by two, okay So I think option number b, yes option number b is correct in this case The first one to answer this was Sai Meer Sai Meer was the first one to answer this very good so guys the Critical point over here was this assumption. This assumption is something which was important That means any vector in space could be expressed as a linear combination of three non-coplanar vectors And after that it was just a mere application of the vector triple product and you know Following the symmetry of the figure and trying to see what which terms You know combine and which terms cancel out So guys any question with respect to this, please let me know right now All right, so moving on to the 12th question for the day The question says decimal parts of the logarithms of two numbers taken at random are found to six places Okay, so I think it's just the decimal places which are six in numbers The probability that the second can be subtracted from the first without borrowing All right, so Rohan says c is correct. Okay I have registered your response Rohan Let's wait for the others again probability and permutation combination very heavily linked chapters But if at all you have not You're not very confident about the use of permutation combination and probability Uh, don't try to slog at the 11th are okay It's fine. It's not necessary that these topics would be strong for everybody Because it's more of an art. I believe So don't waste too much time doing it Yes, of course some basic things like base theorem law of total probability probability distribution conditional probability You can focus on all these concepts Atmesh backs it up with c Shweta says a okay Sai also says c So does a purva. Okay, fine. So I have got enough response now for me to start solving this So guys again focus on one number at a time. So let's say our uh Decimal places the number after the decimal places that is the six places are x 1 x 2 x 3 x 4 x 5 x 6 For the first number and y 1 y 2 y 3 y 4 y 5 y 6 for the second number, okay So what i'm trying to say is that if you're trying to subtract the second from the first Without borrowing that means each xi should be greater than yi Okay, in fact greater than equal to I can say Okay, so let's focus on any one of them. Let's say I focus on this one x 1 and y 1 okay Now it's clear that x 1 and y 1 both can take values You know from 0 to 9 correct 0 to 9 that means they both have 10 10 options each So the sample space for choosing x 1 y 1 would be Uh, nothing but 10 into 10 Nothing but 10 into 10 that is 10 options for x 1 And 10 options for y 1 no doubt about that. So that's 100 correct Now if you want x 1 to be greater than y 1 Okay, then let's say if you choose x 1 as some a Okay, let's say some a which where a belongs to Uh 0 to 9 Okay, a belongs to 0 to 9 Okay Then y 1 can only take values from a plus 1 onwards till 10 Right not 10 9 I'm sorry. Yeah 9 Okay, that means for every option a y has how many number of ways It's a number of ways in which you can choose y 1 would be nothing but 9 minus a plus 1 plus 1 No, it can be a also it can be a also right. So I'm sorry. So it can be a also So it will be a over here Because boring you can put pickup a also Yeah correct Right, so it will be 10 minus a Okay, in this case, you will always realize Oh, I'm sorry. You have Let's do the other way around. Let's say this is your uh This is your let's say Y i then this is your x a yeah, no, it's fine. Is that okay? So very for every y 1 y i as a x i can become a a plus 1 all the way till 9 Okay, so the number of ways in which you can choose your x i is going to be 10 minus is going to be 10 minus a okay So can I say the total number of sample space in the favorable event would be nothing but Summation of 10 minus a all the way from 0 till 9 Okay, so that's going to be 10 plus 9 plus 8 all the way till you reach 1 That's going to be some of this which is going to be 10 into 11 by 2. That's going to be 55 Okay So your probability that your x 1 would be greater than y 1 would be 55 by 100 That's going to be 11 by 20 And there's six such places the six such places. So your answer is just going to become 11 by 20 11 by 20 11 by 20 Okay, hold to the power of 6 which is clearly option number C is correct So I think Rohan was the first one to answer this well done one So with this we move on to the next question over here Question number 13 Yes guys anyone any success? all right It cannot be too for sure It'll either be unique or no solutions or infinitely many solutions Okay, so mostly people are going for one 13a Rohan says 13c a Purva says a Sai also says a okay All right. So can we start the discussion for this? So first of all if t is real and we had we have a equation over here So let me do one thing. Let me generate a quadratic in terms of T So lambda t square plus 3t plus 4 will be a t square minus 3t plus 4 That means a t square lambda minus 1 plus 3 times lambda plus 1 times t Plus 4 times lambda minus 1 equal to 0 Now since t belongs to real numbers, I can say the discriminant would be greater than equal to 0 So the discriminant Will be this minus 4ac will be 16 lambda minus 1 the whole square would be greater than equal to 0 So, I'm sorry, why did I write 14 it should be 16? Yeah So, let me write this as a square minus b square formula.  This should be greater than equal to 0. That means 7 lambda minus 1 and I have 7 minus lambda greater than equal to 0 Okay Which clearly makes me feel that I can write this as 7 lambda minus 1 times lambda minus 7 is less than equal to 0 So lambda has to be a value between 1 by 7 to 7 So far so good Now let me Write down the determinant for this system of linear equations So let me write down the determinant for Or the determinant of the coefficient of the variables in this equation Okay, let's see. What does the value of this come out to be? Let me expand with respect to the first row itself So 2 lambda plus 15 plus 1 lambda plus 18 minus 4 Uh 5 minus 12 5 minus 12 is minus 7 So that would be 7 lambda 45 plus 28 Plus 18. So that's going to be 45 plus 46. That's going to be 91 Why did I write minus over here minus 28? Yeah, so that's going to be 35. That's going to be 35 Okay So now as your lambda lies in this interval We can see that this expression This expression Cannot be 0 Because for it to be 0 lambda has to be minus of 5 Which is which does not belong to this interval which does not belong to this interval Okay, which means the determinant of this cannot be 0 If determinant is not 0, there is no chance of getting no solution and infinitely many solution That means it can only have a unique solution. So option number a is correct So option number a is going to be correct in this case Is that fine guys? All right. So moving on to the 14th question now So this is a question which is Picked up from 3d geometry if p be any point on the plane lx plus ny plus nz equal to small p q be a point on the line op Such that op is equal to op into oq is p square Then the locus of the point q is Then the locus of the point q is Yeah, so these are planes These are planes If they intersect at more than one point that means they would intersect on a line of intersection So they will have infinitely many solution or just one solution just like it happens in case of x y y z and zx planes Or it can have no solution Okay, so rohan has already come up with the answer rohan says It's c Prati kondinia also says c Okay kushal also goes for c I need two more response So i'm very also goes for c kushal also. Okay So most of you are giving the c as the answer. So let's see whether c is the right answer or not Okay, so uh Let the point p Be x1 y1 z1. Okay So since it lies on the plane, I can say the equation of this plane would be So this point will be satisfying this plane. So l x1 My one plus n z1 is equal to p. Okay Now let us write down the equation of The line op so Let's say this is your plane. This is your point p. This is your origin And there is a point q in this line. There's a point q on this line. Okay, so first of all I can always write down Let's say this is alpha beta gamma And this is x1 y1 z1. In fact, I don't have to write the equation of op I can say their direction ratios will be proportional correct So direction ratios of Direction ratios of op Would be proportional to the direction ratios of oq That means x by alpha Will be y by beta is equal to z by gamma Okay, let me call this as k Now the third condition says op into oq is p square op will be x1 square Okay y1 square z1 square this will be op Oq will be under root alpha square beta square gamma square Okay Now since I have to write The locus of alpha beta gamma I can do one thing I can substitute x1 as alpha k y1 as beta k And z1 as gamma k square Under root So this will become k alpha square plus beta square plus gamma square is equal to p square Okay Now k also should not be a part of my answer K also should not be a part of my answer So I can do one more thing I can substitute x as alpha k y1 as beta k And z1 as gamma k In this equation Okay Which clearly means k can also be replaced with p by l alpha m beta n gamma. So let us put it back over here Let's put it back over here.  Let's see which of the option matches I think option number b matches No, no, sorry option number c matches option number c matches So this will be your desired answer So the first one to answer this was rohan So rohan was the first one to answer this well done rohan. So some of you are constantly replying and you're replying correctly, especially I've seen some of yours Strike rate is very very high So guys, uh, moving on To the last question of the day So we have a question over here picked up from properties of triangles It's a column match question. I'm sorry. This is happens to be your 15th question So match the column one with column two should actually draw it in a very So let's say this point is e Yeah, so this is 90 degree, right? This is 90 Okay, now they're asking you the minimum value of cot b and cot c Okay, how do I find that out? So, uh, I can always find this angle. Let's say this is theta and let's say this is phi So tan of b is tan of theta plus Phi which is actually tan theta plus tan phi By 1 minus tan theta tan phi Okay Now from the figure itself, I can clearly say tan of theta will be x by 2 y And tan of phi will be tan of phi will be x by y and here I will have 1 minus x square by 2 y square Okay So it implies It implies that tan of beta will be nothing but Uh x y x y 3 x y and in the denominator, I'll have 2 y square minus x square So cot of b will be 2 y square minus x square by 3 x y, okay I can similarly say Cot of c would be 2 x square minus y square by 3 x y, okay So let's add these two. Let's add these two. So cot b plus cot c Will be nothing but x square plus y square by 3 x y Now here, I will use am greater than gm since both are positive quantity. I can say this will always be greater than root x y Correct. So x square plus y square will always be greater than equal to 2 x y So this quantity will always be greater than 2 x y by 3 x y That means it will always be greater than 2 by 3 So minimum value will be 2 by 3 minimum value 2 by 3 is option number q So a will be mapped to q. So a answer will be q Let's see anybody has been able to give a s q No, Rohan says a is r sai also says a is r Okay, so a is q my dear students Right, let's talk about the b part if capital r is 13 by 2 small r is 2 And small r1 is 3 find the area of the triangle in square units area of the triangle in square units Okay, so how do I do this problem? So the circum circle radius is given to you The in circle radius is given to you and one of the x circles Radius is given to you. So whoever are saying b is equal to s.  I keep on picking up from different different books Okay, I would like to know whether how many of you are aware of these basic formulas which connects R small r r1 r2 r3. Okay. I wanted to know properties of r1 r2 Are you all aware that the sum of all x centers radii is actually r plus 4 r. Are you aware of this? How many of you are not Then let me tell you You need to keep these formulas in your mind Another important property was one by r is one by r1 plus one by r2 plus one by r3, right Now why I'm using this is because I've been given such information. I've been given I've been given. What is the value of capital r? I've been given. What is the value of small r and small r1? Isn't it? So r1 is given to us So what is the value of r1 given to us? Let's check r1 is 3 So let's use this so r1 is 3 r2 and r3 is not known And r is given to us as 2 And capital r is 13 by 2 So 2 and 4 into 13 by 2. So r2 plus r3 is going to be 26 plus 220 a 28 minus 3 which is 25. So that's your first equation Now using the second one. This is by using the first one using the second one. I can get 1 by r is 2 r is 2 Is equal to 1 by r1 which is 1 by 3 And 1 by r2 and 1 by r3. I don't know right So I can write this as 1 6th as r r2 plus r3 by r2 r3 That means r2 r3 is going to be 6 times r2 plus r3 which is 6 times 25 which is going to 150 Which is going to be 150 which clearly gives me These two equations Which everybody can guess that r2 will be r2 or r3. One of them would be 10 and another one of them would be 15 correct Now still how does it help me to get the area of the triangle? Okay Now if I know r2 r3 I can use yet another property which is which says that S square that is semi perimeter square is r1 r2 r3 by r. How many of you are aware of this formula? Semi perimeter square is r1 r2 r3 by r So using this formula I can say s square is r1 r2 r3 which is going to be 3 into 10 into 15 By r r which is 2 that's going to be around 75 to 25 so s is going to be 15 Now area of the triangle is nothing but r s That's going to be 2 into 15. That's going to be 30 square units 30 square units So yes those over those whoever said that your b will be s you are correct b will be s you are correct Okay c what do you think will be the answer for c anybody who has given the answer for c c is q C is not q So those those ever who are saying c is q that's wrong I think working on d is easier. So let's work on d first So let me let me work on d first okay So I have been given that uh sine 4a plus sine 4b plus sine 4c plus 8 cos a Is equal to zero so first of all I can Club up these terms So let me take sine 4a plus sine 4b together And then sine 4c.  This is equal to zero given to me. Okay, so there is a possibility that your a could be 90 degree Your a could be 90 degree And sine a sine 2b sine 2c is equal to 1 Which clearly implies if a is 90 degree Then Sine 2b into sine 2c will be 1 which clearly implies each one of them has to be 1 Because they cannot exceed anyone cannot exceed 1 right Which means b is equal to 45 degrees c is also equal to 45 degree which clearly satisfies this case And hence we can say that Hence we can say that cos of a is going to be zero correct Cos of a is equal to zero. So this is going to map to p So I think This is the only one which is left to map to r. So I think this is map to r But I would leave this as a homework for you to prove that cos a is equal to half in this case Okay, so please do get back on this Okay, so guys we have already exceeded The time limit that we had set for today. So anyways, it was a great having the session with you all So hope you learned many things new especially in the last problem Make sure you know your weaknesses and keep revising it through these problems Thank you everybody for coming online and all the best for your exams which are which you have when your school reopens And of course a very very happy new year to you. So may the new year bring a lot of good marks good ranks and happy faces Okay So enjoy bye bye Take care over and out from syndrome academy