What is going on guys welcome back to the channel and I appreciate you being here as always today I am back with another fortnight video and today we're going to cover the secret rings fad in the game and a couple other Things but before we do that people as per usual. I'm giving away 2800 vbooks to win make sure you are subbed to the channel and have that bar button ticked drop a like on this video Leave a comment down below winners are picked weekly and contacted via direct message. So good luck everybody Okay, so you might not know but there are secret rings within the sky in fortnight battle Yeah, at the moment we don't know what they are for but we have a few guesses which do make sense Now these rings were first spotted a few days ago But actually they're kind of weird in you trying to find them To actually get a glimpse at these you need to go into replay mode Go into that free drone and while you are waiting to board the bus and our mini little island You need to fly over the main island here within the sky should be quite a few rings Now here's the weird part as far as I know these are only visible when doing this on PC I've tried myself twice on Xbox two of my friends have tried on Xbox with no luck My friend tried on PS and had no luck, but everyone on PC seems to see them They do disappear not long after you've left the bus to land also noticed there are two altitudes of rings Some lower than others. So no doubt these will be a new challenge to come I think in the coming weeks, we might see a sky dive challenge of flying through a certain amount of rings cool challenge for sure So yes guys if you are on PC check them out for yourselves as far as I know an Xbox and PlayStation These cannot be seen but if you do do this if you do go into replay mode and you spot these Let me know down below even that comment section That's if you're playing on Xbox or PlayStation because like I said I tried few my friends tried on both Xbox and PlayStation we couldn't get them to appear But everybody on PC seems to have them which is real weird, but yeah, it does sound like a cool challenge So now we wait Moving on and unscrewing now you can see this coming first days challenges These were leaked and I believe by fortnight tracker and they are deal damage to opponent structures You need to do 5000 in total search chests in haunted hills. You need to find seven chests build structures 250 materials used there I believe that is visit the different tackle shops in a single match You need to do three of these follow the treasure map found in moisty Maya shotgun eliminations We need four of these eliminate opponents in a lucky landing. They need three of these now I will mention this past week's challenges were also leaked but few of them were wrong So although these were data mines from the game is it 100% confirmed that these will in fact all be this week's challenges But if we are to go on these challenges, it does seem pretty easy I mean shotgun eliminations you need for these this class is a hard I mean, that's the easy part of stars seriously is eliminate opponents in looking landing three of these easy again Search chests in haunted hills left side map simple deal damage to opponent structures I mean everyone builds all over the place simple visit different tackle shops in a single match You need to do three of these now at the moment these are located you can see them But when this comes I will drop a guide like many of these other challenges I will drop guides on my channel this coming Thursday So make sure you are subbed if you are new to this channel and make sure you have that bell button ticked also So you never miss a video I upload but you're like I said these challenges do seem pretty easy once again So we're wait this Thursday to see how many of these challenges actually are there but guys that is it for another video Sure, but sweet if any more news or leaks drop or have you guys covered here on my channel for sure So make sure you are subbed and hit that bell button if you are new around here So never miss a video thanks as always for stopping by leaving a like really helps me out and hopefully people I will see you on that next one


Greetings, XPrize. My name is Mike Fossum. I'm an explorer and a crew member on an international space station. Exploration is an important part of who we are as a species. A favorite fictional character of mine, Star Trek's Captain Kirk, spoke of boldly going where no man has gone before. Space exploration helps us to learn more about our planet and the wonders of our universe. My missions in space and the ability to look back at the Earth have forever changed how I think of our planet and myself. I cherish and prize exploring to allow us to extend our presence in space and to help create a better future for all of us. What do you prize?


tangents and normals. Tangents and normals. Okay. This is again an easy concept but it is slightly more lengthy as compared to the previous topic that we had taken today. So what do you understand? What do you understand by dy by dx? Geometrically. So we had learned this in our class 11th itself that if you have any curve, okay, and at any point x if you are finding or you can say at any point x comma y you are finding dy by dx. That means you are trying to find out the gradient of the curve at that point. So dy by dx gives you the gradient of the curve at that point. Now gradient of the curve at that point means it gives you the slope of the tangent at that point. So the slope of the tangent at x comma y is basically dy by dx or you can say x1, y1 to be more precise. So dy by dx at x1, y1 gives you the slope of the tangent drawn at that point. Okay. This is what we had understood from our geometrical interpretation of dy by dx last year. So if I know this I can easily get the equation of tangent and not only that I can also get the equation of the normal. Okay. First of all how do we get the slope of the normal? Slope of the normal at x1, y1 is negative reciprocal of dy by dx at that point. So you can say it is negative dx by dy. So if somebody asks you what's the equation of the tangent at x1, y1 you can easily use the formula which is a two point you can say a slope point formula which is y minus y1 is equal to dy by dx calculated at x1, y1. Now I would like to spend a few seconds over here. This process is very important. Many people what they do is they just use dy by dx. No, dy by dx is an expression for the slope. It is not the slope actually. It is the formula for the slope. So unless until you specify the point you will not get this value of the slope from there. People have done lot of mistakes here because sometimes I see they use the exact expression for dy by dx and as a result they get a nonlinear equation of a straight line which is not possible. Okay. Year after year people make such mistakes. They just do dy by dx and put it in place of a slope. No. If you do that your entire equation of a line will become nonlinear. Right. In many cases. So please put the value of the point. Get a value. This is a value that you will get. Some number will come out. That number you have to put over here. So this will be the equation of the tangent. Similarly equation of the normal is nothing but negative dx by dy calculated at that point x1, y1. Okay. And if this curve is given to you in a implicit form please note you have to use your implicit differentiation to find dy by dx. Many books will write it as f dash x. Many books will write dy by dx as f dash x. Please note this will be true only when your function is explicitly given to you. If your function is implicitly defined you have to figure out dy by dx from it by using implicit differentiation. Okay. Are you getting my point now one thing that I would like to highlight over here is that one thing I would like to highlight over here is that the equation of the normal the equation of the normal and the equation of the tangent can also exist even if the function is non-differentiable at that point. Remember differentiability also comprised of the fact that the slope of the tangent at that point is infinity. So you can have a tangent line or you can have a normal line which is having an infinite slope that is fine. So don't get confused between the two scenarios. Okay. So if the function is non-differentiable at that point because it has got an infinite slope there it doesn't mean you cannot find out the equation of the tangent or you cannot find out the equation of the normal. Are you getting my point here? Okay. So many people say that it is non-differentiable at this point. So how will you find a tangent there? And it is non-differentiable because it has got infinite slope. So if it has got infinite slope then you can write down the equation of a tangent. Don't we have lines with infinite slope? Are you getting my point? Okay. Now second thing that I would like to discuss over here is that. By the way what is the normal? That's also one thing that people don't know very well. When I ask about normal they say sir a line perpendicular to the curve at that point. No. It is not a line perpendicular to the curve. It is a line perpendicular to the tangent at that point. Are you getting my point? So normal at a point on the curve is nothing but a line which is perpendicular to the tangent drawn at the very same point. And you will be surprised to know that many a times the same line can behave as tangent and normal also. Can you give an example of a curve such that the tangent and the normal at the very same point is the same line? Mod X. Where? At what point are you talking about? At X equal to 0. Can you draw tangent at X equal to 0 for mod X? No you can't but like from one arm no. The tangent and the normal is the same line at the very same point. Okay. I'll give you such a curve. Look at this curve. Okay. This curve is basically called folium of the cartus. Let me draw this curve for you on the GeoGeballer platform. Why am I writing in caps? Yes. Sir by exactly is the statement that normal is perpendicular to the curve wrong? I mean even if you want to consider that I don't think it'll be technically wrong. How do you define the angle between a line and a curve? Like you take a very small portion of the curve then becomes horizontal or straight line and then you and the given line is already given so you can measure the angle. I know in effect that small portion becomes a part of the tangent itself but. Angle can only be defined between two rays. That's the very basic definition of angle. If you go to your class 11th first chapter first day of trigonometry. What did the teacher tell you? Angle is basically a parameter which tells you by fact by what you know parameter or by what magnitude a particular ray has rotated. So it only exists between two lines. There cannot be an angle defined between a line and a curve. If you want to define an angle between line and a curve you ultimately make a tangent at that point and then see the angle from it. Okay. What the point? Yes sir. Okay look at this curve. Look at this point. Look at the origin. If you see your x-axis. If you see your x-axis. This x-axis is also tangent because it is you know just trying to touch this part but it is also normal because it is trying to cut this part. Are you getting a point here? It is not differentiable here because even y-axis will behave as a tangent. So there are two tangents here and there are two normals also. So x-axis is behaving both as a tangent and a normal at this point and same goes with your y-axis also. It will behave both as a tangent and a normal. Okay. So it is a tangent for this part, blue part and normal for orange part. So at origin this for origin for this curve as you can see it is called folium because a kind of a folium is created here. Okay. There will be two tangents and each of them would behave both as tangent and normal. Okay. This is just an extra information I'm giving you because many people don't know this it's for your information kind of thing or for your knowledge kind of a thing. Okay. So the tangent can also behave as the normal at the very same point. Okay. One such example is this. Folium of discartus. Okay. So this concept is not that great because we have already learned this to a large extent in our class 11th when we were learning the geometrical interpretation. So let's take questions. Questions is what is going to be interesting over here. Let's start with this question. I'll put the poll on. Very good. It's last 30 seconds for this. Five, four, three, two, one. Okay. Let me end the poll. Most of you have gone for option C. This is just a slightly twisted question where instead of giving you a Cartesian equation, this has been given to you parametrically. Okay. And they're asking you to find out the equation of a normal to this curve at the point 2 comma minus one. So first of all, normal is what normal is DX by dy. This is the slope of the normal at that point where you're trying to find the normal. But the question here is we have everything in terms of a parameter. So what I will do the same thing I will translate like this minus DX by dt upon dy by dt. Now this time calculated at that parameter. Now first thing that we need to do is what is the parameter which corresponds to this point P remember every point on the curve will have a unique parameter for it. So we need to figure out what is that T value which generates xs to and simultaneously generates y as a minus one. In other words, I'm asking you to solve this equation and simultaneously I'm asking you to solve this equation. Are you getting my point? So in order to get this T value, I have to do this activity. So this gives me T squared plus three T minus 10 equal to zero, which is easily factorizable as T plus five T minus two equal to zero. So implying T could be two or it could be minus five. Same here I can do our two T square minus two T minus four equal to zero. That means T square minus T minus two equal to zero. That's T minus two T plus one equal to zero. So that means T could be two or T could be minus one. Now which is the overlap? Which two which value of T is common to both of them? Two. Okay. So this point will correspond to the parameter T equal to two. Let me write it in some different color. So this will have a parameter T equal to two. Got the point? Now let us use our parametric differentiation here minus dx by dt will be minus two T plus three upon dy by dt which is four T minus two. Put your T value as two. That will give you minus seven upon six. So whichever has a slope of minus seven by six, that would be your answer which is clearly option C. You don't have to even find the equation. So option C is right. Are you getting our point? Any questions here? Any questions? Any concerns? Please do ask. No questions. Next. If the tangent, if the tangent drawn at x one comma y one to the curve x cube plus y cube is equal to a cube meets the curve again at x two comma y two, then prove that x two by x one plus y two by y one will be equal to minus one. Intersect Venkat. See, I'll just take a rough example here. Okay. Let's say this curve is roughly shaped like this actually. There is a okay. This is how the rough shape of the curve will look like. Okay. So let's say I sketch a tangent at this point x one y one. So this tangent drawn here goes ahead and cuts the curve again. Let's say at this point x two y two. You have to prove that x two by x one plus y two by y one for such a case will be minus one. Got it. Thank you. What is there in chemistry syllabus for tomorrow? So electro dam and solutions. Unfortunately, we haven't done solutions in center. Oh, okay. What do you do? A lot of solutions in maths. No. Okay, this question is not that easy. Actually, you'll be stuck at a lot of points here. So I'll also help you as you are also solving it. Okay. What's there? You're not trusting us only that we'll be able to do it. Okay. I think one more class or two more class should be we should be able to finish application of derivatives and then we can start with definite integrals. A bit of functions will be done in between definite will take some classes, then you have probability, then you have differential equations, then you have area under curves, then vectors and 3d linear programming.  So today only I got a news of there is a corona patient in our apartment. So I was like, let's finish the syllabus early before I but now we are not scared actually now earlier it was like, yes, it was very scary, but now it's just another thing. This is the entire character of human race. It's getting recorded humans will hear it. Okay, let's let's be slightly lazy here and let's apply implicit differentiation here. So for this curve, if I apply implicit differentiation, I'm calling this as your f of x comma y equal to some constant. Okay. So this will give you negative three. Now please note we are finding it at x one y one because you're sketching the tangent at x one y one. So this will be negative three x one square and three y one square. That's nothing but negative x one square by y one square. Okay. Good. Now you know that this tangent slope must also be the slope of the line connecting these two points that we call them as pq. So slope of the line pq will be nothing but y two minus y one by x two minus x one and this should match with your x one square upon y one square. Correct. Let's call this information as one as of now and let's keep it. We'll use it when the right time comes. Second thing that I can say is that both x one y one and x two y two points must satisfy the curve. Right. So x one y one and x two y two. Both must satisfy the given curve, which means x one cube plus y one cube. Okay. Is equal to a square and x two cube plus y two cube is also equal to a square. So what we can do is we can write it as x one cube minus x two cube is equal to y one cube minus y two cube. Okay. And we already know that this factorization is this lot of writing involved. Yeah. I personally believe maths loses its sheen when you have to write a lot. Maths is supposed to be like click. Right. All right. So from here I can say correct me if I'm wrong y two minus y one by x two minus x one y two minus y one. Oh my bad. This is negative sign. If you're subtracting it, it'll become a plus here. My bad. So there would be a negative sign coming here. Yeah, somebody likely corrected me. I think. Okay. All right. So yeah. So can I write this as y two minus y one by x two minus x one as negative x one square x one x two plus x two square by y one square y one y two plus y two square. Just check whether this is correct. Mind you, I made a small modification here, which I think I had overlooked when you're subtracting both the terms will come on the same side. Right. So if you're taking it to the other side, a minus sign will come over here. Just check it out. Okay. Yeah. That is what I'm going to do next. Now you can see this term. This term matches with this term. And hence I can replace it with negative x one square by y one square. Okay. So what I'm going to do is I'm going to replace this with a negative x one square by y one square. As a result, minus sign will go for a toss. So I'm just doing it in the next step. In fact, I'll cross multiply first. So y one square x one square is equal to x one square y one square y one y two plus y two. Okay. A lot of terms will get cancelled. For example, if I expand this x one square y one square will get cancelled off. So I'll not write it down. I'll just write down the terms which matter to me. Here also I'll have x one square y one y two plus x one square y two. Is this fine? Any problem with this term? Please stop me here. If you're not happy with this expression, please stop me here. Now what I'll do is I'll club these two terms together and I'll club these two terms together. So I'll write it down as x two square y one square minus x one square y two square is equal to is equal to on the right side. I'll get x one square y one y two minus minus x one x two y one square. This I can write it as x two y one plus x one y two. And this I will write it as x two y one plus sorry minus minus x one y two. And this side I'll take x one y one common x one y one common. So this will give me x one y two minus x two y one. Now if you see this term and this term are almost the same, of course, they differ in their science. And I know they're not zero. Why they're not zero? See, if they're zero, then what will happen? x two y one is equal to x one y two. Yes or no? Negative. That means x one by x two is equal to y one by y two. Yes or no? And let's say I call this as a k. Okay. Now if this is to be true, if this is to be true, can I say it can be true only when x one and x two are equal to each other and y one and y two are equal to each other? Can I say that? I mean, why sir? You can argue against it also. Yeah, why sir? You tell me why? See, x one is k x two and y one is k y two. And you know that x one cube plus y one cube is equal to a cube, which would be ideally k cube x two cube plus y two cube. And even this is a cube. Even this guy is a cube because you know x two x one x two y two also lies on the curve, which leaves you only with a k cube equal to one. That means k can only be one. Of course, it cannot take imaginary values. Okay, which leaves you with the fact that x one will be is equal to x two and y one is equal to y two, which is not the case because x two y two point is separate from x one y one point. Okay, this is not the case. Now the equations never lie to you. They're actually giving you all possible scenario. They're telling you that the tangent drawn there will cut the curve at the same point where the tangent is drawn, which is actually not a lie. It is actually true. But it is up to us to understand that okay, we are not referring to the very same point where the tangent is drawn. We're referring to some second point or the third, you know, some different point. So if you drop this out, the only possibility that you will be left with is x one y one is equal to x two y one plus x one y two. Okay, of course, with a negative sign. Remember, I did not these terms are not exactly equal. They're negatives of each other. Okay, and if you divide throughout with x one y one, you'll end up getting you'll end up getting x two by x one plus y two by y one is equal to minus one. That's what we wanted to prove. Where is it x two by x one plus y two by y one equal to minus one. Got it. Okay, let me write it. This this is not possible because this is not possible. Any questions here? This may come as a school question, but I think it'll be too heavy for school question to in for your school teacher to ask you in a school exam. But be prepared for such questions. Good enough. Yes, sir. Next question. Oh, do you want me to go back? Do you want me to revisit the previous problem? Okay, you can manage right? Okay, I'll put the poll on for this. Okay, yes. So we have to find out the equation of the tangent to y equal to gx at x equal to three g is the inverse of this function. Only three people have responded so far. Very good. Okay, last 30 seconds. Five, four, three, two, one. Hey, vote. I'm counting and nobody's voting only. Come on. 19 of you have not voted yet. Please vote. Please vote. This is not an election voting. Choose your candidate without any fear. Vote for vote for this question. Okay. So despite all the pleading 11 of you have not voted. Most of you have gone with option A. Okay, let's check. See, guys, when it comes to finding the derivative for the inverse of a function, we don't really need to know the inverse. Okay. We just have to know the fact that f of g of x is an identity function. That means it always gives you x. So if they say f and gr inverses of each other, they'll follow this kind of a relation. Right. Now differentiate both sides. Now what do we need? What is our requirement? Our requirement is the equation of a tangent drawn to g of x at x equal to three. So in order to know the equation of a tangent, we need to know what is g dash three, then only we'll get the slope of the tangent, right? So if I put a three value over here, let's say I write it like this first and I put x is three here, then my problem arises. How do I get this value? Okay, I know the function, but what is this point? How will I figure this out? Okay. Now when I say g three gives me a value x or gives me a value k or this value is equal to k. What do I understand from it? What do I understand from it? It means it means what value of k put in the function f will give me three. Are you getting my point? So these two are synonymous. When you say g three is k, what does it mean? It means what value of k when put into the function, so when this this is fed to the function, it should give you a three. And the only possible value here I can see is k equal to one. In other words, you are trying to find out one by f dash one because g three is k and k is one, which means you're trying to find out what is the derivative of this function at one. So I think it is three x square plus one. And when you put x equal to one, you get one by four. Okay. Now this point k also tells you that three comma one lies on g of x and it is this point that you are sketching the tangent on. So this is the point of tangency. So you know the slope, you know the slope, which is one by four, you know the point of tangency. So I think nothing can stop us from getting the tangent equation now. So it's y minus one is one fourth x minus three. So it's four y minus four is equal to x minus three. That means your equation would be, yeah, equation will be x minus four y, x minus four y plus one equal to zero. This is going to be your answer, which is option number a. Please have a good look at this problem because these kind of problems are very common nowadays in J main exams. So this is something that you should all be knowing by now because you have done your differentiation. Okay. G three means that value of k which fed into the function will give you a three because F and G are inverses of each other. So that can only be one. So G three will be one. So when you know G three is one, that means the point becomes three comma one. And from here, you end up getting slope also. So slope and point both are known. Which, which party could not get? The G dash of x. No, no, before that. This part. Yeah. Yeah. This is after that from that, I think it from that sir, this point that I got sir. Yes, you got. Yeah, that also I got. Okay. And after that, I know the problem is you want G three. What are the meaning of G three? G three means, see, you know this property also, right? Correct? As long as they're inverses. Yes, sir. Yes, they are inverses. That's why, you know, we know this property. Correct. Now if you want your F of x to become a three, what should be the x corresponding to it? G of three. If you want your F of x to be three, what should be the x value? One only. Oh, you're asking like that. Okay. Okay. If you want your this term to be three, then what should be the x value that you write over here? Yeah, one. One. So can I say G three will be one? Yes, sir. That means three comma one is a point lying on the curve where you are sketching the tangent. Oh, okay. This becomes your point of tendency. That's what I wrote over here. This becomes your point of tendency. Correct. And F dash G three, G three already you figured out it's one. So F dash one is what you can easily figure out by differentiating it. So three x squared plus one put x as one, you get a four. So one by four. That will be a tangent. So you know the slope, you know the point, write down the slope point form. Done. Got it, sir. Is it clear to everybody? Let's take a simpler question. Let me put the poll also so that you can reply. Very good. Okay, let's close this in another 30 seconds. Guys, this is easy. Under two minute question. Five, four, three, two, one, go. Please vote, please vote. Okay. End of poll. Most of you have voted for option C again. Let's check. Okay, the tangent at one comma one meets the curve again at P. So first, let us differentiate it. Two y dy by dx is two minus x the whole square minus two x two minus x. Okay, let's calculate the dy by dx at one comma one. So y will be one. This will be one. Okay. So this will give you dy by dx as if I'm not mistaken, it'll be minus half, right? So the equation of the tangent will be y minus one is equal to minus half x minus one. That means y will be equal to minus x by two plus three by two, right? That you can write it as minus three, three minus x upon two. Okay, put it back over here. Because you want to know where does this tangent cut the curve again. So in place of y, I'm going to put three minus x by two. Let us try to simplify this. Okay, this will give me a cubic. Okay, correct me if I'm wrong. Now this equation will have definitely one root as x equal to one. Now how did I predict it so fast? Of course, you'll say, sir, you have put the value one and check that it's coming out to be zero. But other than that, how did I predict it so fast? Remember, the tangent will meet the curve again at the point where it is tangent because that's how equation will treat this as equations will always give you all possible scenarios that will be covered into that case. So since you're drawing the tangent at one comma one, the curve will the tangent will meet the curve also at one comma one, right? So one will definitely be a root of it. So if you try to factorize it, you'll have four x square and this will be plus nine if not mistaken, let's say I take it as kx. Let's compare the coefficient of x square. So x square will be obtained as minus four plus k and that's equal to minus 17. So k will be minus 13, k will be minus 13. So this will become x minus one, four x square minus 13 x plus nine equal to zero. This is factorizable as four and nine itself. Oh, it has a repeated factor that means so it has a repeated factor of this and the other term is other factor will give you x as nine by four. So whichever says nine by four as its x coordinate, that would be your answer. I think only option C says that so C will be the right answer. Clear everyone. Any questions, any concerns, please do let me know. Yeah, the poll is on. Almost two minutes over. Only one person has responded. This is so difficult. Okay, I've got two answers, three answers so far. Last 30 seconds. Last 30 seconds. This is under one minute question, actually. Five, four, three, two, one, 17 of you have not voted yet. Please do that. Please do that fast, fast, fast, guys. I understand NAFLA is not voting because they have a UT tomorrow, YPR not voting because they have a UT tomorrow. At least NPS Rajaji Nagar should vote, right? Go NPS, go. Okay. Chill, end of poll. Maximum people have said B. Okay, let's check whether B is correct or not. See, if you look at the numerator at zero to zero, denominator will also be zero because you're feeding in f of half minus f of half that will anyways cancel out. So it's a zero by zero form, right? So we can apply Lopital. So this limit is of the zero by zero form. So we can apply Lopital. Okay, that means you can apply Lopital. So when you apply Lopital, you differentiate the numerator separately and differentiate the denominator separately. So 2x minus 2x minus one would be your derivative of the numerator term. The denominator will give you f dash e to the power 2x by two into e to the power 2x by two into two. Correct? So I think two two will get cancelled off. No need to write it. Okay. The other term would give you negative f dash e to the power minus 2x by two into derivative of e to the power minus 2x by two, which is half e to the power minus 2x into minus two. Right? Right? So if you write it properly, you would realize you have written limit extending to zero 2x minus one f dash e to the power 2x by two into e to the power 2x plus plus e to the power, sorry, f dash e to the power minus 2x by two into e to the power minus 2x. Okay. Now when you put zero, you will end up getting minus one on top. Here you will get f dash half. This also will give you f dash half. Am I right? That means minus half into two f dash half. Now how do I get f dash half? For that you have been given that this is a tangent to this curve at half, which clearly means f dash half is the slope of the tangent. And that is nothing but negative three by two. Am I right? Okay. So your answer here would be minus one by two into negative three by two. That's one third. So Janta was wrong. Answer is option A. Janta, what are you doing Janta? Could you go down slightly? Okay, I hope you now realize your mistake. And I hope you realize this was not a difficult question actually. Let's take one more question. Can I move on? Yes, sir. Okay. One more question which probably will come in a school also, but in a different way, different format. Show that show that if x cos alpha plus y sin alpha equal to p touches the curve x by a to the power m plus y by b to the power m equal to one, then then a cos alpha to the power m by m minus one plus b sin alpha to the power m by m minus one is equal to one, or sorry, not equal to one is equal to p to the power of m by m minus one. I remember the same question will come in school.  So right now I'm taking a more generic version. Okay, in school, they'll give you a very specific case where your m is two, where your m is. Try this out. Again, in this problem, many students make mistake. They get muddled up. Okay, because of these ugly powers coming up. So let me also contribute towards solving this so that we expedite the process. Okay, let's say, let's say that this this tangent touches the curve at x comma y. Okay, let's say the point of tangency point of tangency is x comma y. Okay. So yeah, so I'm just putting my small contribution here. So yeah, so dy by dx at x comma y, we can write it as minus let's let's treat this as let's treat this as f x comma y let's treat this as an implicitly defined function. So it's negative dou f by dou x by dou f by dou y. Okay, so let's differentiate this. So this will become minus m x minus a to the power m minus one upon one by this will also become my by b to the power m minus one into one by b. If you simplify this becomes minus b by x by y to the power m minus one, and you'll also have b by a to the power m minus one. Oh, why not we combine these two because they're of the same type. So we can say it's minus b by a to the power m times x by y to the power m minus one, right? Okay. Now ideally, what should be the equation of the tangent now? Equation of the tangent now should be not mind you, I'm using capital Y and capital X to represent the variables because I have already made use of small x and small y. So I hope you can understand that change in my equation. Okay, remember, I have already taken the point itself as x y so I can't take the variables also as small x and small y because you will get confused, right? They'll all mix up. Okay, so in order to show my point as small x and small y and the variables differently, I have taken my variables as capital X and capital Y. Okay, I hope you don't mind this. Okay, now let's make some small modifications here. Let's bring this b to the power m down.  Okay, let's do one thing. One more thing we can do is let's multiply with y to the power m minus one throughout. So this will become y to the power m bm. This will become minus minus let me open the brackets also at the same time minus x x to the power m minus one by a to the power m plus x to the power m a to the power m.  If I just bring this term to the site and bring this term to this site, I would realize this will be my equation of the line. Equation of the tangent. Everybody's happy till here. If no, then please highlight. Okay, one good thing is now this is actually one. This is actually one. Why? Because the point satisfies this curve. The point satisfies this curve. So I can say the equation of the line now would be this. Now this line is the same as the line given to me in the question because they're saying they both are tangents. So for the purpose of comparison, I will also use capital X and capital Y there so that you know, you're not getting you're not confused between the variable names. So yeah, so this tangent equation that they have given to me in the question, I'm just replacing small x and small y with capital X and capital Y, because that's what name I'm using for the variables. That's what name I'm using for the variables. Okay. Now if these two equations are same, can I say we can compare their coefficients? Correct. So please pay attention here. So this divided by cos alpha. This divided by sine alpha should be one divided by P. Okay, so x to the power m minus one a to the power m cos alpha should be equal to y to the power m minus one by B to the power m sine alpha and it should be equal to one by P. Right? Yeah. Any questions here? Okay. Now all of you please pay attention. This is what I'm going to do. Can I say from this relation x by a to the power m minus one is a a cos alpha by P. Can I say that? Yes. Correct. Similarly, I can say y by B to the power m minus one is B sine alpha by P. Can I say that? Yes. Okay. Now.  Because first I took the m minus one-eth root and then I raised the power of m to both of them. Similarly, this could be written as B sine alpha by P raised to the power m by m minus one. Can I say that? Everybody's happy till here. Now, if you're happy till here, you must be very happy to see the next step also. Since this is equal to one, it means it means this should be equal to one. In other words, whatever is given to me in the then condition comes out from here. Yes or no? That's what you wanted to prove. This is what you wanted to. Okay. Yes, slightly lengthy question and there is very, very high chances that you get lost. You get lost in the process. Okay. But in the school, they'll give you a very lighter version. They'll give you MS2. They'll give you m value as two max to max. Of course, they can't give you m value as one because then there will be a line only done copied. Yes, sir. Let me give others a bit of time to copy. Let me know if you want me to drag this screen anywhere left, right? Aditya Manjulath? Aditya, see? No, sir. After I turn the page, you'll say sir, sir, sir, sir. Can you go back? Okay. Next question. Next question. Next question is a verbal question, verbal and herbal. The minimum distance, the minimum distance between the curve, y is equal to e to the power x and y is equal to ln x is option a one by root two, root two, two root two, three units. Hole is on. Good. I've got three responses. I'll give you one more minute. Okay, Venkat, we'll check. Done. Okay, last 15 seconds. Sir. Yes. Sir, is there any legal way to solve this or is it only graph? Let's check. Let's see. Oh, okay. Let's see the legalities of it. Okay, five, four, three, one. Okay. Most of you have given the answer as option number B. That is root two, 68%. Okay. Let's check. Okay. When do you think the distance between them would be the least? At one comma one, like that. Yeah. Let's take a point which has got the least distance. Can I say at this point, at this point, the line connecting these two points would be normal. The line connecting these two points would be normal to the tangent drawn at both the positions. Correct. Then only this would be the least distance. So this will be the minimum distance. Correct? No. In other words, can I say that tangent slope at these two points must be equal? Right. So what are the tangent slope at any point? Let's say I take a point t comma e to the power t. So what is the slope of the tangent here? And let's say I take a point p comma lnp. So can I say the slope of the tangent here would be e to the power t and the slope of the tangent here would be one by p and both should be equal to each other. Yes or no? So yes. I didn't quite catch what you did. Sorry. What is the derivative of e to the power x at t comma e to the power t e to the power t only? No. Yeah, yeah.  See, when will be the minimum distance happening between them when the line connecting them is perpendicular to both the curves? In other words, it is perpendicular to the tangent at both the curves. Oh, okay. Yes or no? That tangent drawn at these two points would be equal to each other? Because they have the same number. Yes. Okay. Now try to understand this. When do you think this scenario will feature it? 1 comma 00 comma 1. It's a two variable equation. T is 0, p is 1. T is 0, p is 1. How do you know that's the only situation? Yes, that's the only situation. I mean, you look at the graph can find out. Okay. Now, how do I justify that? There cannot be any other value of t and p. So he's claiming t should be zero and p should be one for the situation to arise. Let's say if I go any further to the right, p will increase and t will also increase. Correct? If p will increase and t will increase, remember, this is an increasing function, whereas this is a decreasing function. Correct? So this is going to rise and this is going to fall. Correct? So how can how can you find another point other than t equal to zero and p equal to one? Because if you increase t from zero to any higher value and p from zero one to any higher value, the right side value will drop and the left hand side value will rise. That means any quality will get disturbed. Same will be true if you go to the left side. Right? If you go to the left side, that means if you go lesser than one, that's a p is lesser than one and t is lesser than zero, you realize that the right side will increase and the left side will fall. Right? So the only consensus that will be built up when you are referring to these two points actually. I mean from the from this diagram, actually I've chosen a neutral position, but actually your t position is here. That is t is zero and p is one. So this is the point of minimum distance you can have. This is the point of minimum distance you can have. So the tangent here will be parallel to the tangent drawn over here. Okay? So the distance will be the same as the distance between zero comma one and one comma zero, which has to be root two distance has to be root two option number B is correct. One last question we'll take before we go for a break. Let's take this question. I'll put the poll on for this has this equation has zero gradient at zero comma one also touches x axis at minus one comma zero. Then the values of x for which the curve has a negative gradient. Hello. Yeah, let's check this out. The first thing that I would like to state here is that if the curve has a zero gradient at zero comma one means when you put y is one x should be zero. That gives you one SD. Second thing is the derivative of this function at x equal to zero should also be zero. So that makes that makes now when you put zero, it becomes C is equal to what does he become equal to zero? Yes, sir. Okay. So D is one C is zero. Next thing what they have given the curve touches the x axis at minus one comma zero. What does it mean? Again, two more information is given to me x axis is tangent at that. So the color first passes through minus one comma zero. So can I say a minus B minus C plus one is equal to zero. Correct. C zero. That means a minus B is equal to minus one. Okay, one equation. Second, the tangent at this point should have a slope of zero. Okay, you can put it over here itself. So minus four a plus three B C is equal to zero. That means four is equal to three B. Let's use these two to find a and b values. Let's solve this simultaneously. So B is four a by three. So a minus four a by three is minus one. So B is going to be three. So if B is going to be three, is this correct? The base four and three. Let's check. A is three. No, sir. Yeah. B is four a by three. A minus B is minus one. And yeah. Yeah. So minus a by three is minus one. So a is equal to three by five. Yeah. So A is equal to three. B has to be four. Okay. So finally, your curve becomes three x to the power four plus four x cube plus one. Remember, C was zero. And they're asking you for what interval of x is the gradient negative. That means dy by dx is negative, which means 12 x square plus 12. Sorry, 12 x cube plus 12 x square is negative. That means 12 x square x plus one is negative. Remember, this guy is always greater than equal to zero. So the only possibility is x plus one is negative. That means x is less than minus one. So this becomes your answer, which is option number C. Option number C. Okay. So we'll take a break right now on the other side of the break. We'll talk about sub tangent, length of tangent, sub normal length of normal. Okay. And we'll also talk about angle between the curves, angle between the curves. So right now I'm assuming the time is 639 p.m. We'll meet at 650. Okay, 654 p.m. Alright, so I hope everybody's back. Yeah. So the next concept that we are going to talk off, talk about is concept of the length of tangent, sub tangent, normal and sub normal, length of tangent, sub tangent, normal and sub normal. Okay. So let me show you what do I mean when I say length of a tangent, because we all know tangent is a line, right? So how do you define the length of a line? Because it's infinitely extending geometrical figure, right? So let me draw a curve for you. Let's say we have a curve like this. Let's say there's a point P. There's a point P, whose coordinates are alpha comma beta. So there's a point B whose coordinates are alpha comma beta. At P, let's say we draw a tangent, we draw a tangent like this. Okay, and we draw a normal. We draw a normal at P. Now this tangent will meet the x-axis, let's say at a point T. And this normal, let's say meets the x-axis at a point N. Okay. Now length of the tangent is basically defined as the distance between the point of tangency and the point where the tangent cuts the x-axis. So the length of the tangent here is basically your length Pt. This is how the length of the tangent is defined. So the length of the tangent is going to be Pt. Okay. Similarly, length of the normal is Pn. Length of the normal is Pn. Okay. So normal length is defined as the point of normalcy to the point where the normal cuts the x-axis. Okay. So if somebody says what's the length of the tangent, you must specify Pt length. And what is the length of the normal? You must specify Pn length. Always x-axis. Yes, always x-axis. That's how it is defined. Okay. Now let us say I drop a perpendicular from P on to the x-axis. Okay, let's say I call this point as, let's say m. Okay. This length mt is called as the sub tangent length. So length of sub tangent, sub tangent is mt. Basically, it is the projection. It is the projection of the tangent length on the x-axis. Okay. Similarly, mn is defined as the length of the sub normal. Length of the sub normal. So please pay attention over here because this concept is not there in your school. So again, I'm repeating Pt is the length of the tangent. Pn is the length of the normal. mt is the length of the sub tangent. Many times we write it as st in short form also. And this is the length of the sub normal. Sometimes they will write it as sn in short form. Okay. Now I would like you to prove the following. I think you can easily prove it. So let me give you as a direct question. Prove that prove that Pt length would be given by mod beta under root of 1 plus dx by dy calculated at alpha comma beta whole square. Okay. Prove that Pt length that is the length of the tangent is mod beta. What is beta? Beta is the y coordinate at that point under root of dx by dy calculated at alpha beta whole square. Just write done once you're done with it so that we can discuss it. Whatever I'm going to give you under this, please remember it. Okay. It'll save a lot of time. Done. Very simple. See, you know that this angle tan is going to be dy by dx calculated at alpha comma beta. Okay. And this length is mod beta. So you know Pt is going to be okay. Pt is going to be mod beta by sine alpha or you can say Cosec alpha. Cosec theta. Isn't it correct? Cosec theta is under root of one plus cot square theta. And since cot theta is sorry tan theta is dy by dx. That means cot theta would be reciprocal of that. That means dx by dy calculated at alpha comma beta whole square. Okay. So this will be the length of the tangent. Whatever I'm writing under this remember, please keep this in your mind because you will need it in problem solving. Similarly, Pn the length of the normal are no need to derive it. It's a simple trigonometry that you have to use. So it is mod beta under root one plus dy by dx calculated at alpha comma beta whole square. Okay. Mt length of the subnormal that's given by under root mod beta dx by dy mod calculated at alpha comma beta. Now there's an easy way to remember this. You just assume that for calculating the sub tangent this one disappears all of a sudden. Right. If one disappears all of a sudden how will this formula become? This formula will become this. Okay. In a similar way, your subnormal length is mod beta times dy by dx calculated at alpha comma beta. The way to remember this is the same as what I told you for sub tangent. Just assume that this one has disappeared all of a sudden. So it becomes mod beta under root dy by dx whole square which is actually mod dy by dx at that point. This can easily be derived. So I'm not wasting time deriving these results. I'm interested in solving questions based on it. Put it down everybody. I'll just summarize it. I think let's call M as dy by dx calculated at alpha comma beta that is the slope of the tangent. Okay. Then just to summarize whatever we had done in the previous page length of the tangent length of the tangent is oh sorry mod beta under root one plus one by M square length of normal mod beta under root one plus M square length of sub tangent sometimes also called sub tangent itself length of sub tangent will be mod beta by M in fact mod M we should write and length of sub normal is mod beta mod M. Please note this in your formula list. Can we take a question now? Yes sir. Could you go back? I can take a screenshot. It's the same thing that I've written in the previous page but I've just used M as dy by dx. Thank you. Yeah. The length of the sub tangent of this hyperbola corresponding to the normal having slope unity is one by root k. Find the value of k. So normal has a slope of unity. Okay. Good. Let me put the pole on for the same. Last one minute guys. All right. Five four three two one. Okay. End of pole. Very few people are voting. Most of you have gone for option C. Let's check. See first of all if the slope of normal is unity that means slope of tangent is minus one. Okay. Let's figure out what is that point where the slope is minus one. So if you differentiate it you get 2x minus 8y dy by dx. dy by dx is equal to zero. Correct. Which means 2x minus 8y into minus one is equal to zero. That means let me put the points here in case we get confused. Okay. That means y1 is x1 negative x1 upon 4. Right. If you feed over here in the given equation so x1 square minus 4y1 square that is going to be 4. That means 3 fourth of x1 square is going to be 4. Correct. That means x1 is going to be 16. x1 square is going to be 16 by 3. Okay. Actually I needed y so it should have been a better call had I replaced. It's not a good move. Actually I wasted some time here. So I could have done x1 in terms of y. x1 is minus 4y1. Then it would have been an easier one. Okay. Yeah. So 16y1 square minus 4y1 square is equal to 4. So 12y1 square is equal to 4. That means y1 is equal to plus minus 1 by root 3. Okay. Anyways what are the length of the sub tangent? Subtangent length is known to be mod of beta by m. Correct. Now m is already 1 or minus 1 as you can see and you have beta is plus minus 1 by root 3. So if you take a mod of this it is as good as 1 by root 3 and you're comparing this with 1 by root k 1 by root k which means k has to be 3. Option number c is correct. Simple. No problem. Easy question. Any doubt? Please ask. Please note in the length of tangent, normal, sub-tangent, sub-normal there's no there's no role of the x coordinate. The entire role is of the y coordinate of that point. Of course the role is there in finding the tangent values in finding m values but in the formula beta makes its appearance along with m. Can we take another problem? This may be more than one option. Correct. Okay. Sharmik good. What about this? That's a two minutes. Please. Okay. Okay. Shristi. Okay. From the formulas that you have LST is mod beta by m. LSN is mod beta m. LT is you can say beta under root of 1 plus m square by m. Okay. And LN is beta under root 1 plus m square. Okay. Now let's look into your options. The first option says LST into LSN is 210 square. 2010 square. So if you multiply both of them you'll actually get a beta square. That means you are squaring the y coordinate of that point which is correct. So option A has to be correct. Next. If you do LT by LN. If you do LT by LN LT by LN you'll end up getting yeah what will you end up getting? 1 by mod m right? So LT by LN square is 1 by m square. 1 by m square you'll also get when you do LST by LSN. Just start dividing it so that beta gets cancelled. LST by LSN. In other words LT by LN is under root of LST by LSN. In other words LT by LN under root of LSN by LST is equal to 1 which is actually a constant. So mod of it will also be 1. That's actually a constant. Okay. That means option number B is also correct. Now LST into LSN is 2010 square and I don't think so 1 minus 2010 square is going to be a fraction here. So C is definitely wrong. Now what can we write over here? What can we comment over here? Option D. We have figured out that LT by LN is equal to under root of LST by LSN. Correct. Okay. If you do component or dividend over here. If you do component or dividend over here it becomes LT plus LN by LT minus LN. Same thing here will be what? LT plus LN means under root LST plus under root LSN upon LT minus LN means LST minus sorry LST minus under root LSN. If you square this I don't think so this is going to give you LST by LSN. I really doubt that will happen. Correct? No. So this cannot be option number. D cannot be your right option. So D is also wrong. So only A and B can be right options. Is that fine? Any questions here? Okay. Last I'm going to talk about angle between the two curves. Angle between two curves. So first of all what's the definition of angle between two curves? Angle between two curves is basically defined as the acute angle between the tangents drawn to the two curves at their points of intersection. Okay. So let me show you by a diagram. Let's say I have one curve like this okay and I have another curve like this. How do I define what is the angle between the two curves? So let's say alpha beta is the point of intersection. So what do we do is we draw a tangent to the yellow curve at that point of intersection. We draw a tangent to the white curve at the point of intersection. Okay. And the acute angle between them is defined as the angle of intersection between the two curves. Okay. So basically you may be given the two curves as fxy comma zero. I'm assuming an implicitly defined function is given to you in the worst case. Okay. So what do we do? We first find out your slope of these tangents m1 and m2 by finding the dy by dx for these two curves at that point of intersection and then we use the formula that we had learned in our straight lines. Tan theta is m1 minus m2 upon 1 plus m1 m2. Okay. Where m1 and m2 are respectively the slopes of the tangents drawn to f and g functions. Okay. So m1 is dy by dx at alpha comma beta for the curve fxy comma zero. And m2 is the dy by dx expressions at alpha comma beta for the curve gxy equal to zero. I don't want to write f dash x and g dash x as m1 and m2 because you may have an implicitly defined curve. So there you have to find dy by dx. Now from here two cases arise case one. If your m1 and m2 are equal then we say the curves touch each other at alpha comma beta. Okay. So if you realize that both the slopes have become the same at that point means there's a touching happening between the two curves and if m1 and m2 product is minus one then we say that the curves intersect each other at 90 degree. The curves intersect each other each other at 90 degrees. Now when the curves are intersecting each other at 90 degrees we use the word orthogonal. So you can say f and g are orthogonal to each other. We don't use the word perpendicular. Perpendicular is used when there are two lines. Okay. When there is a line in a curve we use the word normal. But when two curves are cutting each other at 90 degree we say the curves are orthogonal to each other. So this is the condition that we call is condition of orthogonality. Condition for orthogonality. Similar to circles. Circles. Correct. Remember 2f1f2 plus 2g1g2 equal to c1 plus c2 was the condition for orthogonality in a circle. By the way. Yeah. Yeah. By the way conic section is going to be tested in the next week's s which is on 23rd, 24th whatever. Why sir why conics. No that is not called touching when cut. That is called cutting. Okay. That's why we use the word differently touching and cutting. Okay sir. However mathematically touching and cutting mean the same thing but when you are talking about English words we use the word touching when it is not going past it. But when you're solving it the point of touching will also come as a point of intersection. Yes. Okay. Okay sir. Let's take questions on this. One minute sir. One minute. Yeah. Can we go to the next question? Can you go to a question on this? Yes sir. Okay. Very good. Okay. Last 30 seconds for this. This is not a difficult question. One minute. Yeah. 5, 4, 3, 2, 1. Please vote. Please vote. Okay. First of all let me show you the poll result. Most of you have voted for option number B. Okay. So let's discuss. Where do you think these two curves will intersect? Where do you think these two curves will intersect? Okay. I know solving this is not an easy task because it's having some exponential kind of an expression under it. But can you take a guess for a point which satisfies it? Huh? X equal to 1. X equal to 1 is a very very sensible guess actually because it'll make both the sides become zero. So can I say they intersect let's say I take a point 1 comma 0 the point of intersection. Now let's find dy by dx for first curve at 1 comma 0. Okay. So for this curve let's use product rule. This will be 3 to the power x minus 1 ln 3 into ln x. And when you put x equal to 1 it becomes 1 again. Okay. Let's find m2. m2 is the derivative of the second curve. I should better write f and g in order to make distinction. Let's say the first curve and this is the second curve. Yeah. So x to the power x derivative is x to the power x 1 plus ln x. I hope you remember the result. And when you put x equal to 1 it becomes 1 again. What does it mean? It means they are parallel. I mean the slopes are same. That means the curves are touching each other. If they're touching each other the angle between them is going to be 0. Okay. That means your angle is going to be 0. So option a is going to be correct. So they ask cosine. Oh, okay. Sorry. Cosine. Cosine is going to be 1. So option b is going to be correct. Let's check. Yes, b got the maximum percentage of vote. Janta is absolutely correct. Next question. Which of the following pairs are orthogonal? More than one option may be correct. I think in this case you can assume that between any two pairs that is given in the options intersection is happening. Okay. Because first question can arise whether are they intersecting at all? Okay. So I'm assuming that intersection is happening in this case. So let's assume everybody that intersection is happening in these cases. Okay. Let's start inquiring one by one. Let's do one minute. Okay. Never mind. Okay. Let's look at option number one. If you find out the derivative over here at any point x comma y, let's say they meet both at x comma y themselves, rather than taking x1, y1, let's take xy itself as the point of intersection. So the derivative here will be if I'm not wrong is minus 32x by 2y. That's minus 16x by y. For the other curve, it would be minus k by 16y to the power. Sir, shouldn't it be only k? Minus k. Yeah. This because it comes on the other side will become only k. Yeah. That's why I was wondering. Now, when you multiply these two, what do you get? You'll end up getting minus kx by y to the power 16. And isn't y to the power 16 minus kx? So it would become a minus one. That means, yes, the product is product of the slope is minus one. So one is definitely a pair. But again, I'm assuming here that the intersection is happening. Okay. Let's say I talk about b option. Yeah. For b option, what is the slope? It's one minus c e to the power minus x. Okay. And for the other line, what's the slope? Let's differentiate it. So one is equal to dy by dx minus k e to the power y dy by dx. So dy by dx here would be or you can say m2 will be equal to m2 would be equal to one upon one minus k e to the power minus y. Yes or no?  Because it will satisfy the first curve, isn't it? Yes or no? Here I'm going to write k e to the power y as one minus that will be x minus y plus two. So this side gives me one plus x minus y. This gives me one by one minus one minus x plus y. Check it out. So if you multiply both of them, your product will be equal to minus one. And therefore, yes, option B also is your pair of orthogonal curves. Let's look into option C. For option C, m1 is going to be 2cx. Okay. And m2 is going to be minus x by 2y. Correct me if I'm wrong. So their product is going to be 2, 2 will get cancelled. So it'll be minus x square by y, isn't it? Minus cx square by y. So since cx square is y, I can say it's minus y by y, which is actually minus one. So yes, option C is also correct. Right? Let's look into D. Yeah. So for D, what is m1? Can I say m1 for D will be x by y. What will be m2? m2 will be minus k by x square. So their product will be minus k by xy. And since xy is k, I can say it will also give me minus one. That means all these options are correct. All these options are correct. Does it find any question here? Any concerns? Please do ask. Let me know if you want me to drag the screen anywhere. Sir, can I solve this without using graph only? Like for example, the first one, we know that it'll be a vertical ellipse, right? Also it'll be a vertical ellipse. And second one will be probably some kind of line, like either a set of two parallel lines. So could we probably, without doing all these, in case we knew the basic skeleton of the graphs, could we somehow, is there a method to find out this question graphically? Graphically, how will you figure out that they are intersecting at 90 degrees? Like I can justify for the first one. I'm not sure about the last one, but first one I'll give, sir. Like now I assume, let's say for one particular case, I assume both C and K are one, just for explanation sake. And now this will be a standard form of ellipse, right? The best case scenario. Now there'll be a vertical ellipse also, correct? This will be a vertical ellipse, okay? Let it be somewhere. That doesn't matter because the second one is going to be a pair of infinite lines. So just for explanation sake, I'm going to make that at the center. And even let's say second is not an infinite line. Second one, let me tell you it'll look like parallel lines. No, no, no, no, no. It will be a flat parabola like this. The same way as you get when you put y is equal to x to the power 16, you have to reflect it about y equal to x time. So this is a flat parabola like this, right? Oh, right, right, right. It's a flat parabola like this. It's a, it's not a parabola, I would call it, but it's a structure like this. So yeah, so now how do you, how do you, even if you make the graph, if you make a graph, how would you know that these angles are 90 degrees? Oh yeah, you can't. Your human eye is not capable enough to detect that. Yes, sir. Right, we are not terminators. And 90 degree fire. Okay. Next question. Find the angle between the intersection of the curves. Find the angle between the intersection of the curves. GIF of mod sine x plus mod cos x and a circle. This represents the greatest India function. Find the angle of intersection of these two curves. Okay. Good. Anybody else? Sure. Okay. Nice. Next class would be a very important class because we are going to take up two very heavy concepts under application of derivatives. One is your mean value theorems. Yeah, mean value theorems are actually heavy. Okay. And other would be the concept of increasing, decreasing function and I think maximum minimum we can take in the other class, the class coming after that. Okay. Very good. So three people have answered so far. Trippan, Pratham and Venkat. Okay. Sharmika didn't get your answer. Actually, there's only one answer. Okay, sir. We'll see how there's only one answer. We'll see from the situation. The one answer hasn't the acute angles become the same or something? Yes. Okay. See, let's see this. First thing that I would like you to understand here is that you are finding GIF of some function. So it's better to know in which interval does this function like because GIF is very much sensitive to in which interval are you looking at or what is the range of the function on which you are applying GIF.  No matter you take both positive or both negative, this will always be less than root two. Do you all agree with that? Now, I don't know how many of you are aware of the fact that this or you can say power mean inequality. This is always greater than A plus B by two the whole square. Are you aware of that? That is mean of the squares is greater than square of the mean. Okay. What does it mean? It means that if I take A as this guy, B as this guy, so sine square X plus cos square X by two. Okay. This term would always be greater than. Oh, I think it's not going to give me any separate result actually. Yeah, this is going to give me this. Okay. Anyway, this is not going to help me out. Okay. Y equal to one with that GIF of that. That's going to give me the same term. So it's not going to help me out. Now, I want to see till what, what are the lowest value it can take? Can it, can the, can I say the lowest value that it can take would actually be a one? Can I say one? Yes, sir. Yes, sir. Okay. How can we say that? Sine and cos are not straight graphs. So they're curvy. So the moment you exceeded one's going to heavily compensate for one small reduction. Oh, no, I would try to go to buy a very mathematical way of looking at it. Oh, okay. Sure. Yeah, AMGM. Yes, sir. Okay. But here also, am I getting some kind of response check? Let's say I put a two here and I put a one by root two here. No, I think this is also not helping me out. This term will always be greater than I cannot say because it can be zero also. I need to have a transitive kind of relation. So how else we can prove that its minimum value is one. So we can redefine it and then use like we can multiply by root two, divide by root two and then we'll get one sign term. We have to redefine it and then basically apply maximum or something. If you redefine it, every time you need to get a sign plus cos or a minus sign plus cos like that, right? Yes, sir. How do you end up getting its minimum value? Yes, sir. I thought if we look at only the first quadrant, there are, okay, let's check this method. Okay, so let's say square of this, okay, this will be always be greater than sine square x plus cos square x, right? Because the other term would be two mod sine x plus cos x, which will always be greater than equal to zero, right? This term will always be greater than equal to zero, right? So if I just drop this term, can I say this will always be greater than equal to one? So that way we can justify that this quantity will always be between one and root two, okay? So once we justify that, the GIF of that, the GIF of that, would always be one, because any quantity between one and 1.414 will always be one. Indirectly, you are basically dealing with the line y equal to one. So if you draw a representation of these two graphs, one is a circle, one is a circle and other is a line y equal to one. So let's say this is a line y equal to one and other is a circle like this. Now they're asking you what is the angle of intersection between the tangent drawn at this point and this line. So I think you realize from the symmetric figures over here that these two angles would be the same, okay? So there'll be only one possible answer. Let's figure it out. Now this point would be having y coordinate as one. So what will be the x coordinate? What will be the x coordinate? x coordinate will be two, correct? So what is dy by dx here? dy by dx here would be minus x by y. That's nothing but minus half, correct me if I'm wrong? No, no, no, sir. Correct? Everything is fine? Yes, sir. Okay. So the slope of this line is minus half. The slope of this curve is zero, right? So can I say the angle that that will make will be just, see, this actually gives you this slope. This actually gives you this slope, isn't it? Or you can, you even want the minus two it'll be, yeah, not minus half, minus two, yeah. So if you want the acute angle, it is basically what they have given you is this and you want tan of theta. Okay, that's pretty easy from here itself to get because this is nothing but negative tan theta is minus two. So tan theta is two, correct? So your theta would be tan inverse of two. Theta would be tan inverse of two. Is that correct? Have I missed out anything? Please let me know. So Vibhav was the only person who got this correct. Yes, everybody was wrong. Anyways, so with this, we come to an end of tangents and normals application. Next class is very important because very heavy concepts will be dealt with in the next class. So please do not miss that class. Till then, take care. Bye bye. Stay safe. Thank you, sir.


Thanks, Jared, and thanks for inviting me. So hopefully, oh, let's see, forward back, okay. So I'm going to talk today about the National Hydrologic Molecule. Can you guys all hear me? Okay. Yeah. Okay, good. Good. I've been working on this for a long time. It's been myself as well as a lot of other people that I've acknowledged here, and I will try to call their names out as I go through the presentation. So why did we develop a National Hydrologic Model? Probably 10 or so years ago, Steve Markstrom and I got a proposal funded to look at integrated watershed scale response to climate change for basins across the United States. So we said, hey, this would be really cool. Let's pull together all the modelers in the USGS that have developed a PRMS model. Let's bring them all together, and let's downscale some GCMs to those models. And maybe we can provide the foundation for hydrologically based climate change studies across the nation. But anybody who's at the USGS, the group that I used to run, the model and watershed systems group and the research program, which actually is no longer there, a lot of our work was developing models for people to use on the ground. So every one of these people probably worked with us to develop their model, but they developed it for a different reason. They developed it on a different scale. They parameterized it differently. They calibrated it differently. So when we really actually tried to compare things, we realized we couldn't because they weren't defining things the same. They weren't selecting the same process modules because you can select all different types of combinations of process modules and they all use different calibration strategies. So we decided what we really needed was a nationally consistent, stakeholder relevant, locally informed model for the entire country that people could pull from and use. So that's where this all came from about 10 years ago. So we call this the National Hydrologic Model, and this is the infrastructure of the National Hydrologic Model. I probably can't walk. Oh, I can hold this. So this consists of what we call a geospatial fabric, which are the hydrologic responsiveness, the modeling units. It contains the stream network and it contains the parameter values. So what we did is we took the NHDPlus version 1, because this was quite a long time ago, and we aggregated it based on points of interest, which were gauges, forecasting points, all kinds of things. We decided we're important lakes, confluences, and we took the three million catchments from the NHDPlus and we aggregated them to what we thought were good size modeling units for the country. So about 110,000 HRUs, and these are really defined based kind of on drainage density when you look at them. So these are the modeling units that we start off with, that we model the country with, and then we parameterize it with static spatial information, and we used our best knowledge to develop algorithms to parameterize the model to try to get decent distributed parameters.  And so really, it's probably, you can start off with static parameters, but in a lot of cases, there's change and we need to represent that. Here's just an example of some of the dynamic line cover we use from Eros that I'll touch on later here. We use a lot of dynamic line cover into the future as well, and we combine that with climate change. So the next piece we have is the model input data. So what we are doing is we're testing this national scale model with all the different gridded data sets that we can find. You can probably pull a local model and do a better job with your climate inputs if you actually spend some time with it, but nobody seems to, everybody wants everything done for them these days. And so we're testing, well, that's just my experience. So we're testing all these different data sets in the model so that we can tell you what we think works where and why. And then the last part of this, the physical models, we have daily and monthly models running within this infrastructure. We have the monthly water balance model. We have the precipitation runoff modeling system. And within the precipitation runoff modeling system, we also have a stream temperature model running for the country. So the monthly water balance model is a very simple five parameter model that we developed a parameter regionalization scheme for the country a couple of years back. And we've also developed uncertainty estimates associated with the monthly runoff, which I'll touch on how we're using those later. The water balance model has been run for current and future conditions. And you can access a portal that accesses the monthly water balance database and you can get current and future conditions for 235 climate scenarios. So you can basically poke on a location that delineates the area. You can get local or basin derived outputs for the monthly water balance model outputs. That's the website. This is no longer supported. It's still running, but it's not being supported because it has the word climate and model in it. So maybe that'll change. Anyway, so 235 climate scenarios process for the water balance model. One of the first, one of the things I get asked all the time is, what's climate scenario should I be using? And it kind of really depends on what you're doing. But what we did for this, because putting out 235 of them seemed a little bit crazy to me, was I looked at a two-sample Kalmogorov-Svernov statistical test. And so all the downscaled GCMs that we're using, they're all trained with the same data set. So I ran for current conditions, that trained data set, and then I ran the GCMs for current conditions. And I just wanted to see if they could match the cumulative probability here. So that's about as simple of a test as you can get, because you're not going to match day to day or month to month. But you would hope your distribution was the same. And this is monthly as well. And so here's just some results that we got. If we split up those scenarios into the CMIP3 and the CMIP5, these are the GCMs, zero of them if it's pink, actually represent current conditions, and 100% of them represent current conditions if they're blue. And this is ones that represent precipitation, temperature, and runoff. And you can see it's actually kind of interesting. You can see the resolution of the, oops, that wasn't the right one, here we go. You can see kind of the resolution of the GCMs showing through in here. But you do see a big difference between CMIP3 and CMIP5, which is pretty good. Yay, they're getting better. And really, this is driven by the precip. They're not able to get the precip. And California still has problems. None of the GCMs can represent current conditions there. So you might want to think twice about what you're using, but you can actually subset with the common ground of smear test in the portal to just get the GCMs that represent your conditions. So PRMS, precipitation runoff myelin system. This is a, excuse me, this is a daily time step, deterministic distributed model. And this has been around for a very, very long time. And there's been a lot of changes to it over the course of its life. It's really simply, just if you consider a conceptual, there's a basin. And there's a, you consider a surface subservice and subsurface part of surface, subsurface and groundwater portion of the basin. You divided into hydrologic response units. You've got infiltration to the subsurface. And then you've got recharge to the groundwater. That's a very simplistic representation of this. The combination of the three has given you stream flow. And we do have it running for the country. So I always show this to the people that you believe we have it running for the country. But this doesn't mean it's right. People love showing visualizations. Put this up here, but doesn't mean it's correct. Just looks good. So I thought it looked good. We have a tool that we call Bandit. And this takes, this actually accesses the national hydrologic model infrastructure and subsets models for you. So if you can pick any location of the country, it'll give you a local model. It'll give you the inputs, the parameters, the executable and the control file. So you can be running a model within minutes after getting the outputs. And that's a Parker Norton who developed Bandit with very proudly displaying his Bandit t-shirt. And so what you do then is that you say, okay, I want this location. You can pull the model and say, okay, this is great. But I don't really like this resolution in this area right here. There's been a lot of change going on. And so what usually happens is people want to nest, find a resolution, HREs into the area that's important. So instead of running hyper resolution across the whole country, you run it at a resolution that's handleable, that we can parameterize, and then we actually nest the fine resolution where we need the information. So that's where it's locally informed usually. So one of the things that as an example of calibrating the model and pulling a model, we pulled the model for the prairie pothole area in North Dakota. And we looked at the surface water depression storage. PRMS uses a merge, fill, and spill concept. So within a response unit, all the surface depression is considered as a, is this kind of a representation here, which really bothers the ecologist. But you have to start, if you're parameterizing the country, you can't consider every depression storage by itself, right? So it was kind of cool because I pulled the prairie pothole area model, and I calibrated the model to stream flow. And when I looked at the surface depression, the HRU fractional storage, this is the purple line here, that line just went straight across. My stream flow looked fine, but my storage had nothing to do with really what was going on. So if I calibrated my HRU fractional storage to normalize lake elevations that were nearby, I was actually able to get good stream flow, and I was actually able to follow the normalized lake elevations. And then after the fact, I took the remotely sensed surface water depression area for that HRU and overlaid it, these green triangles. But that's pretty darn cool. So you can get the right answer for the wrong reason. But if you start bringing these alternative datasets, there's a lot we can do with this, and we can get these surface water depression areas dynamically from remotely sensed information to calibrate our models. So that brings me to continental domain parameter estimation, and what are we doing for it? Well, I have always thought about calibration as first looking at volume and then looking at timing. And that's just like for a small basin that you might be calibrating. But that wasn't going to actually work for the cone essence. So this is my scheme. I know it's very colorful. Don't worry. It's actually pretty straightforward. Other people hate it when I say that. But first, I think about volume. So I used a fast parameter sensitivity analysis that actually looks at dominant process and looks at parameter sensitivity for every respondent across the country and tells me which parameters we should be calibrating to begin with. And then I calibrate each respondent as its own models. I have 109,951 models. And I calibrate them to what I call calibration or baseline data sets that I've derived. So I have information for runoff, snow covered area, AET, soil motion and recharge. And these can all be improved upon. And that calibration order is determined by a fast analysis as well. Because I want to calibrate the dominant process in that HRU first. And so I ran the fast sensitivity analysis. And now I know which one of those is the dominant process. And I order my calibration based on that. And this just actually shows you which one of these is dominant. So it does make sense that you've got snow covered area being important here in the prairie pothole region. We have recharge being important. And the rest of the place runoff seems to dominate. So anyway, what I'm looking for is the sweet spot. In these parameter ranges, I'm looking for the sweet spot. And you'll notice I'm not using observed stream flow. I'm looking for the sweet spot. I'm trying to fit all these things at the same time. So here's my baseline calibration data sets. And it would be great if we actually improved upon these. I have thrown these together. I have a lot of ideas of how I would make these better. But basically either I have something where I have uncertainty estimates for a range or I bring in three different model outputs or different remote sensing data sets. And I use their range as my uncertainty bound when I calibrate. So I fit it to a range. I don't fit it to a value. Some of those ranges are pretty darn big. Here's the mean of those five data sets. Here's the mean of that range. And that might not be all that exciting to look at. But here's the mean divided by their range. So one means you're meeting your range or equal. This means that your range is a lot bigger than your mean, which means your uncertainty is huge, right? And it's kind of interesting when you think about that later when you're evaluating your model. Let's take a look at AET here, for example. AET mean divided by their range. We're seeing this big area here where we've got a lot of uncertainty. And that's because we're using Mod 16, CBAP, and the Monte Warbounds model AET. And they all agree here, like if you correlate them to each other, right? Like, you know, they're correlating great. This is monthly. They're actually showing an opposite sign when they're below zero, right? In these areas here. And so it's gonna show that you're fitting this probably great, because your range is huge. So you need to keep that in mind, but that's the information that we have. And we need to improve upon what these baselines are telling us. So here's how the calibration works, or how it kind of, here's just an example of what it does. If you look at maybe one of the parameters, soil recharge max frac, which is a soil zone storage parameter. And let's look at the six calibration steps. This is a randomly chosen HREU. And this is the order of the six calibration steps based on the fast analysis for that HREU.  So the first step, it calibrated runoff. And so these are the objective function values. These are the parameter values. This was the range it started with. And this is the top 25% of those values. Those are the best objective function values. It takes those and it moves to the next step, calibrates to AET. Takes that range again with the red dots, moves to the next step. Soil moisture. Okay, yay. Look, soil zone storage parameter, soil moisture nails it. So it's actually pretty cool because there's not a lot of parameter sensitivity here. There's a little bit, so it's narrowing the range a little bit, but generally when you hit the variable that you want, it narrows it down. So, and you can look at all the different parameters and this is generally what happens with them. So that's how the calibration works. And this is just one of them pulled out, but they are all calibrated together. So, here's what the calibrated parameter values. There's an example of four maps of parameter values.  If it's not a real sensitive parameter, or if the baseline dataset you're calibrating to doesn't have a lot of information, then you're not gonna get a good pattern. So, how many simulations fall within the baseline range? Here are the five baselines and this is the percent of the simulations that fall within the range. And again, of course, this makes a difference as to how big your range was. So, here's the spatial distribution. Pink is 100% of them fall within it and black is none of them fall within it. And again, you have to actually think a little bit about what this is telling you because it's telling you, oh wow, look, we're doing great here with AET, but remember that's the area where the range is gonna be huge because the data was telling you two different things. So, let's look a little closer at AET. There's the baseline mean divided by the range, remember? So, anything less than one, the orange, red, and pinks, that area is going to have a very big range, right? Here's the objective function values, the final objective function values. So, good is lower values, bad are the upper to red to pink values, right? And so, your objective function values are actually looking better in the middle of the country here. But really what you wanna do is you wanna look at a combination of how big is the range and what is your objective function value, right? So, if you then take your mean and your range and you categorize that on the y-axis here and you take your normalized root between the error and you categorize it into these colors and then you plot them, now it's really showing you that you're really actually doing much better in the Southeast. You've got a really tight range and you've got a low objective function value. And in the area that had the really wide ranges, that's now showing you that maybe it really isn't as good because really what you wanna do is be up here. So, you really have to look at more than one thing when you're evaluating this. Another way to look at that is with runoff. Here's another way people like to look at Nash-Succliff. And this is the Nash-Succliff using the baseline dataset from the monthly water balance model, which is the runoff values. Bad is pink. That means you're just doing less than zeros. One is good. And this is actually the Nash-Succliff compared to the value that the water balance model's putting up. But remember, that's not what I calibrated it to. I calibrated it to the range that came out of that with uncertainty, right? So if I actually only calculate error when I don't follow another range, I'm using as much information as I can out of that runoff of that. Now, you mean I would actually think somebody was lying if they showed me this because you don't get ones for Nash-Succliff. But the range is so wide that you can fit it. So we have to narrow the ranges of our runoff when we're doing this. So that's the volume calibration. And this is relatively model agnostic. I set it up for some GS Flow models a month or so ago. The 100,000 and 60,000 grid cells and used the same concept to distribute the parameters for a GS Flow. And it's a nice way. A lot of times people don't, they say they have a distributed model but their parameters actually aren't distributed. And so this is actually a good way to get distributed values for your parameters as a starting point. GS Flow is an integration of Modful and Pyramus. In case anybody was wondering. Brownwater Surface Water Interaction Model. So let's see, next step. So I take the optimized parameters by HRU and I say, okay, that was my volume. Now I wanna look at timing.  And what I do is I pull all the headwater basins that are less than 3,000 square kilometers in the country with Bandit. So that's 7,265 headwaters. And now I have 7,265 models. And then I have daily time series that have developed, that Will Farmer developed using pooled ordinary creaging at every one of those headwater basins. So I have daily streamflow timing with uncertainty associated with it. And then I calibrate the model now to timing and I can look at all the different routing options that we have. I can look at no routing. I can look at Muskegon. I can look at Misera, which is kinematic. And then we can just test a lot of actually really cool things when we have this set up like this. And then from there in the end, I say, okay, well, let's actually look at observed streamflow. And so now I bring in observed streamflow in the end. So here's the headwater basins. And then the headwater basins that have gauges are 1,417 of them. Those are the red areas that are left now. Now those areas I can fine tune with observed if somebody wants to pull a model at that location. So how does this work? Well, it depends where you go. Actually, this was a great example because it worked really well here. This is where we developed. This is the Gulf Coastal Plains in Ozarks, Landscape Conservation Cooperative. We pulled this model out of the national model and we calibrated it with the BHRU by headwater observed. And so there's our national subcliff values for streamflow, daily streamflow, and then the log of daily streamflow. And actually this was a cool application because we developed the model for this area. And then we ran historical and future land cover and climate through these models for that LCC. And so we ran about 15 different GCMs as four different RCPs. And we also looked at a bunch of statistics because these guys wanted statistics for future conditions. And so one of the things that Jacob LaFontaine did was he used that KS test and he categorized these statistics they were asking for into the duration of frequency, magnitude, rate of change, and timing that they're sort of associated with. And those little Xs, these little Xs actually show you whether for current conditions we were able to replicate that statistic. And so all the daily statistics that everybody wants in the future, we're not even able to come even close to replicating them for current conditions. So you really have to actually think about what it is that you're looking at in future conditions before you just start running all this stuff. So they have a landscape conservation planning atlas that they put all this into and you can look at all the statistics, you can subset them by the KS test. So this is actually really nice and they've actually asked us to actually extend this to my official wildest service wants it extended to region four. So we're producing region four for them right now with current and future conditions with a land cover and climate. So that's a good example of that. And here's another example with surface depression storage that we worked on in that area. When you put future climate and future land cover together you think, hey, well, this is great. We're doing the right thing here, but the land cover change that we get does not have anything that represents the change in surface storage. And if you look at land cover change, there's huge changes in impervious, which means they have to build surface depression to capture the surface runoff. And so people do this land cover change with climate change and they don't consider the change in surface storage that's being built to catch that. And so what we did here was actually pretty cool because we ran the model and then we backtracked out how much depression storage would you have to build in order to capture the surface runoff that's occurring because of the change in impervious. So a lot of really cool things you can do with this and you really gotta think about that. You can't just apply this stuff.  Here's another application using Bandit, the park service last year wanted the model pulled for every single national park and they wanted current and future conditions. And we said, sure, we can do that. So Bill Betaglen and Colin Penn in the Colorado Water Science Center developed the tools to take the park boundaries which are weird, right? There's park boundary here and it's here here and here there's little yellow areas. And we give them maps just to show and how does the park boundary look with respect to what our modeling units look like? And this is just one example, one park example. And then we give them all the components of the water balance for current conditions for every single park. And then we also give them a lot of fun graphics. This one is just for Yellowstone and I pulled out the minimum temperature because it's kind of cool to see even back to 1980 you can still see the increase in minimum temperature here. And then we also give them statistics and as I said, statistics always make me a little nervous for future conditions. And then we give them stream flow at gauges. The park service didn't even know which stream gauges were in their parks or near their parks. And we have all that information in our infrastructure. So we pulled model and we give them pages of plots like this for gauges in or near their basins with observers to simulate it. And actually I would and we give them all the information whether it's a reference gauge, it's not a reference gauge with the disturbance index, what's the comparison of drainage area between what we think and what endless things. And some of these match pretty good and some of them don't. This is actually uncalibrated version of the model. So we're redoing this with a longer time series when we'll give them a better output. And then we also give them future conditions from the Mountain Water Balance Futures portal. Greg McCabe and I took all the outputs from the Mountain Water Balance Model and we just developed the P5 through P95 percentile changes for seasonal temperature precipitation and runoff on the geospatial fabric. So when we pull the park service models, we can actually give them for every single national park. We can give them those percentile changes for precipitation seasonally. We can give them the changes for average temperature seasonally and then we can give them the changes in our runoff seasonally. And then we also developed the same thing with trends and the percentiles of the change so you can see the range of the trends and then we also given them the trends for every single park. And so we can pump out a lot of information very quickly with this type of infrastructure. So there's the trends in precipitation seasonally and then here's our trends in runoff. So last but not least is the Stream Network Temperature Model. This was a model SN TEMP developed a while back. It was really the way it was set up was perfect because we can associate temperatures with surface subsurface and groundwater and then we figure out these channel characteristics and then it's easily run for the country. A lot of this probably needs to be calibrated but we haven't worked on that lately. But it is running and there, yeah, well, you can pull a model and locally calibrate it. One of the things that we're doing with this is we're working with the Forest Service and BLM with this A-Remp project which has been monitoring 215 watersheds within the Northwest Forest Plants Domain. So these are all the watersheds they're monitoring. They've been monitoring them for 30 years. They never monitored stream flow so they don't know what the stream flow is. So we're giving them stream flow. They also have super cool information on land cover change at a really fine resolution and we're now incorporating that change into our model with forest and structure so that we can look at that because what they have seen in this area or what they think they're seeing in this area is a big change in low flows and what does the low flow change due to? Could it be that they killed all the beaver? Could it be that climate is changing? There's a lot of things that we can test here and one of the big things they want is stream temperature and so in the following year we're gonna actually incorporate stream temperature at all these as well. So anyway, that's my talk. We have a lot of stuff that we would love to share. We would love to collaborate with anybody who's interested in baseline information, process representation, model evaluation, model calibration. It's all there and we're willing to share it with anybody who's interested. Thank you. Terrific, thanks so much Lauren. So we have a break next. We'll take a couple of questions before the break. Who has questions for Lauren? Hi, she asked if there is information out there for using the monthly water bounce model as a teaching tool and I would say yes. I believe that the Bureau of Rec has it running in R and they use it as a teaching tool. I know that Greg McCabe and Steve Marksham put out a paper, gosh, five or 10 years ago that have it as a GUI with a GUI interface that you can use as a teaching tool as well. I think that's actually a lot of versions of it out there that are being used as a teaching tool. If you wanna send me a note I can probably give you the different references for it.