 Application of derivatives right so much awaited chapter for us so we'll start with application of derivatives. Application of derivatives. Okay. In this particular concept you are going to see various applications of derivatives so whatever concept of differentiation you have learned. Mostly those same concepts will now be coming in an application format. So what are the applications that we are going to study. Let me give a coverage of the chapter. And I will also tell you how much time will it take for us to complete the entire concept. So, the first, let me write coverage. The first chapter that we are going to study under application of derivatives is d y by dx as rate measure, d y by dx as a rate measure. Okay. This chapter was actually skipped in last year's syllabus of CVSE. Maybe I'm not very sure whether your teacher has covered this concept with you in the in the school or she has kept it as well. But we are not going to skip it. They have done it or they have skipped it. Okay, great. Because I think last year they skipped it because they give a reason that it was already covered in physics. So there was no need to cover it again in mathematics. Okay. Next concept that we are going to talk about is errors in approximations errors. Approximations. And we also include differentials and differentials. Differentials. What are differentials? dx. dx is a differential of x. Okay. So how errors approximations and differentials could be, you know, figured out by the application of derivatives is what we are going to learn under this particular sub topic. Third sub topic that we are going to talk about is tangents and normals. Tangents and normals. Okay. Basically this comes from your geometrical interpretation of d y by dx. So that is what we are going to cover under tangents and normals. The next concept that we are going to talk about is mean value theorems. Mean value theorems. Mean value theorems would have a several sub concepts under it. So we are going to talk about three types of mean value theorems. The first one being called the roles theorem. The second one is your lang ranges mean value theorem. Lang ranges mean value theorem. Okay. And the third one is less than one to most of the students who she mean value theorem. Who she mean value theorem. Okay, it is pronounced as cushy, not couchy. Okay, so we are going to talk about RT, LMVT and CMVT under mean value theorems. Oh, did I. Okay. Okay, so we will cover up the concept of these three mean value theorems under the fourth sub chapter. The fifth sub chapter that we are going to talk about is your concept of monotonicity. So monotonicity is a part of calculus where we learned about the increasing and decreasing nature of functions. Okay, so how does calculus help us to know in which interval a function is increasing which interval the function is decreasing. So we'll be covering that concept and then monotonicity. And the last but not the least, we are going to talk about the concept of maxima and minima, maxima and minima. Okay, along with the application maxima minima minima minima minima minima. Okay, and it's practical application. Okay, and it's application. So we'll be trying to solve some questions based on optimization using maxima and minima and also be also giving you some extra applications like trying to know the nature of the roots of cubic equation. And also how calculus helps us in sketching the graphs. Okay, so a lot of concepts related to sketching are also dependent upon calculus or differentials or you can say derivatives so we'll talk about that as well in the application part. So overall this chapter has got six subchapters under it. Normally I take four classes to finish it. So you can expect at least 14 hours, 14 hours, 15 hours of lecture to happen on this entire chapter. Okay, so let's get started with our very first chapter, dy by dx as rate measure now theory wise, there is nothing special in this chapter. So based on your basic understanding of what is the process of differentiation, right, some is based on chain rules, some of them is based on first principles, some of them is based on geometrical interpretations, like some of them is based on continuity and differentiability, some of them is based on the nature of the slope at different points of the function, some is based on stationary points. So all these concepts that you are basically studying theory part is not very much in it. Of course, some of these concepts are very heavy in theory like fourth concept, mean value theorems, it is slightly theory heavy. But rest all the concepts are like, you know, you already know it, we just have to learn it as an application. So let's get started. Without much waste of time, we have four classes to cover up this. So today I'll start with dy by dx as rate measure as rate measure as rate measure. Now what are the meaning of rate of change. What are the meaning of rate of change, I would like to know from you people you have done it in school, what are the many of rate of change, change with respect to time, isn't it. Rate of change of displacement, ds by dt. Okay, that is your velocity rate of change of velocity, acceleration, rate of change of acceleration. What is it of change of acceleration. Check, right, all those kind of thing, not a nice word to use by the way. So dy by dx as a rate measure, basically what are we trying to learn is that when there is a function which relates to variables x and y with each other. So x and y are variables which are interconnected by this particular, you can say function or relation whatever you have it. Then how do you then how do you link dy by dt. What is it related to dx by dt. Okay, that means if there is a change happening with respect to time in x. How does it influence the change in y with respect to time or vice versa if there is a change in y happening with respect to time. How does it change the value of x with respect to time. So in order to find that you will say sir simple we already know this process, you just have to perform the derivative on both the sides with respect to time. Right, exactly. So when you differentiate both sides with respect to time. Remember, you are going to use chain rule here. So first you are going to differentiate f of x with respect to x and then into dx by dt. So guys and girls I would like to recall this particular thing this is just chain rule. Right, so this entire concept is based on chain rule that's it nothing new is basically learned here. Okay, so when you figure it out, you basically end up getting the rate of change of y. As f dash x times the rate of change of x. But mind you one very, very important thing I would like to tell you these all are instantaneous changes these all are instantaneous changes. These all are instantaneous changes means they are calculated at a given instant. Okay, so dy by dt f dash x dx by dt they all are calculated for a given instant. Okay, maybe some x value will be known to you. Okay, maybe some you know another parameter which is helping you to find out this instances is known to you. Let me give you an example to make this part clear what do I mean by instantaneous changes. See, all of you have celebrated birthdays, correct. And all of you have blown air into a balloon, right, everybody has done it. Have you experienced that when you're blowing air into the in the in the balloon, initially the radius becomes larger at a very, you know, less amount of time so the increase of the radius with respect to time is very high. But later on when it becomes substantially big, and you're pumping pumping pumping pumping it becomes slowly slowly slowly slowly bigger right. So you realize that the rate of change of the radius is not uniform throughout. Right, let's say when it was small it was DR by dt was fast, slightly bigger it became DR by dt became slightly lesser. More your radius became it became more less right so what happens is that your DR by dt itself is basically dependent upon what is the radius at that instant of time. Maybe the volume the rate at which you are putting blowing air inside the machine into the balloon that is fixed because of course our lung capacity is fixed right we cannot blow like machines right. So let's say we are blowing at a rate of 100 centimeter cube per second, but the radius initially will go very fast and later on it will just slow down. Okay, so that is what is meant by when we are trying to find out these rate of changes we are basically looking at at some instant of x or some instant of time. Okay, let us do problems based on that and then you will understand it in more detail. Let's take a question, very small question to begin with very simple one. Let's say I have this question. Yeah, let's let's take this question. Yeah, the question says if the radius of a circle is increasing at a uniform rate of two centimeters per second, find the rate of increase of area of the circle at the instant, read the word at the instant. So they're asking you for an instantaneous rate of change of the area when the radius is 20 centimeter. Now before you start solving this question, there are few steps that you should keep in mind. First, identify the two quantities which are changing. Okay, number one, number one, what is the step identify the two quantities which are changing or identify the quantities which are changing. Okay, so in this case, radius and area are changing. Second, write a relation between the two changing quantities. So you know in a circle, what is the relationship between area and the radius, everybody knows the formula of the area of a circle. Second is differentiate both sides with respect to time. Okay, third step differentiate both sides with respect to time. Once you have done this now start putting the instantaneous values provided to you in the question. Now why I'm making a step out of it because in past I have seen people. They normally put the values and then start differentiating it. Please do not do that. You will end up getting zero, not as marks, but as values. Okay, both in fact, so don't do that. So first, identify the quantities which are changing here identified area and radius are changing. Second, write down the relationship between them, which is a is equal to pi r squared differentiate both sides with respect to time. So this will be differentiated with respect to r because this is in terms of r into dr by dt. What rule we use here, chain rule. What rule we use here, chain rule, simple as that. Next, once you have done this, of course, write it down completely dA by dt as this is 2 pi r dr by dt. Now this is the time where you start putting the instantaneous values. So what are the instantaneous values? Now see radius is increasing at a uniform rate of 2 centimeter per second. So dr by dt is plus 2 centimeters per second. Now many people ask me, sir, why do you categorically write a plus there? Because if the radius is increasing, right, you have to write a plus radius is decreasing. That means dr is reducing with respect to time. That means you have to write a negative. So that positive negative sign is very important because if you don't put that, you will end up getting, you can say, you will not get an idea of whether the thing is increasing with respect to time or decreasing with respect to time. So it's very important to follow this sign convention. Something increasing with respect to time, it's a positive thing. Something decreasing with respect to time, it's a negative thing. So what I'm going to do, dA by dt, I don't know. 2 pi radius, I know instantaneously, 20 centimeter radius is there. Dr by dt, I know it is 2 centimeters per second. So overall, if you see, it becomes 80 pi centimeter square per second. Please ensure you are writing the units properly. R has a unit of centimeter. Dr by dt has a unit of centimeter per second. So overall, it becomes centimeter square per second. Now it is a positive quantity means your area is increasing at the rate of 80 pi centimeter square per second. Is this fine? Easy. Nothing difficult about it. Right. Now in this chapter, you will feel as if you are sitting in a physics class. Okay. So those who are very close to physics or physics is very close to their heart. They will really enjoy this sub topic of application of derivatives. So let us take more questions. Good to know that. Let's take another one. So read this question. The question says a man who is 6.1.6 meter tall walks away from a lamp, which is four meters above ground at a rate of 30 meters per minute. How fast is the man's shadow rendering? How fast is the man's shadow rendering? I want everybody to try this out and give me a response on the chat box. We have to use string. Yes. Whatever helps you to solve the question whether it is coming from trigonometry, coordinate geometry, or any part of mathematics. Let's use it to solve our purpose. Okay. Again, remember my steps. First identify the two things which are changing. Then write the relationship between the two things which are changing. Differentiate with respect to time throughout. And then put the instantaneous values. Simple as that. No brainer. This is no brainer actually. There is no need to think. Okay, Gayatri. Very good. Very good. Adwik Sinha. Good going guys. Aditya. Okay. Aditya has a different opinion than others. Okay, Pranav. Pranav, let's keep the unit as the same unit of time, minute. Okay. Guys, one more minute. This is not a difficult question. I'm sure in your school also you would have done a similar question. Okay. So I'm getting almost three to four types of answers. I don't know why. Okay, let's say this is a lamp. Okay. And there's a man who is walking away from the lamp. So let me make a man. Okay, let's say the man is this. Okay. Let me make head. Small arms. Small legs. Okay. This man is moving away from the lamp. Okay. Now lamp is four meters long. And this man is moving away at a rate of 30 meters per minute. And because of that, his shadow is rendering. Okay. So let me first cast the shadow of this man. So the shadow of this man on the ground will be this length. Okay. So let me make in blue color. So this is the shadow of the man. Correct. Now what are the two quantities which are changing? You'll say obviously said the distance of the man from the lamp. Okay. Let's call that as X. And of course the length of the shadow. So this is your distance of the man from the lamp. And this is your length of the shadow. Let me call it as Y. Okay. Now if the man is walking away from the lamp, you realize that DX by DT is given to you as how much? How much? Plus 30 meters per minute. Now Y plus, because when the man is walking away from the lamp, X is increasing with respect to time. Had the man been walking towards the lamp, I would have written minus 30 meters per minute. Get this point clear. Okay. Now second thing is I need to relate these two quantities X and Y with each other. So you'll say sir simple use your similarity. Right. Can we use similar triangles? Can I say triangle? Let me write it down. Lamp post man shadow. So can I say triangle MNS is similar to triangle LPS. Okay. So I can say LP upon MN. LP upon MN is nothing but X plus Y. X plus Y upon Y. Right. And LP upon MN is nothing but four upon 1.6. Correct. Any questions, any concerns? Okay. So this is as good as saying 40 by 16 is equal to one plus X by Y. In other words, 24 by 16 is equal to X by Y. And if I just simplify this X by Y is equal to three upon two. That means Y is two-third of X. Correct. So I have now struck a relationship between the two quantities which are changing. Now, once I have done that, let me differentiate both sides with respect to time. So it'll become two by three DX by DT. Now, after we have achieved it, fit the instantaneous values given to you. What are the instantaneous values? Nothing but 30 meters per minute. That is DX by DT. So I'll put that down. So two by three into 30 meters per minute. That actually gives me 20 meters per minute. That is the rate at which the shadow is lengthening. So the positive sign automatically implies that the shadow is lengthening with respect to time. Okay. Lending shadow. Is this fine? Any questions, any concerns? So those who gave 20 meters per minute. Okay. Please do not change the unit. Some of you have started giving me answers in per second, but keep the unit as the same unit as given in that question. Is it fine? Easy. Some of you did make mistakes. I was quite surprised. Okay. All right. Now, let's take another question. Let's take this question. I hope everybody can read the question. It is everybody is able to see this question clearly. So there is a horse which runs along a circle with a speed of 20 kilometers per hour. A lantern is at the center of the circle. A fence is along the tangent to the circle at the point at which the horse starts. Find the speed with which the shadow of the horse moves along the fence at the moment. At the moment means at the instant. When it has covered one eighth of the circle, your answer should be in kilometers per hour. Please do it. Please try it out. And let me know your response in the chat box. Question is clear to everybody. Is the question clear to everybody? See, the question is, let's say there is a horse which is running on a circular path. Okay. It has been tied by some rope to the center of the circle and there is a light source being kept over it. Okay. So this horse when it is running, it is casting its image on a fence, which is let's say like this. Okay. So this is the fence. F E N C E. Okay. Now let's say the horse is at this position. Okay. So as you can see, because of this light source, it will cast a shadow over here. So this will be the shadow of the horse. No. So let's say it was at this position because of this light source, there will be a shadow cast here. So my question here is, as the horse is galloping along this circular path, the shadow will also be running along the fence. Correct. So shadow will seem to gallop on the fence or along the fence. Correct. So at the instant when this horse has covered one eighth of the circle, what is the rate at which the shadow is basically moving along this fence? This horse is galloping at the rate of 20 kilometers per hour. They have not provided us with the radius of the circular path. R is not known to us. Okay, Aditya. Okay, Gaurav. Anusha, you have answered with its R only. I didn't get that point. You mean to say the rate of the rate at which the shadow is running on the fences are the radius of the circle. I think four of you have answered so far. And your answer is, I think so. It is correct. We will check it out. Shritya also has answered very good. Shritya Pratik. One more minute I can give for the people who are trying hard so that all of you are satisfied with your attempt. And then we can discuss it. Okay, Anusha. Okay, Anusha. Anybody else? All right. So let's solve this question. Let's say I call this angle as theta. Let's say I call this angle as theta. Okay. Now, what has been given to us as our information? We have been given provided that the horse is running at a speed of 20 kilometers per hour. That means the rate at which it is covering R theta that is given to us as plus 20 kilometers per hour. Okay. Now, R is fixed. So that means R d theta by dt is given to you as 20. Correct? Yes or no? Now, what are the two things which are changing here? Of course, theta is changing. And this length, let me call it as X. X is changing. So theta is changing and X is changing. Radius is not going to change, right? So can I link these two quantities? Can I relate X and theta to each other? You will say, sir, yes, very easy. Write X as R tan theta. Okay. Now, having got that, let us differentiate both sides with respect to time. When I differentiate this, this will become secant square theta d theta by dt. Okay. Now, what are the quantities that we need over here? I need R, I need theta, I need d theta by dt. Okay. And dx by dt can be obtained after I plug in those values. But there is no separate R, no separate d theta by dt given, but we have been given a combined R into d theta by dt, which is actually 20. So can I say this into this is actually a 20. So it is secant square theta into 20. And what is theta? Now look at this information. When the horse has covered one eighth of the circle, one eighth of the circle means he has covered 45 degrees. That means theta at this instant of time. So let me write it down. Theta at the instant of time is 5 by 4, which is 45 degrees. So put it over here. So secant square 45 degrees into 20. Secant square 45 degrees is 2, 2 into 20. Answer is going to be 40 kilometers per hour. That means the shadow of the horse is running at twice the speed, 40 kilometers per hour. Is it fine? Correct. And if you realize as theta approaches 90 degrees, the speed of the shadow will become infinitely big. Right? Of course we are not going to calculate for that theta equal to 90. So you won't be able to see the shadow to be so fast. And if it crosses 90 degrees, your shadow will not be visible at all. Okay. Shadow cannot be seen. Yeah. Maybe that relation will now be true if the fence was on this side. Okay. Aditya. So this is one of the questions that comes normally, shadow-based questions. So we are going to take different varieties of questions. Let's take the next one. Yeah. I hope you all can read this question. Yeah. The question is, sand is pouring into, sorry, from a pipe at the rate of 12 centimeter cube per second. The falling sand forms a cone on the ground in such a way that the height of the cone is always one sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is four centimeter? So I hope the scenario is clear to you all. So the situation is something like this. So there is a pipe. Let me make the pipe in white. And through this pipe, sand is being poured. Okay. Sand has been poured. And the sand forms a cone like this. And the cone is also like that. At whatever instant you look at the height is always, the height is always one sixth of the radius. Okay. Oh, too fast. Very good. Oh, okay. Don't worry. We are just doing the basics question first and then we'll slightly complicate scenarios. Okay. Okay, Pranav Burman. Very good. Who else? Gaurav Burman Pranav so far. Guy three. Very good. Very good. Okay. Prka Thake Be Gav. Okay, guys tell, we'll discuss it. Now. First of all, what are the two things which are changing? You'll say said, the volume is changing of course, because the sand is constantly being poured. And because of that radius and height is also getting changed. And so can we write a relationship between the three quantities, which are actually changing yes there ideas. height and volume are related to each other by volume is equal to 1 third pi r square h. The good thing is we know what is the relationship between height and the radius and now look at the question. The question is basically trying to ask you what is the rate at which the height is changing at the instant when the height is 4. So at this instant when the height is 4 centimeter, how far is the height is changing? So it will be advisable that we convert everything in terms of h on the right side. So I will remove my radius and write 6 h in place of that. So volume is going to be 36 by 6 which is 12 pi h cube. Now once you have got that you can start differentiating both sides with respect to t. So this will be 12 pi 3 h square dh by dt. Now start plugging in values. So what are the values? dv by dt. dv by dt is already provided to you in the question. dv by dt is already provided to you in the question and that is 12 plus 12 centimeter cube per second. So this is 12, 12 pi 3 into 4 square into dh by dt I have to find out. So 12, 12 gone. So dh by dt comes out to be dh by dt comes out to be 1 by 16 into 3 is 48 pi. Please write down the proper units. It's centimeter per second. So this is going to be your answer for this question. dh by dt is 1 by 48 pi centimeter per second. Most of you have got it right. People who have got it wrong Pradyun and others Hariharan please have a look at your working. Alright next question. Next question is two men PNQs start with velocities you at the same time from a junction of two roads. Excuse me inclined at 45 degrees to each other. If they travel by different roads find the rate at which they are being separated. Find the rate at which they are being separated. Some of you would like to use vectors over okay Aditya. Right Hariharan. So most of the concepts that you have already done in physics we are learning it right now in mathematics. Okay Shatish very good. Okay Gaurav we'll resolve that. We'll resolve that. Let me give one minute for those who are trying hard. Okay let's discuss it. Let's discuss it. Okay Hariharan did you use a calculator or something for that? Okay since both are moving at the same we can say velocity u you can say let after a time t they would have covered a distance of x and let's say this distance let me first connect them. Let's say this distance is y. Let's say this distance is a y. Of course you have been provided that dx by dt is u okay and this is 45 degree. So because of changing x y is also changing right. So they have been getting separated as x you know they're getting separated more and more as x increases correct. So first of all what is the relationship between x y. So you can use your cosine rule for this and you can say cos 45 degrees is nothing but x square plus x square minus y square by 2x into x correct. That means 2x square by root 2 is equal to 2x square minus y square okay. Yes that is what I was saying most of you would have used vectors over here but not required you can basically solve without that also. So y square is nothing but 2x square minus root 2x square or you can say 2 minus root 2x square okay. Now this is a simple relation which says y is equal to under root of no plus minus business because you're dealing with distances y is equal to under root of 2 minus root 2x. So just differentiate this with respect to time right and just put your dx by dt as u over. So the rate at which they are getting separated will be this. So this is the rate at which rate at which separation is happening okay. So this is your answer. Any questions? Where did that cos 45 come from? The fact that the roads are inclined at 45 degree got it got it. Yes have the relationship been different then basically what would happen you have to use three variables x y and z write a relationship between them and now when you're differentiating both sides you would need instantaneous values of x instantaneous values of y and dx by dt and dy by dt got it. So dz by dt will come from there clear Aditya okay next. Now I have a very special question for you I will not be writing too many things here just listen to this question okay the question is there is a could you guys go to the previous question okay sure yes anything that you would like to ask me done okay thank you yeah so now this is a wall okay this is a wall and this is the ground this is the ground and this is a wall there is a ladder there is a ladder which is actually taller than this you know wall and it is basically like this okay so it is it is having some part protruding over the wall it is having some what part protruding over the wall I'll give you the dimensions this wall is seven meter long this wall is seven meter long and the length of the ladder the length of the ladder is 15 meters okay this ladder length is 15 meters okay now this ladder is slipping against this wall this ladder is slipping against this wall in this way okay and this rate of slipping is two meters per second so I let it down over here another important information is the ladder is slipping the ladder is slipping against the wall against the wall two meters per second okay question is find the rate at which find the rate at which the foot of the ladder is slipping is slipping away from the wall from the wall at the instant at the instant at the instant the projection of the protruded part of the ladder along the vertical ladder along the vertical is three meters okay the question is slightly lengthy so what are they asking you they're asking you what is the rate at which the foot of the ladder is slipping away this is moving down constantly at a rate of two meters per second so what is the rate at which this slipping is happening when when the projection of the protruded part along the vertical at the instant when this guy is three meters so at the instant when this length is three meters okay I'm not writing anything over there because you start thinking that that is a fixed value no that is changing so at the instant when this guy becomes three meters what is the rate at which the foot of the ladder is slipping away from the wall is the question clear to everybody if not I can explain you once again seven meter long wall 15 meter long ladder ladder is basically protruding over the wall and it is slipping like this okay it is slipping like this this contact is never broken okay this contact is never broken so this is slipping against it at the rate of two meters per second and because it is slipping this ladder is slipping from the ground as well the rate at which it is slipping against the wall is two meters per second find the rate at which the foot of the ladder the foot of the ladder is slipping from the wall at the instant the projection of the protruded part is three meters along the ladder Harir the slipping is along the ladder that's why I've shown it categorically the ladder is going down like this at a rate of two meters per second no vectors this is not a vector chapter you have to use derivatives how do you use a vector unnecessarily the process is the same guys the problem may look a bit complicated but the process is well defined just follow the process anybody any success anybody okay enough time has gone okay so first let us identify what are the quantities which are changing so can I say since the ladder is slipping against this wall this length is changing let me call this length as y that length is changing let me write properly yeah let me write on this side okay since the wall the ladder is slipping on the ground as well can I say this length is also changing let me call this as x okay so x and y are changing correct now the question says dy by dt is negative two meters per second because y is reducing with respect to time because the ladder is climbing down and down and down isn't it now how do I strike a relationship between x and y okay now one thing I can do here is I have to also relate this projection so what I'm going to do I'm going to take this angle as theta that means that that means this angle is also theta okay so can I first write down a relation between x theta and y theta and try to see what kind of a connection can happen between them so from this figure I can say x is equal to or x by seven is equal to tan of theta so x is equal to seven tan theta okay now what I want I want to know what is dx by dt at this instant where theta is something now what is that theta which we don't know right now okay at theta equal to something right we'll have to figure it out we'll have to figure it out now what I need here is dx by dt so that means I indirectly need seven six square theta d theta by dt so without knowing theta and without knowing d theta by dt I will not be able to solve this question correct okay now let's apply some more information given to us in the question if this is why this length this length is going to be 15 minus y and can I say can I say 15 minus y 15 minus y 15 minus y is seven seek theta right in other words y is equal to y is equal to 15 minus seven seek theta correct yes or no now I'm looking at that instant where where this length becomes three now if you look at the upper triangle let me name it for your sake of convenience let me call it as p q r s t if you look at the triangle p q r you are looking for that instant where y actually becomes three seek theta isn't it because of the basis three why will actually become seek three theta correct so at least by comparing these two I will get the instant of theta I will get the value at which or theta at which I'm trying to calculate dx by dt so when I relate these two I will get this is equal to seek three seek theta which means 15 is equal to 10 seek theta in short I get cos theta is equal to two by three if I'm not mistaken okay so this theta that I'm looking for this question mark is basically achieved that is actually cos inverse of two by three okay let me just segregate it from the question yeah now I need d theta by dt how will I find that out very simple let's come back to the same expression over here differentiate it with respect to time you will end up getting minus seven seek theta tan theta d theta by dt so for this d theta by dt I can use my this expression because I know dy by dt minus two I know seek of theta which is three by two by the way if cos theta is two by three what do you get as tan theta if I'm not mistaken this will give you a under root of five by two am I right if cos theta is two by three and theta is under root five by two so this will be under root of five by two d theta by dt okay so minus minus gone so d theta by dt will become eight upon 21 root five radians per second sorry I missed out I think the correct so now I have all the wherewithal with me I know the theta I know d theta by dt so what are we waiting for let's do the honors let's complete this so dx by dt will become seven seek square theta seek square theta will be three by two the whole square into d theta by dt is eight by 21 root five okay let's simplify I think 21 will get cancelled straight away here this will cancel two and I will end up getting six by root five meters per second so this is your answer to this question how many of you got it does it come out to be the same I think roughly yes I think your answer is roughly correct is it understood any questions any concerns any questions any concerns now one question that I would like to ask you to find d theta by dt to find d theta by dt why didn't I differentiate this guy it is recommended that you can use that you can if you want to use a parameter to make your life easy you can directly you can use that instead of directly relating x and y basically you are introducing a parameter to link them all right now my question to you is everybody why didn't I differentiate y equal to 3 c theta to get my to get my dy d theta by dt we cannot differentiate with respect to ty so Hariran if we could why didn't we do that why did I chose 15 minus 7 c theta think think answer is very simple we are not concerned after that point in time see this is an instantaneous value 3 is a fixed cost theta is also fixed so this is actually a constant for us because we are calculating it at an instant this is an instantaneous value of y this is a generic value of y get the point exactly front of are you getting my point this is a generic y value that means if your theta keeps on changing this guy will change your y as well but this is an instantaneous value so at the instant when that projection was 3 that is the y value at that time so you cannot differentiate this you should not be differentiating this are you getting my point because you'll get faulty results are you getting my are you getting my explanation why didn't I differentiate this to get my d theta by dt because the projection will not be 3 always so this I use only to get what is my instantaneous theta because I needed it in my in my expression here secant square theta right so this is only useful for finding your instantaneous theta but when you're differentiating why you have to take this because this is a generic y and theta relation no matter wherever is the wherever is the ladder rather point clear everybody okay next question next question is there is a lamp okay there is a lamp which is 50 feet high this is a lamp which is 50 feet high okay this is 50 feet high okay there is a ball which is at the same level as the lamp and place 30 meters away from it 30 feet away from it okay so this is a ball okay now this ball when it is dropped it covers a distance of 16 t square okay so when it is dropped at a time t let's say after a time t it would have covered 16 t square after time t okay my question here to you is my question to you is okay let me not draw anything here my question to you is find the speed of the shadow of the ball on the ground at the instant at the instant t is equal to half a second okay so what has happened this ball is released right at the instant when t is half at that instant of time what is the speed at which the shadow of the ball is moving this distance is 16 t square distance it has dropped distance it has dropped in time t that is 16 t square this distance so let's say in one second it would have dropped 16 meters 16 feet so this is the distance it basically drops in one second okay two second it would have dropped 64 feet by the way before two second it would have hit the ground actually clear Siddish is my question clear to everybody so it is not half gt square like the way you would have thought it to be okay let's say it is a different planet altogether where on free fall it covers 16 t square yes any success it's a big number how big in thousands okay so yeah that could be the answer why not that could be the answer anybody else okay Pranav see t equal to half is for the last okay don't use it initially see what are the quantities which are changing relate them first of all that is the very first step once you've related it then differentiate both sides with respect to time and once you have done with that then start putting the instantaneous values okay before that instantaneous values out of no use okay Pranav okay Pradyum yeah yeah I got it minus means it is coming towards the lamp post okay Aditya all the dimensions are in terms of feet only so I think Pradyum you meant to say that many feet per second right okay anybody else okay let's discuss it out let's discuss it out see if you see the image the image of this ball on the ground will be here correct this would be the image of the ball on the ground so what is changing you say this is changing let's say I call this as X okay and of course this is also changing let me call this as Y so as the ball is falling on the ground the height of the ball from the ground that is changing and the distance X is changing okay now how do I relate Y or X again can I use my similar triangles over here I can say this length I'll just name it P and I just drop a perpendicular from here P Q let me call it as L M N so can I say triangle triangle P Q N is similar to triangle L M N okay so P Q over Q N is equal to L M over M N could you show it with trig so am I not using trigonometry here triangle only no correct so P Q by Q N is equal to L M by M N right so let us try to write down P Q is Y okay and Q N is all of you please note down Q N will be X minus 30 right because this whole thing is X I have taken so this is this is 30 so I've subtracted 30 from X to get Q N okay this is equal to 50 divided by M N M N is X in other words what do we see is that Y is equal to 50 X minus 30 upon X okay that's nothing but 50 minus 1500 by X this is your Y correct now having got this relation let us differentiate both sides with respect to T that gives you minus 15 plus 15 X square DX by dt yes or no now do you know dy by dt so first of all what is Y Y is nothing but it is 50 minus 16 T square so dy by dt will be nothing but negative 32 T and you have to find the instantaneous value of this at T equal to half that means minus 16 feet per second so this is known minus 16 what is X what is X so for that you have to again find your X from this relation itself so Y is equal to 50 minus 1500 by X right now Y at that instant of time let us find out instantaneous Y Y at that instant of time will be 50 minus 50 minus 16 times half square that is actually 46 if I'm not mistaken isn't it so we have got from here let me just put a barricade 46 is equal to 50 minus 1500 by X so X is going to be 1500 by 4 1500 by 4 is 375 so let us try to put here so this is minus 16 this is 1500 this is 375 square 375 square I will write it as 375 375 DX by dt I don't know okay let's try to simplify so this will go by a factor of 4 and this also goes by a factor of 4 so you'll have minus 4 over n so dx by dt is anyways minus 4 into 375 that comes out to be minus 1500 feet per second so the shadow is traveling towards the lamppost or the speed of the shadow on the ground is minus 1500 that means it is moving towards the lamppost so you can say 1500 feet per second towards the lamppost is it fine now most of you could not solve this question I could get answers only from I think Pradyum Shatish and later on Pradam sir is there any easier way than dealing with X square dealing with X square in that case you have to in that case you may take an angle theta and deal with it that is another way to do it but there you have to strike a parametric relationship between X and theta and Y and theta it will it will come with a cost of increasing more variables okay so you can try it out you can try different methods to solve the same problem okay so there you have to find d theta by dt and then you have to relate dx by dt with d theta by dt that's all you need to do okay it's coming like that okay okay should this try it out you should be getting the answer Hariharan I substituted Y is equal to before differentiating and got stuck it's fine you can you can put your Y's here also right and then you can put TS half throughout that is also fine Hariharan you should not be ideally getting stuck okay that should have given you the same result oh okay okay maybe calculation error maybe calculation error okay next question next question okay the question says uh there is a comprehension okay it says that there is a lamp of length 10 meters placed at the end of a ladder AB maybe it is not too scale because as per the question this lamp should be of length 10 meters so this is 10 meters okay it's not too scale by the way and the ladders of length 13 meters which is leaning against a vertical wall as shown in the figure and its base slide away from the wall okay at the instant base is 12 meters from the vertical wall at the instant this is an instantaneous condition the base is moving at the rate of 5 meters per second okay and there's a man of height 1.5 meters standing at 15 meters from the vertical wall so this gap is 15 meters this gap is 15 meters there are two parts to the question there are two parts to the question find the rate at which theta decreases when the base is 12 meters from the vertical wall okay second is the rate at which the length of the shadow of the man increases when the base is 12 meters from the vertical wall so let us solve the ninth part of the question first give me a response for the ninth part very good Shatish anybody else with the first part first part is easy I believe okay Aditya okay last 30 seconds for covering the ninth one and then we can start with the tenth one okay all right so first part ninth part let's look into this see in the ninth question just focus on the ladder part because they're asking you how is the angle theta changing with respect to time when this ladder is slipping away right so what are the two things which are changing here that is the foot of the ladder let me call it as let me write it in green this is x okay yes I know correct and theta and theta correct so how is your x and theta related to each other so we'll say simple x is going to be 13 cos theta correct so dx by dx by dt is going to be negative 13 sin theta d theta by dt now they're asking you for that instant when this x becomes 12 when this x becomes 12 when x becomes 12 this guy is 5 isn't it so dx by dt which is 5 plus 5 because it is slipping away from the foot of the wall minus 13 sin theta in this case will be 5 upon 13 and d theta by dt I don't know okay so automatically from here you can get d theta by dt as negative 1 radians per second now this basically means theta is decreasing at a rate of 1 radian per second since the question is already asking you what is the rate at which theta decreases so your answer will be 1 radian per second option number 8 is it fine so there is no problem with question number 9 anybody having any issue with question number 9 let's work out 10th part the rate at which the length of the shadow of the man increases when the base is 12 meter from the vertical okay Shrutish let's let's draw the situation once again so let's say this is your ladder this is your lamp this is the man okay and there is an image cast by this lamp on the ground okay and this is your ground this is 10 meters this is 13 meters man I think is 1.5 meters okay and this is the shadow length so what are the two things changing x is changing and y is changing right so as the ladder is slipping away this guy x is changing and the length of the shadow of the man is changing correct so we want what is the rate at which the length of the shadow of the man is changing at the instant x is 12 meters okay at the instant x is 12 meters I think this distance was already mentioned to us as 15 this distance is already mentioned to us as 15 meters so what is happening the ladder is slipping because of that the lamp is coming down so you can imagine that the lamp is actually reducing its height so the lamp is coming down and because the man is stationary man is not going moving anywhere man is stationary and lamp is moving down because of that the shadow length is increasing so we have to find the rate at which the shadow length is increasing so I have got only one response so far which is from Shatij so far anybody else okay Pranav okay Aditya okay let's discuss it out this is not a difficult problem so let's say I call this this particular length as z okay so the reason for your x to change is because your z is changing and because of that your y is also getting changing so x is changing that creates a change in z and because of that z there is a change in y correct so let us first try to understand what relation connects x and z and that's a simple Pythagorean relation x square plus z square is equal to 13 square correct in other words if you differentiate both sides with respect to t by the way this is slipping if I'm not mistaken the slipping is happening at a rate of how much 5 meters per second yeah so dx by dt is 5 meters per second given to us plus 5 meters per second so when you differentiate both sides with respect to t you get 2x dx by dt plus 2z dz by dt is equal to 0 in other words dz by dt is negative x by z dx by dt correct now at the instant when x is 12 we know z was 5 and this guy was plus 5 again so it is going to be minus 12 meters per second so the foot of the lamppost is coming down at a rate of minus 12 meters per second at the instant when your dimensions of x is 12 of course z will become 5 by that time and of course dx by dt is also 5 okay now let us try to relate let us try to relate z with y but that is simple you can use the fact that let me name the triangle abc and apq so abc triangle is similar to is similar to triangle apq so I can say ab by bc y by 1.5 is equal to this whole length which is y plus 15 upon upon 10 plus z so 10 plus z is nothing but 1.5 times 1 plus 15 by y correct me if I am wrong okay so 1.5 10 z this is going to be I think 2.25 by y or 10 plus z my my so z is going to be z is going to be 2.25 y minus 8.5 any questions here anybody now differentiate both sides with respect to t 22.5 so this will be minus 22.5 by y squared dy by dt so let's try to put the values so here I know what is my dz by dt dz by dt is minus 12 minus 22.5 y square I do not know that I have to figure it out okay so you can use this fact that at the instant when you're looking for the slipping z was actually 5 okay so y will become correct me if I'm wrong 22.5 divided by 13.5 okay so this will become minus 22.5 divided by 22.5 square and 13.5 square okay this is equal to minus 12 sometimes we'll get cancelled off so your answer oh sorry dy by dt I forgot minus minus will also get cancelled off so your dy by dt will become 12 into 22.5 divided by 13.5 square correct me if I'm wrong I think this will go by a factor of you can write it as 120 into 22.5 by 135 into 135 so this thing will go by a factor of 25 times okay this is I think 9 and this will be I believe 6 no 5 we'll go by 25 or you can cancel off by 5 at least so this will be 45 and this will be 27 okay again let's cancel off by a 5 this will be 24 this will be this will be 27 again okay let's cancel off by 9 no I can cancel off by 315 and this is going to be 9 okay so this will be 3 this will be 8 this will be 5 oh 40 by 27 a bit of calculation towards the end 40 by 27 meters per second this is the rate at which the shadow is going to grow which is option number b which is option number okay so only three of only two of you about this correct shittish and aditya is there any question related to the solving of this part any question any concerns be do let me know if there was any and I knew about that I would have I would have used that but yes this is not the only way to solve it you may use parameters to solve it okay okay so I think we have done enough you know question on dy by dx as rate measure so now we can go on to our next concept