 Excuse me sir. Yes Tirpan tell me. Sir, will you do functions as a separate chapter like composite functions and all that. Yes, yes. You're after this chapter? Yes. I think I've already sent one small part to you which is even odd functions even that is not complete I'll send you one more part. Okay. Okay. Alright, so good evening. Welcome to the fourth session of differentiation. So in today's session we are going to primarily talk about approximations also called differentials. And we are going to also talk about tangents and normals. Okay. So let's get started with the concept of approximations, approximations and differentials. Okay. Now this concept is not new to you because in physics also you would have done it in last year's first chapter of physics, units and dimensions, how to find errors, etc. And the same thing would be used also to do approximations. So both these concepts of errors and approximations are linked to each other. Now how does calculus feature in over here? Let us talk about a function in general, let's say y is equal to f of x, where y is an output which is dependent on the input x. So obviously if you do a small change in the input, let's say you make x as x plus delta x. Okay. Remember we call delta x as absolute error in x, absolute error in x. So if you make a small change in x, which we call as delta x, because of that a small change in y would be introduced. That means an absolute error of delta y would be introduced in y. So delta y is nothing but absolute error in y. Okay. So how do we figure out what is this absolute error without having to undergo the full process of finding what is the value of the new y when you are substituting the new x, that is x plus delta x into your function. Okay. Now for this we are going to use our first principles of differentiation. I hope everybody remembers here that what is dy by dx? dy by dx is nothing but f of x plus delta x minus f of x by delta x, limit delta x sending to zero. Okay. I am assuming that the function is differentiable and hence the left hand derivative, right hand derivative both can be used or anyone can be used to find out dy by dx. Okay. Now this expression if you make some small changes, very, very small changes, instead of saying delta x sending to zero if we say delta x is very small, right? Delta x is a small quantity. Okay. This is a very small quantity. Okay. Rest everything I keep it as it is. Okay. Then your left hand side instead of dy by dx it will become delta y by delta x. So you can say that these two quantities can be compared to each other. You can say it can be approximately used for each other when your delta x is small. Okay. So what we say is dy by dx could be approximated as delta y by delta x. Okay. For very small values of delta x. And remember this dy by dx is calculated at the point where the error has occurred. So x is that value where the error has occurred. I'll give you an example. I'll give you an example for the same. Let us say, let us say there is a metallic cubicle container which is kept under the sun. So I'm just drawing a cubicle container, okay, which is kept under the sun and it's very hot that particular day. Okay. Now, because it is kept under the sun, this metallic container will expand, isn't it? And as it expands, its volume will change. Right. Let's say initially this container was having a dimension of two centimeters. So it's length, breadth and height were two centimeters each. Okay. So what are the volume of this container? It's a volume of the container is not going to be x cube where let's say x is the edge length. So it is going to be two cube, which is eight centimeter cube. Okay. But let's say after being in the sun for let's say an hour or so, it's, you can say new length became 2.012, 2.012 centimeter. Okay. So what will be the new volume or question can be asked in this way, what is the change in the volume? So one way to do it is obviously find 2.0 to whole cube, but that would be a painful process, especially when you are devoid of calculators. Correct. So how do we calculate such things by use of differentiation and how is differential calculus helpful in evaluating this? So because of the expansion in the side length, we can say let the, let the change, let the change in the volume, change in the volume be Delta V. Okay. If you use this formula that we have derived just now, this formula, remember why is playing or we is playing the role of why for you? So we can say Delta V by Delta X will be approximated as DV by DX at the instant when the error has occurred. Now the error has occurred at X equal to two. So initially it was two and now it has become 2.012. So you can say this term is actually your X plus Delta X. Now one small piece of advice that I would like to give you, we always write the change as plus Delta X. There's no minus Delta X because Delta X itself will be positive or negative quantity. If let's say you had kept this metallic container on a winter day outside, obviously it will contract. Correct. In that case, your Delta X would be a negative quantity. Right. But you still write X plus Delta X equal to that new value of the length of the container. Okay. Never put your external sign in place of, in front of Delta X. Don't put minus or plus. It's always plus. Okay. So in this case, if you are looking for what is your Delta V, we need to know the following things. We need to know Delta X. Delta X can be obtained from here. Originally your X was two. Originally your X was two centimeter. Okay. So your Delta X will become or you can say two plus Delta X is 2.012. Okay. Let's not write down the units unnecessarily. So Delta X will become 0.012. Okay. So this will become 0.012. This is approximately equal to the derivative of this function calculated at the point where the error has occurred. So three X square at X equal to two. I think that will give you 12. Isn't it? So Delta V would be Delta V would be 12 times 0.012, which is 0.144. Correct. So this tells you that there would be a change of 0.144 centimeter cube. Right. The positive sign indicates that there is an increment. Okay. So always you need to do plus. Right. If there was a decrement, it would have automatically come with a negative sign. So I'm repeating again. Please do not put any external sign here. Delta V comes with its own sign. Okay. So the new volume of this particular cubicle container would be 8.144 centimeter cube. I would like you to check this out if you have a phone with you or a calculator kind of an object with you. Just check whether when you cube it, how close you are to the answer. You would realize that you are very close to the answer. Okay. But we never had to do a cubing process. We found it out by the process of calculus. Is this fine? Any question here? Oh, it's matching. Very good. And if that tells me that is quite matching. Okay. So I think with this example, the fundamental of approximation and differentials is clear. Okay. Now basically, you can also try to understand this from geometrical point of view. So let's say there is a curve. There is a curve. Okay. And at a certain value of the input, the value of the function is y. Let's say at x it is y. Okay. Now, let's say I decide to change my input. I'm just taking a positive change. It could be any any change. It could be a negative change also. And I want to know what is the change in my output. That means I want to know what is delta y. Okay. Now, what does this process say? It says if you want to know this change approximately, you can draw a tangent at that point. And this change that you have over here, this change you have that you have over here, let me call this as change k. This k will be approximately equal to your delta y. That is what it's trying to say. Right. I know there is always an error. There will always be some kind of an error in your finding because it is just an approximation. That's why I have written an approximation symbol here. This is approximately. Okay. Now remember, more closer you are, more your delta x is tending to zero, more this approximation will become equal to sign. Okay. So if your delta x is tending to zero, then your delta y by delta x will be equal to d y by dx at that point of error. Okay. So smaller this delta x, more close you are to the actual value. More close you are to d y by dx. Getting the point. So here if you see this is your delta x. Okay. This base is your delta x. You can say run. You have to find rise. So what do we do is we use the fact that tan theta will give you k by delta x. Okay. So dy by dx is definitely k by delta x. Correct. If you see the triangle over here, k by delta x will be tan of this angle, which is nothing but d y by dx calculated at that point. Are you getting my point? But remember, when you're trying to approximate it, it becomes delta y by delta x because k is not exactly delta y, but it is very close to it. So as a result, we can use dy by dx is approximately delta y by delta x. Don't forget to put the x value of the point where the error has occurred in your dy by dx. And therefore, this formula will be helpful for you to find your error in y. Is that fine? So with this, we can actually start solving few questions. This topic is not very difficult. Let's jump into our question directly. Find approximate value of find approximate value of value of 127 127 cube root of 127 approximate value. Many a times you will see they use the word this for this both are same thing. Okay, don't don't get confused. Both are same thing actually. Please do not use calculators and all as the purpose will be defeated. We have to find it without the use of any kind of a calculator. Okay, Aditya good. So guys, here we have to take care of two things one. In fact, three things one. What is that output input relation that you're going to build up? For example, when you see this kind of an operation, you know that your output is input to the power of one third. So this is the one thing that we need to understand. Many a times this will not be given to us. So we need to figure it out from the question. Okay, second thing is for what value of x, which is close to this? Can you find why without the use of any problem or without the use of any calculator? So tell me a value which is very close to 127 for which cube root is very easy to obtain. So you will see 125 right? Yeah, 125. You won't take 64, right? I'm sure you won't take 64. Neither will you take 225 or right? Sorry, what is seven cube 343 or 36 into six that is how much 216 that is also be very far. So the best value to start here with will be x equal to 125 because for that your y can be easily be obtained and that is five. Okay, now the problem is we don't have 125. We have a slightly deviated value and that is 127. So they're not trying to ask you what is this new value of the output? This is what we need to find out. So first we need to understand that our delta x here is plus two. Okay, now many people ask me said two is not a small value, right? So how are you trying to use it here because your delta x should be very, very small. It is basically in comparison to the x value. Your delta x is small in comparison to the x value. So two as compared to 125 that is a small quantity. That is what we need to take into account. For example, when you are dealing with, when you're dealing with let's say the frequency of a simple pendulum. Okay, what do you consider? You consider that the length of the string is very small as compared to the radius of the earth. If it is very large, your formula is going to change. Isn't it? Right? People who have done gravitation chapter. So we consider that L is very small as compared to the radius of the earth. So it's not like L is actually small as small as zero. No, it is small as compared to the radius of the earth. So in the same way, this delta x should be small as comparison to x. So now the third step would be using the formula d y by dx at the point the error has occurred. Now where has the error occurred? The error has occurred at 125. This is approximately equal to delta y by delta x. Delta x is already known to you as a two. Okay, so let's solve for delta y from it. So d y by dx, this is your y. So d y by dx will be one third x to the power minus two third calculated at 125 is equal to delta y by two. So this is going to give you one third, one third and if I'm not mistaken, this will give you 25, isn't it? Yeah. So delta y, delta y by two is equal to this. So delta y is going to become two upon 75. Okay, two upon 75. This is approximately 0.026666. You can write 267. Okay. So again, it automatically comes with its positive sign. Right? So if there was a case of subtraction, it will automatically come with a negative sign. So your new value of the output will be the old value, which was five plus 0.0267. That is nothing but 2.0267. This is going to be a new value. Okay. So you don't need a calculator in order to calculate it. Now, it doesn't mean you have become a calculator because such things will only work when your value of the new input is very close to a value which is known to you. Let's say had I given you something like 170, then probably this method will not work because there will be a large amount of error in your answer. So only when your delta x is very small as compared to x, then only you can expect a very accurate, in fact, you can say close to accurate answers to come out. Is this fine? Any questions here? Please do ask. If no, can I move on to the next question? Clear? Okay. Next question is if one degree is alpha radiance. If one degree is alpha radiance, however, if I want, I could give you alpha value that is 0.0174. But that is not required for us. Then, then, cause of 60 degree one minute is approximately equal to option A, half plus, half plus alpha root 3 by 120, option B, half minus alpha by 120, option C, half minus alpha root 3 by 120, option C, half minus alpha root 3 by 120, option D, none of the above. Let me put a poll on. Can you all see the poll? Yes, sir. Yes, sir. So 29th onwards you have your UT, right? What is the syllabus for maths? The full integration and application of integrals. And linear programming? Yes, sir. Oh, they have taught application of integrals also? There is only one exercise in that, sir. It's an easy question. I'll give you another 30 seconds to finish this off. Only 5 people have responded. Okay, at the count of 5 I'll stop. 5, 4, 3, 2, 1. Please vote. Please vote. 12 of you haven't voted yet. Okay. Good, good, good, good. Fine. See, guys, whenever you're dealing with angles, you must always deal in radiance, right? All the calculations actually should be done in terms of radiance. That's why this first information is given to you that 1 degree is 60, 1 degree is alpha radiance. As you can see, most of you have gone for option C. Let's check. So the obvious relation that you're going to have over here is y is equal to cos of x, because you're evaluating cos of something. Okay. Earlier, your x was pi by 3. That is 60 degrees. But now it has, by the way, y for that was half. But now it has gone to 60 degrees plus 1 minute. Now, if 1 degree is alpha radiance, 1 minute will be alpha by 60, correct? Okay. And they're trying to ask you what is the new value of y? Okay. So delta x is very clearly alpha by 60. So let us use the formula. Let us use the formula dy by dx at the point where the error has occurred is approximately equal to delta y by delta x. So delta y will be approximately equal to delta x times dy by dx. That is sine x with a negative sign at x equal to pi by 3. So delta x is alpha by 60 into root 3 by 2. So as you can see, there is a negative answer coming, which clearly suggests the fact that there would be a decrease in the value from half. And we all know that because cos x is a decreasing function from 0 to 90. So there will be a fault. Okay. So your new value of the output would be the old value plus delta y. Delta y will be this. So as a result, you end up getting option number C as your answer. Option number C will be your answer. Is that fine? Any questions here, anybody? Next thing that we are going to talk about is the relative error. Is the relative error. Relative error. So I think in physics also you would have learned about it. Relative error is basically the absolute error divided by the point where the error has occurred. You can also write it as this. Okay. If you multiply this by 100, then the relative error becomes percentage error. This is read as relative error in x. And if you multiply this by 100, or let's say this by 100, it becomes the percentage error in x. Percentage error in x. Okay. Now we'll take up some questions which are based on relative error. These are slightly trickier ones because I've seen people making mistakes in them. Okay. By the way, when I say delta x or when I say this, please note that this is approximately equal to dx. Okay. So when you say delta y, this is approximately equal to dy. Okay. So many a times we replace dx and dy terms with delta x or delta y. So as to make our life convenient. Okay. So this differential change in x can be approximated as delta x many a times when you know that these quantities are very, very small. Okay. But basically it is done in presence of in presence of an x because then only these quantities would be considered to be a very small quantity. Okay. Let's take few questions on the same. Then probably it'll give you a lot of idea. All of us have seen in thermodynamics adiabatic process. Correct. So for a diatomic gas undergoing adiabatic process, we know that the relationship between pressure and volume is given as PV to the power 1.4 is a constant. Okay. My question is, find the percentage, find the percentage change in volume, in volume. If the pressure is increased by, if the pressure is increased by 1%, find the percentage change. Now I'm not saying increase or decrease. You have to tell me what is happening actually when the pressure is increased by 1%. How will you solve such questions? Would you like to try first? Your voice is breaking. Yes. Yeah, please try. Please try. So what is given to us? We have been given that delta P by P into 100. This is 1. This is given to us. And what do we need to find? We need to find this. Correct. Correct, Trippan. Okay. So let's discuss this. Very good, Shruti. Very good. Okay. Let's take log on both the sides. So log P plus 1.4 log V is equal to a constant. Okay. Constants log is also a constant. Okay. Let us differentiate with respect to V. Let us differentiate with respect to V. So when we do that, it becomes 1 by P dP by dV plus 1.4 by V is equal to 0. Now, we have learned that dP by dV can be approximated as delta P by delta V. So we'll do the same thing over here. We'll write this as delta P by, sorry, delta P by delta V. Delta P by delta V. You can write this as negative 1.4 V. Let us take this delta V to the other side. Let us take this to the other side. So it'll give me delta P by P is equal to negative 1.4 delta V by V. Let's multiply both sides within 100. Let's multiply both sides with 100. And when you do that, automatically you'll end up seeing this expression which is actually given to us as 1. Okay. And they want us to find out this expression. They want us to find out this expression. So I can say here that delta V by V into 100 will be negative 1 by 1.4. This is approximated as negative 0.7 if I'm not wrong, or 7-1 whatever. Okay. So what we can say is that there is a decrement. There is a decrement of 0.71 percentage in volume. Okay. So as the pressure rises by 1%, there is a fall of 0.7% in the volume. Understood how it works? Yeah, sure. Any questions? Oh, yes. Correct. All right. Let's take the next question. If the sides and angles of a plane triangle vary in such a way that its circum radius remains constant, find the value of DA by cos A plus DB by cos B plus DC by cos C where DA, DB, DC are small increments in the sides of the triangle. I'll put the poll on for this. Please put your response on the poll. Okay. Excellent. Very good. Three people have responded so far. Nice. Yeah, yeah. Take it. I'll give you 45 more seconds from here. Please vote, guys. Last 15 seconds. 5, 4, 3, 2, 1, go. Please vote, please vote. End of poll. Almost equal responses have been given to A, B and C. Of course, the reason for D not getting any response, it is none of these. And many of us, we know very well that none of these being correct is very less. Okay. Now let's check here. What is the thing given to us? If the sides and angle both vary, but they vary in such a way that the circum radius remains constant. Right. So we all know by our sign rule. Sign rule says A by sin A is equal to B by sin B is equal to C by sin C. Okay. And this is equal to 2R. So A is equal to 2R sin A. B is equal to 2R sin B. And C is equal to 2R sin C. Okay. Now can I say DA would be 2R cos A DA. Can I say the differential change in A will be 2R cos A, the differential change in angle A. Similarly, DB will be 2R cos B DA. And DC would be 2R cos C DC. I should write a B here by mistake. In other words, which is very rhyming with the expression that we are supposed to get. DA by cos A is actually 2R DA. DB by cos B is 2R DB. And from here I can say DC by cos C is equal to 2R DC. Okay. If you add them, if you add them here, that given expression will become 2R DA plus DB plus DC. Isn't it a differential change in the sum of the angles of a triangle which we very well know that there is not going to be any change because A plus B plus C is a constant. Right. It's a fixed value. So the change in the value here would be 0. There cannot be a change in the value of the sum of the angles of a triangle. The given expression to U will be equal to 0, which is option number A. Which is option number A. But I thought it was tomorrow Venkat. Oh, it was today. Yes, any question anybody? Why are your voice coming very feebly to me? Which one sir? No, I thought your bio was tomorrow. Oh, tomorrow is GEM. Is this fine? Any questions here? Okay, looks like an easy concept, but many people make mistakes here. Let's take one more. In a acute angle triangle ABC, A and B are constants. A and B are constants. Right. But angle A and B are varying. But angle A and B are varying. Okay. Prove that. Prove that D A D capital A by under root of A square minus B square sine square A. Is equal to DB by under root of B square minus A square sine square B. Try this out. Just right done. Once you're done with the proof, just right done. See here, DA by cos A, DB by cos B, DC by cos C is clear. What is your given expression? Your given expression is the sum of these, right? So when you add this means you're adding this. DA, DB, DC means differential of A plus B plus C. Correct? Okay, that partly I doubt sir that why you took D out. Okay, when I say D by DX of sine X. Plus D by DX of X square. Can I not write it as D by DX of sine X plus X square? Yes, sir. The same way. A change in the error, a change in A value and a change in B value and a change in C value. The change in the A plus B plus C value. Okay, sir. Let me go back to the question. When is your semester exam starting? I didn't tell sir. No tentative date also? No. Okay. Okay, as soon as you get your dates, do let me know because I have to plan for your next monthly test. Very good, Tirpan. That's a one minute. Yeah, very good Siddharth, very good Ayush. Okay, let's check this. From our sign rule, we can say. Xiaomi also done, Patam also done. Okay, I'll just give you a hint from the sign rule, we can say A by sine A is equal to B by sine B. In other words, A sine B is equal to B sine A. Okay. Differentiate both sides. That means do this. Somebody's trying to say something. Hello. This will become A cos B db. Remember, there is no change in a value. So A will be like a constant. So only will differentiate sine B. So the differential of sine B is cos B db. In the same way, this will become B cos A dA. Correct. In other words, dA divided by a cos B will be equal to db divided by B cos A. So basically I brought it down to the left and I brought this down to the right. Any doubt so far? Okay, so now we can write cos of A as sorry, cos of B as under root of one minus sine square B. And we can write your cos of A as under root of one minus under root of sine square A. Correct. Let us introduce this A inside. Let us introduce this A inside. This will become A square minus A square sine square B. And this will become under root of B square minus B square sine square A. You can write this as a square of A sine B actually. And this is square of B sine A. Okay. If you go a little up, you realize you have written here A sine B is equal to B sine A. Is it? Yeah. So I can replace this A sine B with a B sine A. So I can write this as under root of A square minus B square sine square A. And on the same way, on the left hand side, I can write D B by under root of B square minus A square sine square B. Right. So this is what we actually were looking for and hence proved. So I think this is more than enough required for approximation and differentials. You will not find many problems on this concept. So without much waste of time, we will move on to the next concept now.