 Hi, I'm Zor. Welcome to Unisor Education. Today I would like to talk about definite integrals in application to calculating the lengths of a curve. This lecture is part of the course of advanced mathematics for high school students and teenagers. You can find it on Unisor.com. If you watched this lecture from YouTube or any other website, I would suggest to actually to switch to this one because every lecture has very detailed notes and you can take exams for those topics where it's available. So it's much more functional and there is some other functionality on the website. Okay, so lengths of the curve. First of all, we have to define our curve. What is the curve? Well, from a calculus perspective, the curve, let's consider the curves on the plane because the curves in a three-dimensional space will be exactly equivalent to this. So let's talk about the curve on a plane. So you have some kind of a coordinate plane and you have some kind of a curve. How can we define it? Well, in this particular case, the curve like this cannot be defined as a function. Now, this curve can be defined as a function of x. Now, in this case, a much better way to represent the curve is through two functions which depend on certain parameter. So these two functions define a point on a curve based on certain parameter t where t is from some kind of an interval from a to b. Now, this particular case is actually a particular case of this case. Why? Because you can always say that in this case, my x is actually t. So x of t is equal to t and y, since y of t is a function of x and x is the same as t, you can say this. So this is a particular case of a general dependency. Maybe I will talk about this a little later, but in any case, it doesn't really mean much if we will consider this, if we will consider that. So we are talking about the curve defined parametrically as two functions, and we obviously assume that these two functions x of t and y of t are smooth enough, at least differentiable. Next. Next, we are talking about the lengths of this particular curve. So as t is changing from a to b, my point is moving from here to here. Now, as in the case of the area under the curve, which actually was the way to introduce definite integrals, we will do very, very similar thing. So let's divide this particular interval from a to b into a different, into n different parts. So we start from this point, and then we will have these division points from a to b. So my first is a, my last is b, and everything in between. Now, as we define these break points on interval a, b, correspondingly, we define break points on our curve. So this point would be x of ti, comma, y of ti. So every ti corresponds to a point on the curve parametrically defined by these two functions. So breaking my interval of argument t from a to b into n pieces, we break our curve along its going into independent little pieces. Okay. Now, if the pieces are small enough, so if this n is large enough, and the distance between different break points is small, each particular thing from here to here can be approximated with a straight line. And obviously, approximation is better if these points are closer to each other. Now, if my functions x of t and y of t are smooth enough, this curve would be smooth enough. And for a smooth curve, it makes sense to replace the length of this arc with a straight line. So we are talking about a straight line from point x of ti minus one, y of ti minus one. So this point to a next point, which is x of ti, comma, y of ti. So this is a small piece. It's a straight line between these two, between these two points. All right. Now, the total length of the curve is obviously a sum of these. So first, let's just calculate the length of this small piece. Let's call it li. So what's the length between two points? Well, they obviously know this from the previous. If you have one point and another point, the length is by Pythagorean theorem is the square of the length. So let's put the square is equal to sum of this, which is difference between these two abscissists, which is between this and this. So it's x of ti minus x of ti minus one square plus this square, which is different between ordnance. And li by itself, obviously, is the square root of this. It would be easier for me if I would do it slightly differently. Equals two. Let's divide and multiply by ti minus ti minus one square. So what will be is x of ti minus x of ti minus one divided by ti minus ti minus one square plus of ti minus y of ti minus one divided by ti minus ti minus one square. And multiply by ti minus ti minus one square, right? So that would be the right thing. I just multiply by this and divide it by the same thing. So now, as my number of intervals is increasing and the distance between break points is decreasing, what is this? What does it remind us? Well, it obviously reminds us the definition of the derivative at point i, right? And if I will summarize, if I will have total length of the curve, which is approximately equals to sum of li, i from one to n, that would be sum of i one to n square root. Now, this is square, right? So I have to do the square root. Square root of what is this? This is delta x of ti divided by delta ti square plus. This is the increment of the function divided by increment of the attribute, right? So that's what derivative is. That's what we're going to. Delta y of ti divided by delta ti square. Now, this is the end of the square because this, since this is the square and I'm doing square root, it goes ti minus ti minus one, which is actually delta ti. So as our, now this is actually a Riemann sum, if you remember the definition of the definite integrals, this is the function, this is a function, right? Some kind of a function, and this is the delta ti, and we are summarizing them. This is the definition of the integral, which means that as n goes to infinity and the length of this largest interval is diminishing to zero, the limit of this is equal to real length, which is equal to integral from a to b. Now, all the different delta ti's are extending from a to b, right? Of this function, which is square root of derivative of x, right? This is derivative of x square plus derivative of y by t square dt. So this is basically straight from the definition of the definite integral, where under definite integral we have this function. Well, basically that's it, that's the formula for the lengths of the curve defined parametrically by functions x and y, depending on parameter t, which is going from a to b. Now, by the way, let me return back to case when the function, when the curve is actually a graph of the function, and we can have something like this. As I was saying, we can replace it with x of t is equal to t and y of t is equal to f of t, right? Well, t is actually the same as x, so to speak, in this case. So, this derivative of t by t is equal to 1, and this derivative is obviously whatever it is, in which case our formula would be, instead of this, I will have 1, and instead of this, I will have this derivative of the function y by x actually, same thing as y. We just change the argument from t to x, but that's the same thing. x can serve as a parameter in this case, that's what it is. Or you can rename it to t, doesn't really matter. Letter doesn't matter, because these letters are all the same. I can put x here or anything or any other letter, right? Okay, so we covered this and we covered that, and we have a formula, a universal formula for all parametrically defined curves, and I would like to have a couple of examples right now with this formula and see if I will have some reasonable answers. Now, my first example is really trivial. Let's consider I have a circle of a radius r. Now, I would like to define it parametrically and calculate its length and see if I will get 2 pi r, all right? Okay, now, what can be a parameter? A parameter can be an angle, let's call it t, and t would be from 0 to 2 pi, right? To get the complete circle. Now, what is my x of t and y of t? In this case, well, x of t is r cosine, so the length would be integral from 0 to 2 pi, square root of... Now, my x of t is r cosine, so derivative would be, okay, r cosine of t, that's my x of t, and derivative would be r sine of t with a minus sign, right? So, I need square, so it will be r square sine square of t. Now, my y is r sine t, so y is r sine t, and its derivative is r cosine of t. Square would be r square cosine of t, square, dt. And what is it equal to? Well, obviously, r square can go outside of these and outside of the square root. It's a constant, so I will have to put r. So, what will be inside? Under square root, there will be sine square plus cosine square, which is 1. So, I will have integral from 0 to 2 pi of dt. Now, what's the... Well, 1 times dt. Now, what is the indefinite integral? We will use the formula in Newton-Lagnitz. What is the indefinite integral of just dt of 1 dt? Well, that's t, right? So, we have to put t from 2 pi to 0, which is... When we substitute 2 pi, we will have 2 pi. When we substitute 0, it will be minus 0. That's what formula of Newton-Lagnitz is, and we have 2 pi r, which is not a surprise, obviously. Now, can it serve as a proof that the length of the circle is equal to 2 pi r? No, that's not a proof. It's just a confirmation, if you wish, because we are talking about 2 pi. What is 2 pi? I mean, we cannot really use all these trigonometric properties without first defining what is a circle, what is the length of the circle, what is the angle, etc., etc. So, all that machinery should still stay. This cannot be a proof that the length of the circle is 2 pi. It's a confirmation, okay? Now, let's go to another problem. A little bit more interesting and less trivial. My function is... Okay, let me change my... Okay, my function is x is equal to r cosine cube of t and y is equal to r sine cube of t. Well, first of all, yeah, and t is from 0 to 2 pi. Well, first of all, what is this curve? How does it look? Well, let's just first determine four main points when t is equal to 0, pi over 2, pi and 3 pi over 2. So, it will be like this. If my angle is 0, then x is equal to r times cosine of 0 is 1. So, we will have r sine of 0 is 0. So, we will have 0. So, this would be r. Now, when angle is equal to pi over 2, cosine is 0, sine is 1. So, it would be 0 r, which is this. Now, if t is equal to pi, my cosine would be minus 1, right? Cosine. So, 0 pi over 2, 3 pi over 2 and 2 pi. So, that's my... No, that's wrong. This is pi. This is 3 pi over 2 and this is 2 pi. So, cosine is minus 1 and cube would be minus 1. So, it's minus r. That's my x coordinate. And sine of pi is still 0. So, it will be this point. And finally, 3 pi over 2, 3 pi over 2, cosine is 0 and sine is minus 1. So, it would be 0 and minus 1. So, it would be minus r. So, these are four points on my curve. Now, how about in between? Well, if you do it a little bit more carefully, you will see that the curve would be something like this. And it's called asteroid. So, we need to find the length of this asteroid. Well, let me simplify the job a little bit. I will find only one quarter of this and then multiply by 4. So, right now, I will do it not from 0 to 2 pi, but I will do it from 0 to 2 pi over 2. That's just easier because I don't have to be concerned about minus sign, plus sign, that's too much bother. It's much easier to do this and then multiply by 4. So, the derivative is equal to r, 3 cosine square. That's because it's a power function. And the chain rule, the cosine is derivative of cosine is minus sign. So, it would be minus 3r cosine square of t sin of t. Now, this is 3r sin square of t times cosine. Okay. Now, remember our formula, square root of this square plus this square, right? So, obviously, 3r square would be 9r square. We can factor it out and get it out from the square root. So, it would be 3r still. So, I'm talking about integral from 0 to pi over 2. 3r would be outside of this integral. And under the square root, I will have cosine in the force of t sin square of t plus sin to the force of t cosine square of t dt. So, that's what our formula does. This is the square of the first derivative by x. This is the square of derivative of y. And 3r would be just outside as a factor and then outside of the square and outside of the integral. All right. Fine. Equals 2. 3r integral from 0 to pi over 2. Now, obviously, if you will factor out sine square cosine square from both of them, what would be remaining cosine square plus sine square, right? Which is 1. So, that's why I can omit it and that's what will be here. Now, what is this? Now, since I'm from 0 to pi over 2, sine and cosine are non-negative. So, I can just cancel this square root and these squares and I will have just this, right? And there are many ways to do it. For instance, one of the ways sine times cosine is a half of the sine of the double angle. And that's how I do it in notes. Here, I can just a little, I can change it a little bit because notice that the derivative of sine is a cosine. So, this and this is equal to d of sine of t, right? And now I can replace actually this integral with integral 2r. Let's substitute sine of t is equal to u, let's say. So, it would be u times du. And what's the limits of integration? When 0 is, if t is equal to 0, sine is equal to 0. So, u is equal to 0. When t is equal to pi over 2, sine of pi over 2 is 1. So, it's 1. This is equal to 3r. What's my derivative of u? It's u squared divided by 2, right? So, it's u squared divided by 2 from 1 to 0. So, the result is 3r and when I'm substituting instead of u1, I will have 1 half. 0 will have 0. So, I will have 1 half minus 0. So, it's 1 half. So, it's 3r divided by 2. 3r divided by 2. Now, this is only a quarter. So, I have to multiply it by 4. And the total length would be 6r. And that's the end of it. So, these are a couple of examples where the formula like this one for the lengths of the curve can be used and it basically produced very easy result. Well, easy in case you can obviously integrate it. In this case, it's easy. Obviously, I came up with relatively easy integration. In some cases, I mean, if you want, for instance, to calculate the lengths of, I don't know, some complicated curve, like a sign, for instance, it's a curve and you can obviously parametrically give it. That might actually get into a little bit more problems because the integral would be not so nice like in this particular case. All right, that's it. I suggest you to read the notes for this lecture on Unisor.com. Other than that, well, that's it. Thanks very much and good luck.