Hi, everyone, and welcome to our lecture on arc length. So recall we've seen before, so if we have a graph of a function, so y= f (x,) we're going to keep things finite here. So a less than or equal to x less than or equal to b, so in your head, you should imagine the xy plane and there's some graph, and y= f (x). And I want to know what the length of this curve is, not the area under the curve, not the slope of the tangent line, but the actual length of the curve, how long the string is, if I tried to measure it, as I traverse it once going from left to right, then we saw that this is called arc length. And in this particular case, we denoted by L, we're going from the definite integral, have a to b of the square root of 1 plus dy, dx quantity squared. And this is all dx, all right, great, so this takes care of the case where the fine arc lengths if I'm given a graph, however, says before, so now what if I'm given something that is not necessarily a graph? So for example, a something that fails the vertical line test, so maybe I have some path and a particle and it cuts back in on itself, well guess what? This is not the graph of a function, this will fail the vertical line test. So I have to describe this to you as a parametric curve, so we'll describe it as a parametric curve, defined by x(t) and y(t). So I'll let t be the parameter now, and I'll use fancy Greek letters alpha and beta. Really to distinguish them thinking of this curve as drawn by an artist over time. So if you give me a time, so t=0 is where I start, or maybe t equals alpha, and t equals beta is where I end. Let's call this alpha and t equals betas were in at any particular time, then I'll give you back the x and y coordinates. Well, it turns out in that case, I can still find arc length by substituting and cleaning up the dy, dx and the differential dx for my x(t) and y prime to t's. It's going to use lots of derivatives with respect to x and y, so, let me give you that formula the arc length for now, this is just a generalization of the prior one turns out to be the definite integral from alpha to beta. Now this is important is the bounds on the parameter t of the square root of dx dt squared, plus dy, dt squared number my parameter is time. So this is no longer dx, this is now dt. Again all derivatives are defined and we're traversing see, just one time as I move from alpha to beta. So this is our new formula and I want you remember this is now for, this is the generalization, this is for parametric curves. Hopefully you have a nice formula sheet going, so please add this one to your list. All right, let's use this as an example, so our famous example of a parametrized curve of course is the unit circle over the unit circle lives right at (0, 0). It's a radius of course, this one, we're going to traverse it in the counterclockwise direction. This parametric curve clearly is not a function fails the vertical line test, but this parametric curve, given of course by the values of trig, this is x equals cosine of t and y is sine of t. And we're going from zero to two pi, right, so as I see this, I'm thinking that x(t) is cosine(t). I'm thinking that y(t) is sine t and around I go in the counterclockwise direction. So I want to find the arc length. Now we don't normally call for a circle, this value that I'm after the Arc Line, this is better known as the perimeter, or perhaps even better the circumference. So, we should know this is one of these formulas that they teach you from high school geometry. But we should know that in general, hopefully this looks familiar C= 2 pi r. So for the unit circle, when I plug in r is one, I'd better get back a value of 2 pi here. But let's just go through all this and show that this actually works. So here comes our formula, it is the absolute, sorry, the definite integral from 0 to 2 pi of the derivative. I'll just write it in new notation here x prime squared, plus the derivative y prime of t quantity squared, dt. I'm going to need some derivatives, but that's not a problem. So let's work within the integral sign. We're going from zero to two pi. The derivative of cosine of course is minus sine of t quantity squared, the derivative of sine of t, my y value is cosine of t, that is all squared and this is all under the square root, so big square root and then dt. Let's clean this up right in a perhaps more conventional way. When I square a negative goes away, so this becomes sine squared of t, plus cosine squared of t. Of course this all falls under the square root, hopefully hopefully hopefully that looks familiar to you. This is the one trig identity that I want you to know sine squared plus cosine of t is of course just one. And so this becomes the square root of one, which is just one, so I have one dt. Whenever you integrate one, this is just the difference or the length of an interval. So of course it is in fact two pi. So this all works out quite nicely. Nice example here, just going through the formula taking derivatives and integration using all of our skills of calculus. So now what happens if I have a curve that lives in space, so sometimes it's called a space curve. And I want to let's give you a picture, imagine some rocket ship or some spaceship, it's off doing its thing. It is not necessarily defined by the graph of a function, but it's maybe it loops back on itself, maybe it has some depth and dimension. And once again, I'm going to parameterize this curve, so it's a function of time as x of t, y of t, now remember, I'm in 3D, so guess what I need a third component, z of t, it looks scary with this new component, but it's not going to be any worse than before. We're going to still bound t with alpha and beta, so we're going to use Greek letters to represent the bounce on the time parameter, and then we're going to define or maybe generalize our formula for arc length. So again, you can think of this as like if I laid a piece of string down along this path, how long would the string be? How far has the rocket ship traveled? So it's still the definite integral just a single integral from alpha to beta. Single integrals reflecting the single dimension nature of this curve, and I just generalized my formula. So I have a square root still of x prime of t quantity squared, plus y prime of t quantity squared, plus z prime of t quantity squared. Now in this formula, I'm using the prime notation, and the other one I use live this notation where I wrote dx over dt. Of course they're the same, right, so take a derivative square, add them up. Same idea, the only thing that's new is now that I have a z component, the formula generalizes quite naturally as I go through this, so I'm going to try to give a little more advanced examples. So we can go through and just see like what is a tough one look like. So let's do an example again, add this formula to your infinite list of formulas for this course. And here we go, so I want to find the arc length of a curve that's defined by, let's do X of t, y of t, and z of t, equal to cosine of t. Sine of t and t squared. And I want to do this for zero to t to two pi. So first off whenever people start seeing sine and cosine I know they always freak out they get weird, but just bear with it. It's good practice to go through this. So if you want, pause the video, we know the formula or write the formula here for convenience. We're going from zero to two pi. We're going to take the square root of x prime of t quantity squared plus y prime of t quantity squared plus z prime of t quantity squared, all d t. That's what I want. Pause the video before I go through it and then check the work as I go. Okay, ready? Alright, let's get all the information we need first. So let's see. So I have a definite integral from zero to two pi and a big square root, I need the derivative of x squared. So the derivative of cosine is negative sine. We did this before. So let's square this. You're going to get sine squared of t. Hopefully you're okay with me doing a derivative and a square in one step. Then I have y which is sine of t, take its derivative to get cosine square that. So there's cosine of t. So there's a identity sine of t plus cosine t, but I have a z prime of t going on t squared, its derivative courses 2t square that and you get two t squared dt. I'm going to leave it like that for a second. You'll see why. But of course, if I clean this up, you get the square root. So here's your nice identity. This is just one, one plus 2t squared. If you wrote this as 4t squared, you are perfectly fine, you are not wrong. Don't panic, you're doing great. Now, we can do one of two things over here. I debated on not working this out; this is a little messy, so if you got stuck on this, this is okay. If it's an applied problem, and you really wanted to know how far the particle went, go get a computer and go work this out. This is a definite integral, that computers are great at. Setting up the problem, knowing the formula, that's all the human component. If you want to get the answer here, there's definitely a shortcut where you can get the computer and work it out and get like 10.63. So if you got 10.63 that's excellent. However, some people when you're trying to learn integration, they want to see the steps like how do I get this. If you just need the number, go get it from the computer, if you want to see sort of the steps well, this is a little tricky, and that's why I picked this one, but let's go through it just to see it. Anyway, first we'll start off with u substitution with u equals to t. And of course d u is two dt. So now we switch the variable from t to u, and this becomes one plus u squared, and then dt becomes one half to u. So there's like a big one half out front. Put u over here. Be careful when you use substitution with the definite integral you got to switch the bounds. So when t is zero, then u become zero. And when t is 2Pi, my Upper bound becomes four pi that I have this integral of one plus u squared is why this one gets a little tricky. This is a nother student for colour. This is another substitution. So now let's do u equals tangent of theta. Again, this is why the computer is kind of nice here. So why would I use u equals tangent of theta? Well again seasoned veteran here if I take u squared, it's just a little trick. Then you get one plus tangent squared theta now one plus tangent squared is another one of your identities. This becomes secant squared. So if you forgot that identity, you can go look up Pythagorean identities. One plus tangent squared is in fact secant squared. Why do I like that substitution? You can see the value of it here. When I take the square root of secant squared, I just get back secant. So this falls under trig substitutions. If you've seen trig substitutions, this is a pretty standard one to know. And I get back secant squared. Secants, which is great. So that's going to swap out my, I don't want to lose that one half, that's going to swap out the square root of one plus u squared. That just becomes secant of theta. I am not going to go through the trouble of finding the new bounds because I'm going to put them all back in terms of view in a second. So I just know that these are not zero and four pi. At least in terms of u, so I'll just going to put dots here for now we'll deal with them later. So I swapped out the undergrad, one plus u squared, I still need to swap out my du. Good practice here. The derivative of tangent is secant squared theta, d theta. So I get secant times secant squared theta, d theta. So this becomes one half definite integrals on these bounds of theta that I'm not aware of yet. I'm not going to find them of secant cubed theta. Now, I'm going to punt a little bit on this, to actually work this out would make this video probably twice as long. This is good exercise. It's a good practice in integration by parts. It's not impossible is just a little annoying. So we're going to punt a little bit. But if you use integration by parts and you get if you really want to see all the steps, you can just Google. What is the integral of secant cubed? It turns out to be definitely known as definitely possible but in the interest of time and space, it is the excuse like brackets here and natural log of the absolute value of secant theta plus tan theta. Close of the absolute values plus one half secant tangent again, integration by parts will give you these two pieces. So if you want to work that out, feel free. Again remember with the computer we are kind of already know the answer, but I'm just going through the, I guess, algebraic steps to show you. Alright, let's see if we can go finish this off here. Now remember, I've evaluated these add my values of theta. I don't want theta. I kind of want to put this back into terms of u and the reason why I can do that quite nicely is because you remember if I see Tan theta, I can replace that with a u. If I see secant theta, I can replace that with square root of one plus u squared. So squared of one plus u squared, and square root of one plus u squared. So I have all the back substitutions I need. So I have this one half out front, maybe I bring it in maybe I don't ln of the square root of one plus u squared plus u. And then I guess I'll bring it in so you gotta hit the one half and then don't forget, you got to hit this other one half to get a fourth plus one fourth square root one plus u squared times u. And now because I'm back in the variable u, I'm going to go ahead and put back my zero to two pi. I just didn't want to deal with arc tan to go find like our tan of four pi and stuff like that. So I'll just put it back into u. Now it becomes a substitution is not as bad as it seems. So if you put one half ln of one plus four pi squared plus two pi, I guess that's not super pretty, but here's the nice part. If you plug in zero for this part, that stuff starts to cancel. If you plug in zero for the right side, definitely get a zero over here that goes away. And then you get lN 1, which is also 0, so you get plus a fourth of the square root of 1 plus 4 pi squared times 2 pi. That's the close form of this number. If you take the calculator, work this out, I promise you'll get 10.63. So whatever makes you happy if you want to go through the long way or the short way, they're both correct depending on what you need. Usually when you see the definite integral, feel free in a tough case to grab the answers you need. A couple things. So when you have different parameterizations of a curve, the arc length of a curve, how long that string is does not depend on the parametrization. You can reparametrize a curve, and as much as you know that it doesn't change your answer. We call this thing like it's independent of parameterization. Arc length is independent of the algebra that's used to describe it. You'll always get the same answer, in particular, if you traverse the curve in reverse. So if [INAUDIBLE] backs up along the same path, you'll get exactly the same arc length. So now I want to introduce this notion of an arc length function. So now we're going to introduce something called the arc length function. So suppose that we have a curve in space. We're going to define it by a vector function. So let's say R of T is X of T, Y of T, and Z of T. And we're going to again bound to our time parameter, A less than or equal to T less than or equal to B. And I want the derivative of all these pieces to be continuous and I want C to be traverses that who wants coming as T increases from A to B. And then from that we can define the arc length function. We're going to define it as, F of T, so it is in fact a function of T. It'll be defined as the sort of accumulated magnitude or the sum of the definite integral from A to T. So T becomes our upper bound of the magnitude of the derivative vector. And we're going to have to use a new variable a dummy variable, so we'll call it U with respect to you. It should look like the fundamental theorem of calculus a little bit. So what is the magnitude of the derivative vector? So friendly reminder, if I take the derivative vector, that's just the derivative of each of the components. So you have X prime T, Y prime T, Z prime T and if I want the magnitude of that, so how do you find the magnitude or the length of a vector, you take the square root of each of the components squared. So you can see the relationship, everything should look the same. So each derivative squared, since our dummy variable is U, when I write that out, that's going to look like The integral from A to T have the square root of a right in both notations to get used to it, DX DU quantity squared plus DY DU quantity squared plus DZ DU quantity squared. So I find what this is I integrate, then I plug back in T, and I plug back in A and you get a function in terms of time. That's why this is called the arc length function. And the idea of this is the length of the part of the curve between the starting point and just some other point in time. And so if you differentiate both sides, you sort of returned back the fundamental theorem of calculus. It's a really nice relationship. It's just another way to sort of study curves in space. And the idea is like it's often useful to parameterize curves with respect to arc length. Because arc length is a natural thing, natural property that arises from the shapes of these curves. In particular, everything about this depends only on the curve. Nothing here depends on the particular coordinate system. So if you start changing coordinate systems, this will not change. So if a curve RFT, when that's given, is given in terms of a parameter T and in terms of S of T, this new arc length function then I can solve for T as a function of S. I can write this down and we can totally reprimatize the curve and make a substitution and we get back this nice way to describe a curve in terms of its own intrinsic property and not on the ambient space. That was a lot of words to sort of go through. I'm pretty sure your forward calculation. Once again though, add this definition arc length function to your evergrowing formula sheet. Let's do one example. And then we'll we'll pause this video to let you guys reflect on some of this stuff. Okay, here we go. So I drew the picture, but I didn't give you the formula. So let's do the example here. This is the helix. Sometimes these curves have names. This is a curve, we're going to define it as a curve in space. It is given by cosine of T, sine of T and T. So if you graph this using a pewter, you get this sort of helix thing. And I wanted to re parameterize is what I want to do. I want to re parameterize with respect to arc length. From its starting point of 1 0 0 in the direction of increasing T, okay, so that's what I want to do. This is the definition I want to read parameterize with respect to arc length. When you see this and they give you a point. First thing you do is you think are what is the value of T at this point normally is pretty simple. Like if the Z component is going to be 0, then I guess we'll T has to be 0. And again, you can totally check that by just plugging in T is 0 to cosine and sine, and so you get T is 0. So we're starting off at T is 0 and then we're going to write down that DS DT. Derivative of S with respect to T, which is our prime magnitude. So what is this going to be? This is the square root of each derivative squared. So this is the derivative of cosine of X, all right? I will write one more time component squared plus the derivative of Y of T squared plus Z prime of T squared. It's more notationally annoying than like computationally difficult. Just gotta write it all out. So the derivative of cosine squared, that's sine squared of T. The derivative of sine is cosine squared and then 1 squared. Of course, this is my trig identity, cosine, sine squared plus cosine squared is 1 you get the good old squared of T. So this is constant. So nice easy function. It's got a nice constant sort of differential here. And therefore when I write out my formula, so I'm going to write out S of T, just like before. This is the integral from the starting point in terms of T though, so 0 to the variable T of the magnitude of the position vector with respect to U DU, DU's my dummy variable and again, that's why this example is kind of nice. The magnitude is just a constant. It's also positive. So I don't even need the absolute values, but that's perfectly fine. So we have 0 to T. of route 2 D U. And of course, that just works out to be root 2 of T. So I have my arc length function written as square to T. Now what you can do once you have this is you can write back T, a sofware T so T becomes S, have S, or sometimes just over the square to 2. And so you can re parameterize writing this thing in terms of S. So that says that R now of T of S it really is a re parameterization becomes the vector cosine. So we're used to be T, but now it's s over 2. Sine of S over 2. And last but not least just t itself. So S over 2. And so I have re parameterize the curve, in terms of its arc length. This has deeper consequences that we're going to get to hear, but just realize I've removed like time is the parameter. This is in terms the curve is defining itself. It is completely independent of X, Y, and Z. It is it doesn't matter what X, Y, and Z are at this point, the time determines the curvature, okay? So it's like at every point, I don't know where I am, but I know how I'm curving. It's a very nice property to have. All right, go through the steps, make sure you can do the computations, and then we'll keep working through these in future videos. All right, great job on this one, and I'll see you next time.