 Welcome back to our lecture series Math 1220, Calculus II for students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Misseldine. In this video, we're starting lecture 20 in our series. And before we get started with that, I just want to remind the viewers here that throughout any of the videos in this lecture series, if you have any questions whatsoever, don't hesitate to post your questions in the comments here, and I'll be glad to answer them for you. So in lecture 20 here, we're going to start our discussion about arc length. How can we find the length of a curve in the plane here? And so let's start off with that plane. We'll have our x-axis and our y-axis, which we'll label there. And consider two points. The first point we'll call point one, whose coordinates are x1, y1. And the second point, we'll call it the second point, whose coordinates are going to be x2 and y2. So let's consider the distance between these two points. So we connect the dots that we have this line segment. What's the distance between these two points? Well, we can compute this using the classic distance formula, sometimes referred to as the Euclidean distance formula. So the distance is going to be computed by first taking the difference of the x-coordinates. So take x2 minus x1, and we're going to square that. Now notice here, it doesn't matter who is x1 and who is x2. If you switch up the order, it doesn't make much of a difference. Because x2 minus x1, when you square, it will be the same thing as x1 minus x2. So in this situation, we have the literal same difference going on here. Continuing on, we have to also take the difference of the y-coordinates, y2 minus y1. Likewise, we'll square it, so don't worry about who comes first. And this will all sit inside of the square root. Again, this is a classic distance formula that many of us have seen before. Now just as a quick reminder, where does this distance formula come from? Well, given any two points in the plane, there is a natural right triangle associated to these points, which you can see right here. Now this third point, which now we're going to color in red here, its coordinates are going to be the following. You're going to notice that the y-coordinate of this point is going to match up with the first point, but its x-coordinates going to match up with the second point. And so if we start measuring the horizontal distance from the blue point on the left to the red point, it's going to be the difference of their x-coordinates, because it's just a horizontal line. This gives us x2 minus x1. If we want to measure the distance along the vertical line, this would be the difference of the y-coordinates, which is y2 minus y1, like so. And so if we call the hypotenuse of this triangle d, as it's the distance between the first point and the second point, applying the Pythagorean equation, we take the adjacent side, x2 minus x1. We're going to square that. We're going to take the opposite side, y2 minus y1. We're going to square that as well. And then that should equal the hypotenuse squared, which is, in this case, d squared. So the typical Pythagorean equation gives us this distance form. We're going to play around with that for a little bit longer. Now when you look at these values, x2 minus x1, we've seen something like this before. This right here is a difference of the x-coordinates as a change of the x-coordinates. And in the past, we typically called this delta x. And so in this form, we have a delta x squared. Similarly, if you take the difference of the y-coordinates right here, y2 minus y1, this change of y, we typically refer to as delta y. And so in this Pythagorean equation, we get this delta y squared. So the left-hand side here looks like delta x squared plus delta y squared. So what we're going to do is we're going to make a slight modification to this problem here. We're going to define the distance from x1, y1 to the origin. We're going to call that s1 for distance there. And then if you take the line segment that connects the origin to the second point, x2, y2, we're going to call that s2. And so be aware that the distance, there's a little bit of geometric argument here, but the distance from x1, y1 to x2, y2, that is d, is actually going to be the difference of s1 and s2 right here. So the d value could be replaced with s2 minus s1. We get this squared. And then the reason we introduce this new variable here is that if we have this variable si, then s2 minus s1 would be the difference of s's. And so we get this delta s squared. And so this equation right here is going to be very much what we build upon to derive our arc length formula. Notice here this is telling us that the change of this distance s squared is equal to the sum of squares of the change of x and change of y. Solving for delta s here, we get that delta s equals the square root of delta x squared plus delta y squared. Again, this equation should not be interpreted as anything different than just the Pythagorean equation with the introduction of these delta symbols right here. So we can use the distance formula to calculate the length of any straight line who has a finite length. Can we use the distance formula to calculate the length of a curvy line or what we would call an arc? Hence why we call this the arc length problem. And the answer to this question, of course, is going to be yes, we're going to use calculus here. And so let us consider the following problem, right? We are going to have again our x-axis and y-axis. Here's our x-axis and y-axis. And consider we have some continuous, yet curvy function f like so. So we have these two points in the domain. So we're going to have a starting point A. We're going to have some ending point B. And so what we're interested in is we have this point A comma f of A. We have some other point over here, B comma f of B. Can we measure the distance from A to B along the curve f? What is going to be that distance right there? Now, the way that we're going to handle this problem is going to be identical to how we solve so many other calculus problems using this technique of accumulation. That is to say we're going to take our interval A to B and we're going to subdivide it into smaller pieces. So taking this domain, we subdivide it into smaller pieces. Each of those pieces we want to be equal distance, right? So we want each of these sediments right here to be a delta x thickness. So we have an x one, we have an x two. You get the idea, we're going to have some xi in the middle, xi minus one, continue on all the way down to B. B of course is xn. We're going to have n subdivisions, B is the last one, and then xA as usual, sorry, xA is just x0 as usual. Now we're not going to be drawing rectangles under the curve per se, but what we want to do is we're going to use the subdivision along the x axis to subdivide the curve. So there's an A point on the curve. Let's get a point for x1. So we get x1 f of x1. We're going to get a point for x2. We're going to get a point for xi minus one for xi. For all of these points along the curve, we're going to get these points. So we get this like point p0, p1, p2. This would be pi minus one. This is pi, you know, keep on going, right? All the way down to pn, something like this. Now for the sake of illustration, I think I'm going to move things just a little bit. We're going to put, because again, it looks a little bit cleaner on the picture. I'm going to move pi to be right here, and this one's going to be pi minus one. So this is my pi minus one. Like I said, just slightly modifying the picture. This is going to be our xi, and we'll just clean this up right here. All right? No harm, no foul there. All right. Now the reason why I kind of moved this again is I wanted a little bit more space so we could clarify what's going on here. What I want to do is if you take two arbitrary points along the way, pi and pi minus one, we can estimate the distance between these two points just by using the usual distance formula, right? That is, what if we calculate, what if we calculate the linear distance between pi and pi minus one? Not the distance along the curve, but if we take the linear distance, that is, if we play connect the dots, we connect P zero to P one, P one to P two, P two to P three, and keep on going, right? If we draw straight line segments, and maybe my hand is too wiggly here, so let's actually do some straight line segments. So we do a straight line, a straight line, a straight line, straight line, straight lines. You get the idea here. Straight line, straight line, like so. We could approximate the curve using these straight lines. And we can kind of see, like in this illustration right here, if the points are close enough together, the approximation of using a straight line is actually not that bad. And the closer together we get, the better the approximations get. You might be able to see where we're going from here. So associate any two points, pi and pi minus one, we get this right triangle, like we saw before. And so we're gonna take this picture right here, and we're gonna zoom it, zoom in, right? So if we expand this picture a little bit, we're gonna get something that looks like the following. We have this right triangle, like so, with the points, right here. Here is our pi minus one. We have this point right here, pi. The function is doing something a little bit different than the line, right? It might be doing something like this. But the idea is for a small enough interval, the hypotenuse of this triangle will be approximately the same thing as the blue curve. So we see that with respect to pi minus one, the adjacent side is gonna have a length of delta x. The opposite side will have a length of delta y. And then the hypotenuse will have a length of delta s. So this triangle relationship we saw earlier, delta s squared equals delta x squared plus delta y squared. This is what we get from these little triangles right here. And in which case delta s is the type, this delta s is the arc length that we're trying to compute. Now again, the length between pi and pi minus one is only, delta x is gonna be approximation of s here. But as we take smaller and smaller delta x's, delta s will be smaller, smaller, smaller as well. And so we will see here, we're gonna see here that if we take the sum as i goes from one to n, we're gonna get the sum of these delta s's, delta si, we should put a subscript here because the delta s does depend on which interval we're at here. If we call s little s the true arc length of the curve, this will be approximately the same thing as this sum where i goes from one to n of the delta si. Now our goal is to be able to calculate an integral from this. And so we have a sum, this isn't quite a Riemann sum yet, but we're getting closer here. Using the substitution from the Pythagorean equation, this sum could be broken up as the sum from i equals one to n and then we take the square root of delta x squared plus delta y squared. Now again, we should probably put a subscript here on the delta y because delta y, the height will depend on where we are. Like with this trinora here, it's a much bigger delta y than like this one right there. It's just de-de-de-de. Delta x is gonna be uniform throughout this whole calculation, but the s and the y's do kind of depend on the location. So let's see, we will continue to make the assumption that delta x is constant for each subinterval, like we said before, but y and s, the delta y, the delta s are not necessarily, they'll depend on where we are in this segment. That's okay. It won't make much of a difference because as we take the limit of this thing, these things will, will, these differences will kind of go away. You know, to get the true arc length, which we're again, we're calling s right here, we need to take the limit, right? Because this is just an approximation. To get the true arc length, whoops, s, we need to take the limit as the number of subdivisions goes to infinity of this sum, try that again, of this sum i equals one to n delta si, which this will equal the limit as n goes to infinity of this sum i equals one to n, the square root of delta x squared plus the square root, oh, sorry, the square root of delta x squared plus the square delta y i squared, like so. That's what I meant to say. Now taking the limit of these deltas here, so the sum of the deltas, these can be written as integrals right here. So for the first one, the limit as n goes to infinity of the sum of delta si, this thing would be the same thing as the integral of ds. It has some specified bounds here. You know, there should be stuff going on there, but I'm not gonna worry about what those are right here. And likewise, the second one is to take the limit because basically we're using the fact here that as n goes to infinity, delta x is gonna approach dx, delta y will approach dy, and then in this case, delta s is gonna approach this ds. So s is equal to this integral of ds, but then we get, whoops, then we're gonna get the limit as n goes to infinity of the sum of the square root. This will look like the integral of the square root of dx squared plus dy squared, like so. And so this one on the right is starting to get, this one right here, is starting to get a little bit closer to integrals we're used to. It does have a dx and it does have a dy in it, but it has multiple, it has both of them. How does one deal with such a situation like this? Well, let's summarize what we've done so far because after all, our variable y, our variable y here, it was given as a function of f. So y equals f of x right here. And so we're gonna actually use this to our advantage. You might remember from Calculus I that if we take the expression dy over dx, this is equal to the derivative of a function. And this notation is used for a reason, right? These differentials dy dx, if you take their quotient, you do get a derivative because after all, derivatives are quotients. Limits of different quotients. And we're gonna use that to our advantage here. Taking back s equals integral of ds, which equals the integral of the square root of dx squared plus dy squared. What we're gonna do is looking at this expression dx squared plus dy squared, we're gonna factor out a dx squared, right? So we factor out this dx squared. Well, if you factor the dx squared away from the dx squared, you're gonna leave behind a one. If you factor a dx squared away from the dy squared, you're gonna get a fraction dy squared over dx squared. And so rewriting this one more time, you get the integral of the square root of one plus. Remember, well, if you take away the squares, you're gonna get dy over dx quantity squared. And if you take the square root of the dx squared, well, by usual arithmetic rules, this would simplify just to be this puppy right here. That we can calculate the arc length of our function by the arc length s is gonna equal the integral of the square root of one plus f prime f of x squared dx. That is, we can calculate the arc length by taking the square root of one plus the derivative of our function squared dx. And this gives us a nice formula for arc length for which we'll compute an example of this in the next video. So take a look for that one in the link you can see right now.