Welcome to our video lecture on Arc Length. This is a nice little break from what we've done. It's one of the shorter and simpler sections that we're going to do in the entire course. So let's get right to it. Our goal here is, we are given a function. Let's say it starts at some point A and ends at some point B. We have some graphs from curve. We'd like to measure how long this is. We'd like to know the distance and object travels how far the rocket ship travels. If it were a straight line, we'd have the distance formula. However, it is not a straight line. So we call this distance arc length. What is the length of the arc? But don't confuse it, it is the distance traveled if this thing represents time, so if it were a line, it's the distance formula. Now you have some curve defined by some function, it's doing whatever it does. We have to find this exact length and to do that. Back in the day, percale algebra was enough to get the distance formula. But now we need some calculus. So the idea of this function is that we're going to try to break just give you a little idea of where this come from all those more important tha the actual formula. You take two points and you imagine that they're really close. You lay down like these toothpicks, you lay down these toothpicks that are flat along with it, right? So you assume these two close points, you have toothpicks. And then the idea of well these toothpicks are straight. So like Riemann sums when you start with an easier object, and then you know, you calculate the total. We're going to start with toothpicks and lemon law. Now, of course, they're going to give us an error. Okay? But with these little toothpicks, I can use the distance formula because each little toothpick is going to give me a little baby triangle. And I can use the distance formula and the base of that triangle is delta x. The height of that little triangle is delta y. And then the length of the segment the length of the toothpick that I just lay down, which we'll call L, is going to be by the Pythagorean Theorem, delta x squared. Plus was really L squared plus parentheses, delta y squared. So you lay down to pick you draw a little baby triangle, and you compute these things. And now we want to finally delete the entire curve so I want to add all these little to fix. How many do I have usually the smaller they are, the more I have this should feel like Riemann sum Little bit. So I want to sort of like sum, right? So I'm just going to write like an integral. But really, this is like we're going to sum from a to b, these lengths, so l, but I want a square root of one L squared. So I have the square root of x, or delta x squared plus parentheses, delta y squared. And then we'll do all these things. So this is the idea, right? So we can write, we can do some manipulation to this. So I'm going to write this as the integral from a to b, but really, I'm just still something's like infinitely thin toothpicks, as many as I can lay down here. I'm going to factor out an x squared under the square root. So I'm going to pull out a delta x squared. Okay, and when I do that, I'm left with a one in the first place. And then I get delta y over delta x, and then that quantity is squared. So I've factored out the delta x squared. And then if I simplify square roots break up over multiplication, and what that does is that leaves out a delta x squared of delta x squared puts a delta x out here. So we're still like summing these things up. And I'm left with 1+ now what's dy over dx? That turns into like, well, let's write up in a more familiar form, dy over dx squared. So when I finally like take the limit of these lengths, these toothpicks go to infinity and my delta become these and all these good old like in the limit. I actually get the true integral from a to b of one plus now dydx is the derivative if you could write a dydx, but you tend to see it with f prime of X. This whole derivative is squared and then DX. So this is our formula. This is our formula for arc length. And it has the same flavor and feel of Riemann sums and in sort of approximating and then taking a limit to get the exact answer. In this case, we're laying toothpicks down and finding their thing. So we have arc length and in our case, this is the formula to know. The idea behind it, it's nice to have an intuition behind it, but it's more important to have oops, the formula. Okay, and now once you have the formula, that's all you need to know for this section. After you have that we go off and use the formula. Let's before we do an example, let's just think of the steps for a minute. What do we need? So step one, we have to find f prime. And then two, I have to square it. So I have to like take this thing, whatever it is, I have to square it. And then I use the formula to set up the integral a to B. Usually a and b are given, maybe there's like just some extra stuff I got to do here so you got that. So then I put it in the formula. And I use all my prior steps. And then I integrate. So 123. Okay, ready? So keep this formula handy. We're going to go do an example. Here we go. Let's do an example. Let's find the arc length of the function. Let's say f of x equals x cubed over six plus one over 2x. This is some graph and I'd like to know what's going on how far. Far does the object travel what is the length of the curve from one to two? If you were to graph this thing from one to two, if you plug in one, you get some number here if you plug in two goes up a little bit. So you have some bendy part, what is the length of this curve? Okay, so let's go through our steps. First thing first. Compute. The derivative. When you do that, you get x squared over 2, minus now be careful here. There's a minus one over 2 x squared. Fantastic. Step one, find the derivative. Sometimes that's easy, more tricky, but dude, you got to do step two. Square the derivative, take this whole derivative you just found and square this thing. This is algebraic in nature. So I want to find x squared over two minus one over two x squared squared. There's a couple ways to do this. But I think the easiest way to do it is sort of clean up the fraction first, I think let's subtract fractions. What do you do that because they're so close to having a common denominator, let's put x squared on bottom of both. In that case, you get so two x squared as your common denominator, and then you get x to the fourth minus one. So that's the whole thing squared and now the reason why I do that because instead of foiling. I square the numerator and I square the denominator, the square sort of hits both pieces, square the numerator that's a foil right. So that's actually a minus two x to the fourth, and then plus one square the nominator and that's four x to the. Okay, so now we're going to go ahead and use our. Formula. So the arc length call it l from one to two integral from one to two of, I'm just going to write the formula down. So I get some pity points in case everything goes terribly wrong. It becomes from one to two of the square root of one plus, and then x to the eighth minus two x to the fourth plus one over four x to the fourth. It's all under the square root. And here we go. Okay. So once we have this, I have to do some algebra. I really need to simplify this. I don't want to integrate this as is, I got to do some algebra. So why don't I put another like common denominator? All right? Four, all right one as four x to the fourth over four x to the fourth. So let's write this again. Everything is happening under the square root, that's okay. So you get x to the eighth. Then you have four to the fourth minus two x to the fourth, that's plus two x to the fourth. Plus one, and then my common denominator is 4x to the fourth. Now, why is this nice stare that numerator for a second? Do you notice anything? That numerator factors, the numerator factors, it's x to the fourth plus 1 squared. And that denominator. You can also write that as 2x squared, squared and that all lives under a square root. Look for this on problems, look for algebra where you can write it as something squared under square root that will clean up nicely. And then of course the square root of stuff squared is just the stuff you started with. So it's the integral from one to two of x to the fourth plus one over 2x squared. D. Okay, this is good. Now you just have a simpler integral without the square root. Of course, you still have to solve this thing. To remember how to solve this thing was the trick for this one. Stare this for a minute. Do you see it? Do you see it? Split the fraction up. So right it is x to the 4th over 2x squared + 1/2 x squared dx. So don't try anything fancy here. We're using algebra the whole way through, and then this cleans up quite nicely. I'm going to run our room here, from 1 to 2 of x squared / 2. And I'm going to write, this in a way that we can use the power rule. So let's, write this as x to the minus two. So now I have a nice easy integral that I can evaluate and we're going to use the power rule on both pieces. X cubed minus six over one on one half, one over two X, evaluated from one to two. And in the interest of space, if you plug this in do the arithmetic, you get 17 over 12, which is about 1.4. So the length of this curve is 1.4, to the units of the line for me the units of the x axis is given in whatever that is, but you can just leave the number as 1.42 or 17 over 12. But the key to this is it, usually looks pretty bad because you have all this like square root stuff going on. But if the questions are cooked up correctly, and they usually are for this section in particular, then look for things to work out to be squares, look for things that will clean up nicely algebraically. Don't get scared if you're staring down the barrel of some horrible, horrible in a row here. Okay, let's do one more. So find the arc length fell always tell you to find the arc length and find the distance. They're very clear when it's this kind of problem of y equals the natural log of secant of x and will travel between zero and pi over four. Okay, so let's go find our function or derivative here. So step one says find the derivative. logarithm says one over, so we'll do one over secant and then times the derivative of secant which is secant x, tangent of x. See cans cancel and we're just left with tangent of x. Okay, that's kind of nice. I have to square this thing. So notice by the way, the notation is now y prime. That's okay. So this just becomes tangent of squared 10 squared of x. And we can set up our integral. So the arc length is going to be the integral from zero to pie over four of the square root of one plus derivative squared dx. Plug it all in and then we'll worry about the integral but that's okay because that's familiar territory. We like that. And I have one plus tangent squared of x. dx. I've never said like, look for stuff to happen under that square root. What is 1 + tangent? Do you recognize that identity, now it's come with a bunch. That is the square root of the secant squared Have x dx. And then because we're on zero to pi over four that works out to of course this where two secant squared, good old secant of x. Now if you remember secant of x is one of our sort of classically difficult ones. This is one that you have to multiply by a tricky value of one to get that we have this one. We've done this one. Look back at your notes if you're not sure about it, but we have this one. It's the natural log of the secant plus the tangent. Notice, by the way, how quickly this just turned into an interval problem. So the the whole like new thing of just using the formula is just that it's just using the formula and that's why we like these because, well, it's just one more formula. We can handle that. Anyway, plug in, do your thing. And in the interest of time, you can check me you get natural log root 2 plus 1, some number. So the length of this curve from that segment is natural log root 2 plus 1. Okay, we'll keep this video short and sweet. It is a new formula, you have to know square root of 1 plus the derivative squared. And, going from A to B. This is the arc length. Lots of examples and we'll see you next time.
