 In this video, we'll do one final example of computing arc length, and specifically we're going to calculate the length of the arc along the hyperbola x equals y, sorry, x times y equals 1 from the point 1, 1 to the point 2, 1 half, as is illustrated right here. Now with the equation x, y equals 1, we could solve for x or y. We could get y equals 1 over x, or we could get y equal, oh, I just did that one. We could do x equals 1 over y. And so in terms of formula, it really makes no difference whatsoever between solving for y or solving for x. The arc length formula can adapt to either situation. Remember s, the arc length, which is equal to the integral of ds. This is the integral of the square root of dx squared plus dy squared. Because of the symmetry of the arc length formula, we can comfortably integrate with respect to x or y. I'm going to choose to integrate with respect to x just because that's the format we're more used to. So we need to use the formula, the integral of the square root of 1 plus y prime squared dx. That's the formula we're going to use here. Now y prime, by the usual derivative rules, we're just using the power over right here, you're going to get negative 1 over x squared. And so when you square y prime, you end up with 1 over x to the fourth. And so for our arc length integral s is going to equal the integral, we're going to go from x equals 1 to x equals 2, because we're going to integrate with respect to x. Then we get the square root of 1 plus 1 over x to the fourth dx right here. And so if you scratch your head on this integral, that's not a big deal for us. This is a hard one, right? It turns out that this integral that we see in front of us has no elementary anti-derivative. Tries you might using some type of algebraic manipulation, u-substitution, integration of parts are some of the variations of those things. You're never going to find an anti-derivative that's going to turn out elementary. And basically what that means is for our purposes, we're going to need to approximate this thing. And so we're going to approximate this using Simpson's rule with 10 subdivisions. All right. And so remember, Simpson's rule, we could take, if we do it by hand, we'd calculate delta x, which would be 2 minus 1 over 10. That is one tenth or point one. And so then we have all of our x's, you know, so we have x zero, which is a one, then the next one would be a 1.1, 1.2, 1.3, et cetera, down to two. We then have to take delta x, which is one tenth over three times one time. We take one times f of one. And of course, when I say f right here, we have to be careful what we're talking about. In this context, the function f would be the square root of one plus one over x to the fourth, like so. So we'd be taking f of one with this function in mind. Then we'd be taking four times f of 1.1 plus two times f of 1.2, et cetera, right? So this is what Simpson's rule looks like. I'm just going to use a calculator for this one we've seen previously, a Simpson's rule calculator from the past. You can actually find a link to those things in the video description below. If we use a Simpson's rule calculator, we're going to see that this calculation turns out to be 1.132104, right? So use a calculator to get us an estimate. We can tell you how good of an estimate there is, but we'll be content with just finding this estimation here. Arc length calculations generally turn out to be very, very hard. And even for very benign-looking functions like this hyperbola, the associated integral that comes from the arc length formula is often non-elementary. That is, you're not going to be able to find an explicit antiderivative that avoids using calculus notation. That is, we can't use the funnel theorem calculus to compute this. Therefore, Simpson's rule, it's going to be our go-to move, or other numerical techniques are going to be necessary to do these things. So don't shy away from them. Feel free to use a calculator when it does ask for an approximate solution here.