 In this video, we're going to calculate the length of the arc of the parabola y squared equals x from the point 0, 0 to 1, 1. Now you'll do notice that yes, this is y squared equals x, right? This is a concave right parabola. We want to go from 0, 0 to 1, 1. So we actually only need the upper half of the parabola. What we can do is we can solve for y here and get y equals the square root of x. Normally, there should be a plus or minus here when you take the square root. But like I said, we just want the upper half, which is the positive side. So we'll just keep it as y equals the square root of x. And this will be our function describing what's going on here. We want to find the length of the arc, a.k.a. the arc length. And so therefore, by the arc length formula we saw before, you get s equals the integral of ds, which equals the integral of the square root of dx squared plus dy squared. And you can go from there. We saw previously that if you factor out the dx squared in this equation right here, this will end up with the integral of the square root of 1 plus f prime of x squared dx. You can integrate this thing with respect to x if you want to, right? And that works nice and dandy. But in this situation, our function f would be this guy right here. We could take the derivative y prime to be 1 over 2 times the square root of x. We could plug that in there. That would be perfectly nice and kosher, but let me mention to you that I keep on throwing this formula out here because it's actually quite versatile. We went from here to here because we factored out the dx squared. Well, what if we factored out instead the dy squared? What if we did that instead? What would happen is your integral would look something like the following. If you factored out the dy squared away from the dx squared, the dx over dy quantity squared plus 1. You think it's square root of dy squared, which this thing would simplify to be the following. You're going to get 1 plus g prime of y squared dy. That is to say you could integrate with respect to y just as easily as you could integrate with respect to x. And because of that, you can actually make a choice. Do I prefer to integrate with respect to x or do I prefer to integrate with respect to y? And in this option here, if we take y to equal to square root of x, they're going to have to deal with a 1 over 2 square root of x inside of a square root. So it's a nested square root. It's like, hmm, I'm not sure I like that one so much. On the other hand, if we just take the original expression and say that g of y, which is x, is equal to y squared, that situation seems much more tame to me because g prime of y would equal just 2y. And again, that doesn't seem so horrible whatsoever. And so actually this format where we're going to integrate with respect to y is going to be a little bit more preferable here. And so the thing is I emphasize this so that you know there's an option. You can choose do I want to integrate with respect to y or integrate with respect to x very easily with arc length. We ran across this problem we dealt with volume problems in the past where we had two methods, the washer method and the shill method. The reason we had two methods is so that we could always integrate with respect to x or y. You know, we could choose the variable that was best for the situation. We don't need two formulas for arc length because the arc length formula symmetric in that regard, it can be easily adapted to dx or dy. So our arc length is going to look like the integral for the square root of 1 plus 2y squared dy. Our boundaries are going to be y-coordinates y equals y equals. Going back up to the original picture right, y will range from y equals 0 to y equals 1. Good news here if you actually didn't wrote the x-coordinates you would be right for the wrong reasons. That's still correct. I guess that's good news, right? So we're going to integrate from 0 to 1 right here. So rewriting this integral because we can square the 2y, integrate from 0 to 1, the square root of 1 plus 4y squared dy. And we've seen integrals like this before. This is actually one where a trigonometric substitution might be quite appropriate here. We could set, because notice we have a sum of squares inside of a square root, we could set 2y, 2y equal to tangent theta. This would then tell us that 2dy would equal secant squared theta d theta. So solving for dy we get dy equals just 1 half secant squared theta d theta. We also know that the square root of 1 plus 4y squared is going to equal a secant theta. So make those substitutions. You can draw a triangle to help you out on this last one if you want to. But using those substitutions, we're going to see that we're going to integrate the square root of 1 plus 4y squared becomes a secant theta. We then get for dy 1 half secant squared theta d theta, like so. If we change the bounds as y ranges from 1 and 0, what's the corresponding theta values? We'll plug this into this equation right here. When y equals 0, the left-hand side becomes 0. So we get tangent of 0. So sorry, tangent of theta equals 0. The arc tangent of 0 gives us a 0. So that's nice right there. On the other hand, if you plug in a 1 into this equation right here, you're going to get just a 2 on the left-hand side. And so when does tangent theta equal 2? That one might not become to us immediately. That's just going to be arc tangent of 2, which again, that's not one of those special angles we had memorized from our trig class. For the sake of brevity, I'm just going to call this arc tangent of 2. I'm going to just call this alpha for short. We'll know that it's arc tangent of 2. But just for, again, simplicity's sake, our integral range from 0 to alpha right here. So notice what our integral has now become. This thing is now one-half the integral from 0 to alpha of secant cubed theta d theta. And so we can proceed to compute this anti-derivative like we've done with our trigonometric integrals in the past. If secant cubed might look a little bit familiar to you because we actually did do this one in a video previously, it was actually someone of a doozy. We're not going to do it again. If you are curious how the calculation goes, please view that video. But we're just going to use the conclusion we had from that previous calculation that the anti-derivative of secant cubed is going to be one-half times secant theta tangent theta. Plus the natural log of the absolute value of secant theta plus tangent theta. Oh boy, ran out of space there. And this will go from alpha to 0 right there. And there's an absolute value showing up right here. So again, there's a little bit of a calculation going on, a little bit's an understatement. So there is a difficult calculation there that we're kind of scurrying over because we've done it before. So the one-halves will combine to give us a one-fourth. That's a coefficient that sits out in front. Let's stick these numbers in everywhere else. So we're going to get secant of alpha times tangent of alpha. Then we're going to get the natural log of the absolute value of secant of alpha plus tangent of alpha. That's the first bit. Then we're going to subtract from this secant of 0 tangent of 0. Also we need to subtract from this the natural log of the absolute value of secant 0 plus tangent of 0. Like so. So that's a little bit of a doozy, but let's try to simplify this thing. We're going to start off with the 0 portion first. So remember, tangent of 0 is equal to 0. We saw that already. So you get a tangent 0 there. So this whole thing is going to go to 0. You're going to get tangent of 0 right here. So that's also 0. Secant of 0 is not 0. So don't make that mistake here. Remember, secant, if you're not sure, you can just consult the calculator, but secant of 0 will equal 1 over cosine of 0. And hopefully we do remember this one. Cosine of 0 is 1, so you get 1 over 1, which is 1. So because if we had the natural log of 0, that would be a no-no. Natural log of 0 is not a real number. But secant of 0 is actually going to equal 1. This actually then gives us the natural log of 1, which natural log of 1 is 0. So it does turn out that all of these portions, this whole right-hand side here, is all going to go down to 0. So what else do we have here? We're going to have a 1 fourth, a secant of alpha. Remember, what was alpha? Alpha was arctangent of 2. Now, when you take tangent of arctangent of 2, that one's pretty clear. That's going to be a 2. And when you look at the second one, you're going to get the natural log of secant of alpha again, right? Plus 2. Tangent of alpha is not so bad, but how does one do secant of alpha? If you had a triangle diagram listed earlier, well, I guess we didn't need one for this one. But let's draw a triangle real quick to kind of see what's going on here. So if we have a right triangle associated to the angle alpha here, right? What do we know about alpha? Alpha is arctangent of 2. So we know that tangent of alpha equals 2. That is 2 over 1. That's our opposite over adjacent. And so by the Pythagorean equation, we get the square root of 5 over here. 1 squared plus 2 squared is 5. So this is our right triangle. And so if we look for the secant ratio here, secant of theta, sorry, secant of alpha. Well, secant is 1 over cosine if you forgot. So cosine is adjacent over hypotenuse. So secant will be the reciprocal. You're going to get the square root of 5 over 1. And so therefore secant of alpha is equal to the square root of 5. We can make that substitution in above. So we're going to get 1 fourth secant of alpha is the square root of 5. You get 2 there. So you get 2 root 5. Then you're going to get the natural log of the absolute value of 2 plus the square root of 5. Like so. Now 2 plus the square root of 5 is already a positive number. You can distribute the 1 fourth through if you so choose. And so to finish this thing off, we're going to get the square root of 5 over 2 plus 1 fourth the natural log of 2 plus the square root of 5. Like so. And so again, we get this interesting irrational number as the length of the curve along this parabola, right? And this would be approximately, again, we probably won't need an estimate to know how big this is. We get an approximation of 1.478943. Like so. This is a pretty nice estimate here. And we get this pretty cool number. This calculation here did require a trigonometric substitution. It was a little bit harder than the previous example we did. But this is starting to show you that arc length formulas can get very complicated, that is the arc length integral can get very complicated very quickly when it comes to trying to calculate these arc lengths here. So you have to be very cautious. The square root of 1 plus the functions derivative squared. That is our formula, but it doesn't generally lead to a simple anti-derivative. And we'll see that in our next video that things can even get worse than we are right now.