 Welcome back everyone. In the previous video we talked about how we can adapt area under the curve problems for parametric curves And then how this really seems like it's just a special type of trigonometric substitution or other types of substitutions We want to do a similar thing for arc length And so remember from arc length we saw before that arc length is s is equal to the integral of ds not a dx there Sorry about that In which case ds as we certainly we learned before from the from the Pythagorean equation Ds is equal to the square root of dx squared plus dy squared And so when we dealt with this in the case of Cartesian functions We learned that if we factored out the dx squared of the dy squared we could get the following situations I guess I can just put it right here. This can be written as the square root of 1 plus dy over dx Squared dx or we could write this as the integral from the square root of dx over dy Square plus 1 dy Like so so by factoring out the differentials dx squared of dy squared We could adapt this so we could integrate with respect to x or with respect to y Now if we're in the parametric setting we want to integrate with respect to dt, right? We want to integrate with respect to t and so the idea is can we take this this expression and factor out the dt? Well, if you do that when you factor out dt squared away from I guess we should factor out a dt squared We're going to factor that away from the dx squared in which case we're going to get dx over dt squared And then you're in fact that away from the dy Squared so you get a dy of dt squared and then you take the square root of dt square you get a dt So ds can also be adapted for this parametric setting in a very very simple way. It's really really nice And so let's apply this to some parametric curves We've seen before take the parameterization of the circle. We've seen many times x equals r cosine theta y equals r sine theta Let's adapt that let's adapt it to this setting so the r-length s We're going to integrate one complete cycle so zero to two pi That's our parameter there theta And so we're supposed to take the square root of Dt so we're going to take the square root of the derivative of x squared plus the derivative of y squared And so if we do this the derivative of x squared we're going to get a negative r sine theta We took the derivative of our Cosine, but we also want to take the derivative of our sine with respect to theta and So we're going to get our cosine Theta squared and so trying to simplify this thing. We're going to end up with Zero to two pi inside the square root. We're going to get r squared sine squared theta plus r squared cosine squared theta Take the square root dt So we can factor out the r squared and So they're going to get a sine squared theta plus cosine squared theta This is side side the square root. We're integrating from zero to two pi Dt and sine squared sine squared plus cosine squared. We've seen that before we know what that is. That's equal to Three. Oh, I'm just kidding. That's equal to a one, right? And so you're going to get the integral from zero to two pi you get the square root of r squared dt That'll just become the r squared the square root of r square will just become an r and So you integrate the specter t you're going to get our t from zero to two pi and So that's going to equal two pi r the usual Formula for there's the circumference of a circle and we did this example previously when we did our claim We calculated the circumference of a circle. We did that in Cartesian coordinates. This is basically the same thing It's just in Cartesian coordinates because you have to take the you have to take an integral the square root in it That leads itself very naturally to a trigonometric substitution and that trigonometric substitution without realizing it is a Parameterization every trig sub that we have done in this series is a special type of parameterization You're parameterizing a function using trigonometric functions like we've seen many times in this chapter already Let's look at another function. We know very well the cycloid would be keep on seeing it It has a very nice trigonometric Parameterization the x-coordinate is given as r times theta minus sine of theta y is given as r times 1 minus cosine theta Can we find the length of a single arch of a cycloid which would be this arch right here? So one arch will correspond to theta will range from zero to two pi and So by the formula the arc length s will equal zero to two pi We have to take the square root of what we're going to take the derivative x with respect to theta That's going to be r times 1 minus cosine theta That's squared we add to it the derivative of y which is gonna be r Times sine theta squared d theta and So multiplying out the things inside the square root We're gonna end up with An r squared times 1 minus 2 cosine theta Plus cosine squared theta and then we're also going to get an r squared r squared times sine squared theta this all sits inside the square root D theta So some observations to make here notice that everything is divisible by r squared We can factor that out and then take the square root of r squared That's gonna give us an r that's in sits in front of this thing r times the integral from zero to two pi Then we're gonna get the square root of 1 minus cosine squared theta plus Cosine squared theta plus sine squared theta This one we know and love Cosine squared plus sine squared of course is equal to one So this right here is just equal to one which we can combine this one with the other one Oh, there should be a two right there One plus one is two and so notice now we have a two. Well, I'll write it out We then have our integral from zero to two pi We're gonna have the square root of two minus two cosine theta D theta So there's a factor of two which we can factor out from both of these things and since it's inside of square root We'll actually be a square root of two so factoring that out We get r times the square root of two times the integral from zero to two pi Square root of one minus cosine theta d theta Like so and so now we have to deal with this one the square root of one minus cosine There's a couple ways one could try to handle that it turns out the half angle identity is going to be a saving grace in this situation Because what we can do is we can actually replace I mean because from what we've seen before We've been using a lot the following identity one half one minus cosine Of two theta equals sine squared theta. We use this all the time. This is the so-called half angle identity Now if we make a slight modification If we cut the angle in half we get theta That'll cut this angle in half So we get theta Over two which is why it's usually called the half angle identity and if we times both sides by two We end up with the following one minus cosine is equal to two sine squared theta Over two and so if you make that substitution in right here, you get a very nice observation in which case you're gonna end up with r root two times integral from zero to two pi of The square root of two times sine squared theta over two D theta and so this is going to give you another square root of two Which will combine with this square root of two just to give you a genuine two So we get to our integral from zero to two pi and then you take the square root of the sine square You're gonna get sine of theta over two D theta and that's a pretty nice integral to handle. I think I can handle that one anti-derivative of sine Right, you're gonna get a negative cosine of pi over two But we have to divide by that prince of the period change so we actually get negative two cosine of pi halves So we go from zero to pi halves right there And so plugging these things in you're gonna get a negative for our times when you plug in pi halves Sorry, if you plug in two pi into the theta over two you'll get a pi cosine of pi is a Negative one right and then you're gonna subtract from it when you plug in zero cosine of zero is a one So we're gonna get a negative two right there. And so we end up with eight r and This gives us the circumference of one arch of our cycloid and this one I always find really fascinating Right, we've now discovered that the arc length of this arch right here is is eight r It does depend on the radius of the circle It's eight times the radius are four times the diameter of the cartwheel if you prefer and doesn't involve pi whatsoever It seems kind of weird that things that are round need to have a pie in it But the circumference is gonna just be Eight times the radius here pretty cool, isn't it and we can find out arc length of other parametric curves in a very similar manner We just have to adapt the DS so that it integrates with respect to our parameter and we're good to go