 So recall the notion of the cycloid that we had introduced before. So if a cartwheel is spinning and a fixed point on that circle is determined, what path does it trace throughout space? You can see that path right here illustrating green it makes these arches over and over and over again. We had saw previously that we can parameterize the cycloid using the equations x equals r times theta minus sine theta and y equals r times 1 minus cosine theta. So theta is our parameter here in play. And so what I want to do is do some type of curve sketching analysis of the cycloid using this parameterization, right? So can we find the tangent line of the cycloid at the point pi equals or sorry theta equals pi thirds. So at the parameter pi thirds, what would the tangent line look like? And so we first begin by computing the derivative. Remember y prime, which is just shorthand for dy over dx. This is equal to dy over dt, or I guess in this case it would be theta, divided by dx over d theta. So taking the derivatives here, the derivative of y with respect to theta, we're going to treat r as a constant. The radius of the circle doesn't change throughout this problem. We're going to get r times the derivative of 1 is 0, the derivative, so I just will disappear, the derivative of negative cosine will be a positive sign. So you're going to get an r sine theta on top. For the denominator, you're going to get r times the derivative of theta is 1 and the derivative of negative sine will be a negative cosine. So you're going to get 1 minus cosine theta for x on the bottom. The r's cancel out, in which case you get y prime is sine theta over just 1 minus cosine theta. So then if we evaluate y prime at pi thirds, we're going to plug in pi thirds in here. We get sine of pi thirds, and then 1 minus cosine of pi thirds. Sine of pi thirds, that's going to be root 2, or sorry root 3 over 2, and right here, root 3 over 2. For cosine, it's just a one-half. So 1 minus one-half is, of course, is one-half, and then simplifying this this compounded fraction here, we get root 3 over 2 times pi the reciprocal 2 over 1. 2 over 1, the 2's cancel, and we just get the square of 3. That is the slope of the tangent line at that point. So the equation of the tangent line will look like y minus k, the y-coordinate at the point of tangency. So we should actually, we can write this as our function g at pi thirds. Right? So this function right here is our g of theta, and this one right here is our f of theta. So we have to compute y minus g of pi thirds. This is equal to y prime at pi thirds times x minus f at pi thirds. Some of these we already know, of course, so the derivative is going to be the square of 3 times x. If we plug in, if we plug in pi thirds into the function g, that's one, this one right here, you'll end up with cosine of pi thirds as we saw before. That's a, that's a one-half, one minus one-half is one-half. So you're going to get r-halves over there, y, y minus r-halves. Like so. And then for the x-coordinate, we have to put pi thirds into this guy over here. Sine of pi thirds is root 3 over 2, theta will just be pi thirds. So you're going to get minus r times pi thirds minus root 3 over 2. That's a delicious irrational number. And this right here would give us the equation of the tangent line. Now, if you don't like this because it's not in slope intercept form, then of course you solve for y and you end up with y equals square root of 3 times x minus r times pi over the square root of 3 minus 2. Is that right? That seems a little bit funky. How did the square root of 3 get in the bottom there? I think I must have distributed something. Oh yeah, that's because when you distribute the square root of 3 onto this part. Square root of 3 on top can simplify with the square root of 3. The 3 on bottom gets the square root of 3 on the bottom. So no, that's legit. That is the real deal, everyone. And so it's, you know, we can approximate it, but that's what the tangent line would look like, or the equation of it. And we could draw it above. Working on one that's a little bit easier. What if we looked for horizontal and vertical tangent lines? Like we mentioned before, the horizontal tangent lines will coincide when dy over dx is equal to 0. That'll happen when dy over d theta equals 0. And likewise, the vertical tangent lines will be when there's a 0 in the denominator of dy over dx, aka dx over d theta equals 0. So using the derivatives that we had calculated before, so dy over dt, remember this is equal to r sine theta from above. We want that to equal 0. And for the vertical tangents, we have to take dx over dt, set that equal to 0. That was an r times 1 minus cosine theta. We want that to equal 0. And so we'll work with the horizontal tangent lines first. Divide both sides by r. We get sine theta equals 0. And so when is sine equal to 0, that would happen when sine is 0. It happens at pi. It happens at 2pi. It happens at 3pi. Or more specifically, this will happen at k multiples of pi. Because this also includes negative numbers as well. All right. Let's see. I'm sure I wrote that down correctly. And we have r sine right there. And then for the other one, if we have r times 1 minus cosine theta, that should equal 0. Divide both sides by r. We get 1 minus cosine theta is equal to 0. So we want to know when cosine theta is equal to 1. When is cosine theta equal to 1? That occurs when, think about our unit circle, cosine starts off at 1. And that then cosine theta starts off at 1. So you get 0pi, 2pi. No, I take that back. Cosine is equal to 1, not at negative pi, but at 2pi and 4pi. So you get things like that. So when theta is a multiple of 2pi, so 2pi k, that's when you will get vertical lines. I see. I guess the issue that I forgot to mention earlier is that when we're looking for a horizontal and vertical tangent, we need the numerator to be 0. But we also need that dx over d theta is not 0. They can't both simultaneously be 0. And then to find a vertical tangent, we have to investigate when we want the denominator to be 0, but the numerator not to be 0. And so there's actually a little bit of a mix up right here. The denominator will be 0 at multiples of 2pi. The numerator will be 0, like we said before. So when we set r sine theta equal to 0, we got multiples of pi. And so this is any multiple of pi. So this is even or odd multiple of pi. So in particular, there's an overlap, right? We're going to get something that looks like 0 over 0 at even multiples. Even multiples, you know, these 2pi k that we saw over here. So I mean this k pi, it could be broken up into two pieces. There's going to be odd multiples. So we get things like 2n minus 1 times pi versus just 2n pi, like so. And so these ones are completely kosher, right? So because there's no overlap right here, these ones will definitely be horizontal tangent lines in that situation. But what happens at the even multiples of pi, it turns out that a L'Hopital argument is going to be necessary for this situation. So we're looking at the limit as say theta approaches an even multiple of pi, right? It's going to approach an even multiple of pi. We're going to take then sine theta on top and then 1 minus cosine theta on the bottom because the r's cancel out this time. And so again, if you just plug in 2n pi, you're going to end up with 0 over 0. So by L'Hopital's rule, we take the derivative from top and bottom. We're going to take the limit now. We're going to get cosine on top and then we get sine on bottom. Again, now we're approaching still 2n pi, an even multiple of pi. On the top, we will end up with a 1. On the bottom, we get 0. And so 1 over 0 here represents we have some type of vertical asymptote. Really what this is describing here is that the 0 and the bottom wins the fight. And so it turns out that for even multiples of pi, there aren't horizontal asymptotes, not horizontal tangents. They're actually vertical tangent lines. And so we come back up to our graph. This actually matches up with what we see that if we look at odd multiples of pi, so we take pi, 3 pi, negative pi, we end up with horizontal tangent lines on our cycloid. And when we take even multiples of pi, like 2 pi and 0, we actually get vertical tangent lines. So again, using our curve sketching type analysis we learned in calculus one, we can use the derivative to analyze parametric functions such as the cycloid, which be aware the cycloid you look at this thing, it passes the vertical line test. We could describe this function using y equals a function of x. It's just if you're if you're daring enough, go look it up online sometime. It is a beast. It's very difficult to use. This parameterization that you see on the screen right here is much, much more natural to use. But you have to be used to doing calculus with parametric functions. How would we be able to find these tangent lines like we did here? We could do it. We could also identify and show that this graph is always concave down by taking the second derivative and showing that it's always positive. I won't do that right now, but I'll challenge you that you should try to do that after this video is over. And this is going to conclude lecture 30 for us in our series. In the next video, we're going to talk a little bit about how integration is affected when we work with parametric curves. So stay tuned for that one.