Curvature of a cycloid

- [Voiceover] So let's do another curvature example. This time, I'll just take a two-dimensional curve, so it'll have two different components, x of t and y of t and the specific components here will be t minus the sine of t, t minus sine of t, and then one minus cosine of t. One minus cosine of t. This is actually the curve, if you watched the very first video that I did about curvature introducing it, this is that curve. This is the curve that I said, "Imagine that it's a road and you're driving "along it, and if your steering wheel gets stuck, "you're thinking of the circle that you trace out "as a result and it various different points "you're gonna be turning it various different amounts," so the circle that your car ends up tracing out would be of varying sizes. So if the curvature's high, if you're steering a lot, radius of curvature is low and things like that. So here, let's actually compute it. And in the last example I walked through thinking in terms of the derivative of the unit-tangent vector with respect to arc length but in this case, instead of doing that, I just want to show what it looks like when we take the explicit formula that looks like x prime times y double-prime, minus y prime times x double-prime, and then all of that divided by x prime squared plus y prime squared. And then I'm writing x prime and y prime and such and all of these you should think of as taking in the variable t. I'm must being a little too lazy to write it. And you take that to the three-halves power. So, this was the formula and I'm not a huge fan of memorizing formulas and then hoping to apply them later. I really do think the one thing you should take away from curvature is the idea that it's the derivative of the unit-tangent vector with respect to arc length and if you need to, you can just look up a formula like this but it's worth pointing out that it makes some things easier to compute because finding the tangent vector and everything can be kind of like reinventing the wheel when you already have the results here. So, first thing to do is just find x prime, y double-prime, y prime and x double prime. So let's go ahead and write those out. So the first derivative of x of t if we go up here that's t minus sine of t. So its derivative is one minus cosine of t. And the derivative of the y component of one minus cosine t, y prime of t, is gonna be, derivative of cosine is negative sine so negative derivative of that is sine, and that one goes to a constant, and then when we take the second derivatives of those guys, so maybe change the color for the second derivative here, x double prime of the, so now we're taking the derivative of this, which actually we just did because by coincidence the first derivative of x is also the y component so that also equals sine of t. And then y double prime is just the derivative of sine here so that's just gonna be cosine, cosine of t. So now, when we just plug those four values in for kappa, for our curvature, what we get is x prime was one minus cosine of t, multiplied by y double prime is cosine of t. Cosine of t. We subtract off from that y prime, which is sine of t, sine of t, multiplied by x double prime. So x double prime is also sine of t, so I could just say sine of t squared, and the whole thing is divided by x prime squared so x prime was one minus cosine of t, minus cosine of t, squared, plus y prime squared so y prime was just sine, so that's just gonna be sine squared of t. And that whole thing to the power three halves. And that's your answer, right? You apply the formula, you get the answer. So for example, when I was drawing this curve and kind of telling the computer to draw out the appropriate circle, I didn't go through the entire "find the unit tangent vector, "differentiate it with respect with arc length" process, even though that's decently easy to do in the case of things like circles or helixes but instead I just went to that formula. I looked it up because I had forgotten and I found the radius of curvature that way.