Curvature of a cycloid

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. 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 watch 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. If your steering wheel gets stuck, you know you're thinking of the circle that you trace out as a result. At various different points, you're going to be turning at various different amounts. So the circle that your car ends up tracing out would be of varying sizes. If the curvature is high, if you're steering a lot, radius of curvature is low, and things like that.

So here, let's actually compute it. 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 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 just being a little too lazy to write it, and you take that to the three halves power.

So this was a formula, and I'm not a huge fan of like 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. 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.

Finding the tangent vector and everything can be kind of like reinventing the wheel when you already have the result here. So the 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. The first derivative of x of t, if we go up here, that's t minus sine of t, so its derivative is 1 minus cosine of t.

The derivative of the y component of 1 minus cosine t, y prime of t, is going to be derivative of cosine, which is negative sine, so negative derivative of that is sine, and that one goes to a constant. Then, when we take the second derivatives of those guys, maybe change the color for the second derivative here.

x double prime of t: 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. Then y double prime is just the derivative of sine here, so that's just going to 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 1 minus cosine of t multiplied by y double prime is cosine of t. We subtract off from that y prime, which is 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 1 minus cosine of t squared plus y prime squared. So y prime was just sine, so that's just going to be sine squared of t, and that whole thing to the power three halves, and that's your answer, right?

That 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 to arc length" process, even though that's, you know, 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.