Curvature formula, part 2

In the last video, I started to talk about the formula for curvature. Just to remind everyone of where we are, you imagine that you have some kind of curve in, let's say, two-dimensional space, just for the sake of being simple. Let's say this curve is parameterized by a function ( s(t) ), so every number ( t ) corresponds to some point on the curve.

For the curvature, you start thinking about unit tangent vectors at every given point. What does the unit tangent vector look like? The curvature itself, which is denoted by this sort of Greek letter ( \kappa ), is going to be the rate of change of those unit vectors—kind of how quickly they're turning in direction, not with respect to the parameter ( t ) but with respect to arc length ( ds ).

What I mean by arc length here is just a tiny step. You could think of the size of a tiny step along the curve would be ( ds ), and you're wondering, as you take a tiny step like that, does the unit tangent vector turn a lot or does it turn a little bit? The little schematic that I said you might have in mind is just a completely separate space where, for each one of these unit tangent vectors, you go ahead and put them in that space saying, "Okay, so this one would look something like this. This one is pointed down and to the right, so it would look something like this. This one is pointed very much down."

You're wondering, basically, as you take tiny little steps of size ( ds ), what does this change to the unit tangent vector? That change, you know, it's going to be some kind of vector. Because the curvature is really just a value, a number that we want, all we care about is the size of that vector—the size of the change to the tangent vector as you take a tiny step in ( ds ).

Now this is pretty abstract, right? I've got these two completely separate things that are not the original function that you have to think about. You have to think about this unit tangent vector function and then you also have to think about this notion of arc length. The reason, by the way, that I'm using an ( s ) here, as well as here for the parameterization of the curve, is because they're actually quite related, and I'll get to that a little bit below.

To make it clear what this means, I'm going to go ahead and go through an example here where let's say our parameterization with respect to ( t ) is a cosine-sine pair. So we've got ( \cos(t) ) as the ( x ) component and then ( s(t) ) as the ( y ) component. Just to make it so that it's not completely boring, let's multiply both of these components by a constant ( r ). What this means—you might recognize this cosine-sine pair—is that in the ( xy ) plane, you're actually drawing a circle with radius ( r ).

So this would be some kind of circle with a radius ( r ). While I go through this example, I also want to make a note of what things would look like a little bit more abstractly if we just had, you know, ( s(t) ) equals not specific functions that I lay down but just any general function for the ( x ) component and for the ( y ) component. The reason I want to do this is because the concrete version is going to be helpful and simple and something we can deal with, but almost so simple as to not be indicative of just how complicated the normal circumstance is.

The more general circumstance is so complicated I think it'll actually confuse things a little bit too much. So it'll be good to kind of go through both of them in parallel. The first step is to figure out what is this unit tangent vector. What is that function that at every given point gives you a unit tangent vector to the curve? The first thing for that is to realize that we already have a notion of what should give the tangent vector—the derivative of our vector-valued function as a function of ( t ). The direction in which it points is in the tangent direction.

So if I go over here and compute this derivative, and I say ( s'(t) ), which involves just taking the derivative of both components. The derivative of cosine is ( -\sin(t) \cdot r ), and the derivative of sine is ( \cos(t) \cdot r ). More abstractly, this is just going to be any time you have two different component functions; you just take the derivative of each one. Hopefully, you've seen this; if not, maybe take a look at the videos on taking the derivative of a position vector-valued function.

This we can interpret as that tangent vector, but it might not be a unit vector, right? We want a unit tangent vector, and this only promises us the direction. So what we do to normalize it, what we do to normalize it and get a unit tangent vector function—maybe a different color—and get a unit vector tangent function, which I'll call ( T(t) ). That's kind of confusing, right? Capital ( T ) is for tangent vector, lowercase ( t ) is the parameter.

I'll try to keep that straight; it's sort of standard notation, but there is the potential to confuse with this. What that's going to be is your vector-valued derivative, your vector-valued derivative but normalized. So we have to divide by whatever its magnitude is as a function of ( t ). In this case, in our specific example, that magnitude—if we take the magnitude of ( s(t) ) gives ( r ) times ( \cos(t) ), ( r ) times ( \sin(t) )—so we're taking the magnitude of this whole vector.

What we get, making myself even more room here, is the square root of ( \sin^2(t) \cdot r^2 ) plus ( \cos^2(t) \cdot r^2 ). We can bring that ( r^2 ) outside of the radical to sort of factor it out, turning it into an ( r ) and on the inside we have ( \sin^2(t) + \cos^2(t) ). I'm being too lazy to write down the ( t's ) right now because no matter what the ( t ) is, that whole value just equals one. So this entire thing is just going to equal ( r ).

What that means is that our unit tangent vector up here is going to be the original function but divided by ( r ). It happens to be a constant, usually it's not, but it happens to be a constant in this case. So what that looks like, given that our original function is ( -\sin(t) \cdot r ) and ( \cos(t) \cdot r ), we're dividing out by an ( r ). The ultimate function that we get is just ( -\sin(t) ) and ( \cos(t) ).

For fear of running a little bit long, I think I'll call it into this video and continue on with the same argument in the next video.