Curvature of a helix, part 1

So let's compute the curvature of a three-dimensional parametric curve. The one I have in mind has a special name; it's a helix. The first two components kind of make it look like a circle. It's going to be cosine of t for the X component, sine of t for the Y component. But this is three-dimensional, and what makes it a little different from a circle is that I'm going to have the last component be t / 5.

What this looks like, we can visualize it pretty well. I'm going to go on over to the computer here. So this shape is called a helix, and you can sort of see how, looking from the XY plane's perspective, it looks as if it's going to draw a circle. Really, these lines should all line up when you're facing it, but it's due to the perspective where things farther away look smaller. It would just be drawing a circle; however, the Z component, because Z increases while your parameter T increases, you're kind of rising as if it's a spiral staircase.

Now, before we compute curvature, to know what we're really going for, you kind of imagine yourself. You know, maybe this isn't a road, but it's like a space freeway. Right? You're driving your spaceship along it, and you imagine that you get stuck at some point. Or maybe not that you get stuck, but all of your instruments lock, and your steering wheel locks, or your joystick – or however you're steering it – all just locks up, and you're going to trace out a certain circle in space, right? That circle might look something like this.

So if you were turning, however you were on the helix, but then you can't do anything different, you might trace out a giant circle. What we care about is the radius of that circle. If you take 1 / the radius of that circle you trace out, that's going to be the curvature. That's going to be the little cap of curvature. Of course, the way that we compute it, we don't directly talk about that circle at all, but it's actually a good thing to keep in the back of your mind.

The way that we compute it is to first find a unit tangent vector function with the same parameter. What that means, you know, if you imagine your helix kind of spiraling through three-dimensional space—man, I am not as good an artist as the computer is when it comes to drawing a helix—but the unit tangent vector function would be something that gives you a tangent vector at every given point. You know, kind of the direction that you, on your spaceship, are traveling.

To do that, you take the derivative of your parameterization. That derivative is going to give you a tangent vector, but it might not be a unit tangent vector, so you divide it by its own magnitude. That'll give you a unit tangent vector. Ultimately, the goal that we're shooting for is going to be to find the derivative of this tangent vector function with respect to the arc length.

So as a first step, we'll start by finding a derivative of our parameterization function. When we take that derivative, luckily, there's not a lot of new things going on. Let's see… derivative. So S prime… um, from single variable calculus, we just take the derivative of each component. So cosine goes to negative sine; negative sine of t. Its derivative is cosine of t, and then the derivative of t / 5 is just a constant; that's just 1/5. Boy, it is hard to say the word 'derivative' over and over. Say it five times.

Um, okay, so that's S prime of t. Now, what we need to do is find the magnitude of S prime of t. So what that involves, as we're taking the magnitude of S prime of t as a vector, we take the square root of the sum of the squares of each of its components. So, uh, negative sine squared just looks like sine squared of t plus cosine squared of t and then (1/5) squared, and that's just 1/25.

You might notice I use a lot of these sine-cosine pairs in examples, partly because they draw circles, and lots of things are fun that involve drawing circles. But also because it has a tendency to let things simplify, especially if you're taking a magnitude because sine squared plus cosine squared just equals one.

So this entire formula boils down to the square root of 1 + 1/25. You might be kind of thinking off to the side that that's 25 over 25 + 1 over 25. Making even more room here, what that equals is the square root of 26/25. Just because 25 is already a square, and it kind of might make things look nice, I'm going to write this as the square root of 26/5.

So this whole thing is the magnitude of our derivative, right? We think to ourselves, it's quite lucky that this came out to be a constant. As we saw with a more general formula, it's often pretty nasty, and it can get pretty bad, but in this case, it's just a constant, which is nice.

As we go up and we start to think about what our unit tangent vector function for the helix should be, we're just going to take the derivative function and divide each term by that magnitude, right? So it's going to look almost identical. It's going to be negative sine of t, except now we're dividing by that magnitude.

That magnitude, of course, is the square root of 26 over 5. So we go up here and we say we're dividing this by the square root of 26 over 5, that whole quantity, and then similarly, the y-component is cosine of t divided by the quantity of the square root of 26 over 5. The last part is (1/5). I'll put that in parentheses, divided by that same amount, square root of 26 over 5.

So we're just taking the whole vector and we're dividing it by the magnitude that it has. We're lucky again that even though this vector is a function, and it could depend on t, the magnitude doesn't. The unit tangent vector function we get as a result is relatively simple. I'll color it in here and then continue on with this same line of reasoning in the next video.