Curvature of a helix, part 2

So where we left off, we were looking at this parametric function for a three-dimensional curve and what it draws. I showed you was a helix in three-dimensional space, and we're trying to find its curvature. The way you think about that is you have a circle—you're thinking of the circle that most closely hugs that curve, or you could imagine, you know, you're flying a spaceship and all your instruments lock up when you're turning. You're wondering what circle in space you're going to trace out, and you're looking for one divided by the radius of that.

If you prefer my, you know, kindergarten or drawing skills, you can just look at the helix over here while we work it through. The part where we left off, we have this tangent vector function, this unit tangent vector function for our curve. At every given value t, whatever point that corresponds to on the curve, this function is going to give us the vector that is of unit length and tangent to the curve.

The ultimate goal for curvature is to find the derivative of that unit tangent vector with respect to arc length. Specifically, we want its magnitude. What that typically requires—and I talked about this in the videos on the formula—is you take its derivative with respect to the parameter t because that's the thing you can actually do. That might not correspond to unit length; right? If you nudge the parameter t, that might not nudge you a corresponding length on the curve, but you correct that by dividing out by the derivative of the parameterization function with respect to t.

That's actually arc length, this derivative, the magnitude of the derivative of the parameterization with respect to t. But is that a hard word? So, with our given function, let's go ahead and start doing that. I think first, given that we have this kind of fractions of fractions form, I'm just going to start by writing it a little bit more simply. So, our unit tangent vector function, that first component, where we're dividing by the square root of 26 divided by 5, instead, I'm just going to write that as negative 5 times the sine of t. I'm just kind of moving that 5 up into the numerator divided by the root of 26.

I'll do the same thing for cosine, where we move that 5 up into the numerator, cosine of t divided by the root of 26. Then that last part, 1 5 divided by square root of 26 fifths, gives us just 1 divided by square root of 26.

All right, so the first step in using our curvature formula is going to be to take the derivative of this guy, right? We need the derivative of the tangent vector function, so we go ahead and start doing that. We see that d big T d little t, so tangent vector function parameter is equal to—and we just take the derivative of each component—so negative 5 sine goes to negative 5 cosine, and we divide out by that constant square root of 26.

Similarly, 5 cosine goes to negative 5 sine since the derivative of cosine is negative sine. Negative 5 sine divided by 26 or square root of 26, and that last component is just a constant, so the derivative is nothing—it's zero.

Next step, we're going to take the magnitude of that vector that we just found. So we’re trying to find the magnitude of the derivative of the tangent vector function. We say, all right, the magnitude of what we just found, d big T d little t, involves—so this magnitude will be the square root, make the little tick there, square root of the sum of the squares of these guys. So the square of that first component is going to be 25 multiplied by cosine squared, cosine squared of t, all divided by 26, right? Because the square root of the square—the square of the square root of 26 is 26.

Then we add to that, um, 25 times sine squared, sine squared of t, also divided by 26. From that, we can factor out 25 over 26 inside that radical because both terms involve multiplying by 25 and dividing by 26. What we're left with is a nice and friendly cosine-sine pair. The reason we love things involving circles? This always happens—nice cancellation. This just becomes one.

So what we're left with on the whole is root 25 over 26—pretty nice, root 25 over 26. For our curvature equation, we go up and we can start plugging that in. We just found that numerator and found that it was the square root of 25 divided by 26, the entire thing, 25 over 26.

We already found the magnitude of the derivative itself—that's one of the things we needed to do to find the tangent vector. That's where this 26 over 5 came from. I deleted it from where we did last video to make room, but if you look at the last video, you can see where we got that square root of 26 divided by 5.

I'll actually write that as square root of 26 divided by 25—just, you know, that's how we originally found it. I'm just putting the 5 back under the radical. It's tempting, if you aren't looking too closely, to think these guys cancel out, but it's actually they're the opposite of each other, right? One is 25 over 26, the other is 26 over 25.

So, if we put everything under the radical there, I'm going to say equals the square root, and we're going to have 25 over 26 divided by 26 over 25. When we flip that bottom and multiply, what we get is 25 squared divided by 26 squared, and the root of that whole thing just gives us 25 over 26. That is our curvature; that right there is the answer—that's the curvature.

So, it's a little bit greater than one. No, no, sorry, that's a—so it's a little bit less than one, which means that you're curving a little bit less than you would if it was a circle with radius one, which kind of makes sense if we look at the image here. Because if the helix were completely flattened out, right, if you imagine squishing this down onto the x-y plane, you'd just be going around a circle with radius one.

But by kind of pulling that spring and pulling it so that there's a z component, you're making it a little bit more straight, so the curvature should go down a little bit because it's becoming a little bit more straight. The radius of curvature will go up, so that's the curvature of a helix. That's a pretty good example of how you can find the curvature by walking through directly the idea of finding dT/ds and, you know, getting that unit tangent vector, getting that little unit arc length.

In the next example, I think I'll go through one where you just use the formula, where it’s something a little bit more complicated than thinking about this, and you turn to the formula itself. I'll see you then.