Hi everyone, and welcome to our lecture on curvature. So first off, we're going to start off with a definition. And we're going to say that a curve is smooth, is a parametric curve defined by some position vector RFT. We're going to say that this curve is smooth if the magnitude of the derivative of this vector is not equal to zero. Well this is avoiding is sort of places where you have costs or sharp corners, or perhaps crossings, we want a nice smooth curve, just like we know and love. And this will be defined by some RFT. You give me a time t and I'll give you x, y or xyz coordinates. If I fix a point on the curve, let's say t equals some fixed a, then I have my position, R of t or R of a at that point. I can talk about the tangents to that curve, some vector that lies tangent, and we'll denote this as r prime of t. And again, this is a vector and defined this vector, we take the derivative of each of the components. But as I move along the curve at different points, my tangent vector could perhaps be different lengths. So what I'd like to do is normalize this vector. And that leads us to the definition of what's called the unit tangent. So friendly reminder, a unit vector is a vector of length 1. We'll denote this unit vector as capital T. It's a function of time, so it's a little awkward, it's capital T of little t. And this will be the normalization of the derivative vector. So what does that mean? I take the vector, and I divide by the magnitude of itself. This is a unit vector is of length one, and it's called the unit tangent vector. Now a couple of things about this unit tangent vector, it gives you the direction of the curve. So it's just going to normalize all these little arrows giving blank one, and it changes direction slowly when C is straight. If you can imagine a curve that slowly bends over the horizon, this tangent is going to change direction very slowly as well. And when C curves, right, you can sell that when C is curving, the vector itself will also change direction. So it's a nice way to measure how a curve curves so weird to say, but how the arc of a curve is changing. This is going to introduce the notion of what's called curvature. So the curvature, and we denote this scalar as a Greek Kappa K. So for those of you who haven't seen Greek before, this is a Greek Kappa is the common letter. So the curvature Kappa, some numbers, a real number of the curve C, you gotta be careful here we're using a lot of the same words, at a point p is a measure of how quickly the curve changes direction at p. We're trying to get the distinction between what's the difference between say, a parabola or a cubic or a sine curve? How do we measure their change? How do they differ from a straight line? So this is a formula that will do just that. And it turns out to be the magnitude of the derivative of the unit tangent divided by the magnitude of the derivative of the position vector. This is a fraction. This is a number on the top and a number on the bottom. Add this formula to your ever growing list. This is a number that will always tell you how things change. So let's go through and find the canonical example. The first example you always have to do once you introduce curvature, and that is of a circle. So we're going to let C be a circle of radius a. So imagine some circle and we have a fixed radius of a. So then I can assume a without loss of generality that it's centered at the origin. And so, C will be defined by the position vector given by a cosine of t, and then a sine of t. This is how we parametrize the circle of radius a centered at the origin. Of course to take one mark around the circle, we're going 0 to 2 pi. Okay, so this is all we need, and I can find the curvature. Again, there's a little Greek Kappa to find this. Okay, so here we go, we have our formula. Good old plug and chug. I'm going to write it down and then we're going to go off and find all the pieces here. So it's the magnitude of the derivative of the unit tangent vector divided by the magnitude of the derivative of the position vector, T prime of t and then R prime t. Both of these are vectors when we take their magnitude. All right, I have a lot to find. So let's be careful as we go through. So first thing I need to do is take a bunch of derivatives, let's find R prime of t, derivative of a cosine t, of course is negative a sine of t, derivative of a sine of t is a cosine of t. I want its magnitude, so let's go ahead and get that. So the magnitude of this vector is the square root of each component squared. This becomes a squared sine squared of t plus a squared cosine squared of t. From here, we can factor out the a squared. And I get a squared parentheses sine squared of t plus cosine squared of t, of course, sine squared plus cosine of t, we've seen this a few times. That of course is just one and you get the square root of a squared, which since a is positive, just a. So that's our magnitude. Now let's get the unit tangent vector. So the tangent vector of t is the derivative of the position function divided by its magnitude. Just a lot of little things, nothing too difficult in itself. There's a lot of little things. So this becomes minus a, sine of t, a cosine of t. And I'm dividing by the scalar. Basically, 1 over a, so I'm going to bring that up front as 1 over a. When I bring the 1 over a in, my head both components, the a's cancel and I'm left with just the vector minus sine of t, and then positive cosine of t. However, I don't want the unit tangent. I want the derivative of the unit tangent and my formula for cosine. So I take one more derivative, this is like the second derivative here, which should make sense from calc one which is second derivative measures concavity. So I'm taking the derivative of the unit tangent, this is like the second derivative. So here we go. Take a derivative again, derivative of minus sine, of course is minus cosine, and the derivative of cosine is minus sine of t. All right, and last but not least, in my formula for my numerator, I want the magnitude of this vector, let me squeeze it in here on the bottom. So I take the square root of the component squared. We've seen this before, get sine squared plus cosine squared which is once again, the squared of one, which is of course, just one. Boy. Okay, so we've put it all back. I gotta remember what I have here. The numerator, the magnitude of the derivative of the unit tangent is just one and the magnitude of the directional derivative way back over here. This was just a, now remember, a is our radius, not calling it R. A is our radius. So let's stare this for a second. The curvature of a circle is 1 over the radius. This is important. This is impressive. This is amazing. Why? If the radius is big. So if the radius is big, then 1 over the radius is small. In a small curvature, think of the earth. If I have a large radius of a sphere or a circle, it seems flat that has low curvature, it feels flat. You see this as we look out the window every day. And if the radius is small, well then the other thing occurs. If I have a small curvature, then I'm going to feel it, then the circle is curving a lot if I have a small radius like a marble or something like that. We'll see that, we'll feel that, okay, and then move along. This gets into the notion, if you take a limit here, you can see that the curvature of a line or the curvature of a plane is in fact zero. This is an important formula to know, you can calculate this every time although it gets kind of tedious to take care of. What you should know that the curvature of a circle is measured as one over it's radius. This is a pretty important formula and result to know. There's another formula for curvature that I want you to know. So this is going to be two ways to find it. They're both the right way, just sometimes one is easier than the other. You can derive this formula from the other one, but if you take the magnitude of the derivative of the position vector cross, this is a cross product, the second derivative of the position vector. So remember cross product is your vectors, we take the magnitude of that vector to get a scalar. And then we do all this by the magnitude of the position vector cubed, works out to be this way, I'll leave it to the details for you. But there's just another way to do it depending on how things are given if you want to go through it. So as an exercise, go back to the other example and show that this formula works for the circle to show that its curvature is one over the radius. Let's do another example, I want you to find a curve by R(t) to be t, 3 sine of t and 3 cosine of t. And for this curve, you can graph it at some curve. I want to find the curvature. All right, so once we know the formula, I'll write the formula down and we'll use our new version of the formula. We can go through and find this thing. So r prime of t cross r double prime of t, all in the magnitude divided by r prime of t cubed. Wow, it's a lot to say. All right, but each one of these is easy to find. We have our position function one at a time. Let's go get them, take a derivative, and go slow and be careful derivative t is 1, derivative of 3 sine t, of course is 3 cosine of t, the derivative of 3, cosine of t is negative 3, sine of t. We're going to need a second derivative in here as well, and we're going to need this thing's magnitude, doesn't matter which one you get first, for no good reason. Let's get the magnitude that'll help us get our denominator. So we take the square root of each component squared. We have 1 squared of course is 1 plus 9 cosine squared plus 9 sine squared of t. Clean this up a little bit, you get 1+9. Again, cosine is sine squared, of course it's just one, so you get 1+9 squared of 10. Great. Let's find the second derivative, r double prim of t. Now, this will be a vector, double prime of t is going to be, now take a derivative of the derivative, we have 0 minus 3 sine of t minus 3 cosine of t, all right? So I have my derivative, my second derivative is vectors, and now I want the cross product. So let's go ahead and do that on this side. So we want r prime, cross r double prime. And this will work out to be set it up here. So we have i, we have j, we have k, 1 3 cosine t minus 3, sine t zero minus 3 sine t minus 3 cosine t. All right, here we go. So we have i, do a cross product, we get minus 9 cosine squared of t, and then minus 9 again sine squared of t minus j minus 3 cosine of t minus 0, and then plus k is going to be minus 3 sine of t, and then minus zero. The piece on the I cleans up a little bit, so we have i minus 9, we have the negative hits the J, so plus 3j and then minus 3 sine t on the k. All right, now for the scalar component, grab its magnitude. Let's do that, and then we're finally all done. So square root of each of the components squared, we have negative 9 squared, which is 81 plus 3 squared, which is 9. And then minus 9 sine, not 3, I've missed the cosine, sorry about that. Cosine t, so cosine squared t and then sine squared t. Clean this up and you get 81 for all the same reasons, plus 9, works out to be the square root of 90. Finally, our [LAUGH] answer after all this work, I know these are not short. The numerator is the square root of 90. The denominator was this number cube. So it's like root 10, root 10, root 10, three of these. Root 10, root 10 of course is 10, and then root 90 over 10 is just 3. So our final answer, three tenths, a lot of work to get here, this is the number. This is how we measure the curvature. So good job on this one, if you got through it. If you could do this example, you can do any example that's going to come at you, just remember there's two formulas, I use them both sort of check my work if I need to go through it. Once in a while, you want to do some algebraic manipulation, so just pick the one that's a little easier for you. But we have two ways to think of our formulas for curvature. All right, great job on this one. Add this to your formula sheet, keep working hard, and I'll see you next time.
