Determining a position vector-valued function for a parametrization of two parameters
Determining a Position Vector-Valued Function for a Parametrization of Two Parameters. Created by Sal Khan.
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- I really dont get why b+acos(s) is a hypotenus? Watching the video i understood it like b was a fixed distance from z on the y-plane? And that s is an angle from the y-plane. So for me b+ acos(s) seems to be y(s,t)....but apparrently is not? Please help:)(5 votes)
- It might help to think of b + a*cos(s) as the distance from the origin of the xy-plane. If so, you would need to break it into its x- and y-components by multiplying this distance by sin(t) and cos(t), respectively.(6 votes)
- I found it very confusing the changing of sign on the x-axis. We should keep the right-hand orientation system.(6 votes)
- Right. This choice of X direction makes the cross product of i and j equal -k
i x j = -k
This is a no-no in physics.(3 votes)
- Wouldn't it be more natural to define the coordinates from a right-hand rule coordinate system?
If I do that then I get these coordinates:
x = [b + cos(s)] · cos(t)
y = [b + sin(s)] · sin(t)
z = b · sin(s)
I think it looks nicer if the x- and y-axes have cosine and sine values respectively like that.(6 votes)
- Yes, you could have oriented the coordinate system like that if you wanted.(2 votes)
- Amazing tutorial! are there videos of other surfaces apart from the taurus?(5 votes)
- it would be really nice if there were some examples that werent the taurus(4 votes)
- Shouldn't the unit vector i go into the other direction of the x-axis? (Minute12:44) Otherwise we set up a left-handed coordinate system, didn't we?(1 vote)
- Yes, Sal decided to change the sign of the x coordinate on minute9:58, turning the system into a left-handed coordinate system.(4 votes)
- This is the first time I have heard about a left-hand coordinate system. Are there situations when it would be an advantage to use such a system? It just seems confusing to me since I am not used to it.(2 votes)
- Why the subtitles are not visible in all the videos these days?(2 votes)
- What would be other notable examples of surfaces defined by two parameters?(1 vote)
- Spheres, open cylinders and ellipsoids.
A more general example would be any surface defined by a scalar function z = f(x,y). The parametrization here would be:
x(s,t) = s
y(s,t) = t
z(s,t) = f(s,t)
This allows you to parametrize paraboloid, hyperboloids, etc.
Then there are 'surfaces' that can be continously described with two parameteres and a distance function but cannot be realized in three dimensional Euclidean space. So you go from:
A surface is a two dimensional segment of the three dimensional space.
A surface is whatever that can be described by two parameters and an appropriate rule for distance.(2 votes)
- Couldn't you just use the disk method rather than parametrize a complicated shape, such as a torus?(1 vote)
- The disk method (of integration, I'm assuming) only gives you a volume, not a function. So... It might give you the volume of an object of the same shape, but it won't give you a set of points based on some input variable which fall on that shape. That, and the torus he outlined had no interior points, and was just a surface, so it wouldn't have volume to begin with. So I guess you could at least get the surface area of the shape with a little messing around, but the disk method isn't meant to give a function, it's only meant to give a number.(2 votes)
In the last video, we started to talk about how to parameterize a torus, or a doughnut shape. And the two parameters we were using, and I spent a lot of time trying to visualize it, because this is all about visualization. I think this is really the hard thing to do here. But the way we can parameterize a torus, which is the surface of this doughnut, is to say say hey, let's take a point let's rotate it around a circle. It could be any circle. I picked a circle in the z-y plane. And how far it's gone around that circle, we'll parameterize that by s, and s can go between 0 all the way to 2 pi, and then we're going to rotate this circle around itself. Or I guess a better way to say it, we're going to rotate the circle around the z-axis, and it's all at the center of the circle, so we're always going to keep a distance b away. And so these were top views right there. And then we defined our second parameter t, which tells us how far the entire circle has rotated around the z-axis access. And those were our two parameter definitions. And then here we tried to visualize what happens. This is kind of the domain that our parameterization is going to be defined on. s goes between 0 and 2 pi, so when t is 0, we haven't rotated out of the z-y plane. s is at 0, goes all the way to 2 pi over there. Then when t goes to 2 pi, we've kind of moved our circle. We've moved it along, we've rotated around the z-axis a bit. And then this line in our s-t domain corresponds to that circle in 3 dimensions, or in our x-y-z space. Now given that, hopefully we visualize it pretty well. Let's think about actually how to define a position vector-valued function that is essentially this parameterization. So let's first to do the z, because that's pretty straightforward. So let's look at this view right here. What's our z going to be as a function? So our x's, our y's, and our z's should all be a function of s and t. That's what it's all about. Any position in space should be a function of picking a particular t and a particular s. And we saw that over here. This point right here, let me actually do that with a couple of points. This point right there, that corresponds to that point, right there. Then we pick another one. This point right here, corresponds to this point, right over there. I can do a few more. Let me pick. This point right here, so s is still 0. That's going to be this outer edge, way out over there. I'll pick one more, just to define this square. This point right over here, where we haven't rotated t at all, but we've gone a quarter way around the circle, is that point right there. So for any s and t we're mapping it to a point in x-y-z space. So our z's, our x's, and our y's should all be a function of s and t. So the first one to think about is just the z. I think this will be pretty straightforward. So z as a function of s and t is going to equal what? Well, if you take any circle, remember s is how the angle between our radius and the x-y plane. So I can even draw it over here. Let me do it in another color. I'm running out of colors. So let's say that this is a radius, right there. That angle, we said, that is s. So if I were to draw that circle out, just like that, we can do a little bit of trigonometry. The angle is s. We know the radius is a, the radius of our circle, we defined that. So z is just going to be the distance above the x-y plane. It's going to be this distance, right there. And that's straightforward trigonometry. That's going to be a, I mean, we can do SOCATOA and all of that, you might want to review the videos. But the sine, you can view it this way. So if this is z right there, you could say that the sine of s, SOCATOA is the opposite over the hypotenuse, is equal to a z over a. Multiply both sides by a, you have a sine s is equal to z. That tells us how much above the x-y plane we are. Just some simple trigonometry. So z of s and t, it's only going to be a function of s. It's going to be a times the sine of s. Not too bad. Now see if we can figure out what x and y are going to be. Remember, z doesn't matter. Doesn't matter how much we've rotated around the z-axis. What matters is, how much we've rotated around the circle. If s is 0, we're just going to be in the x-y plane, z is going to be zero. If s is pi over 2, up here, then we're going to be traveling around the top of the doughnut. And we're going to be exactly a above the x-y plane, or z is going to be equal to a. Hopefully that makes reasonable sense to you. Now let's think about what happens as we rotate around. Remember, these two are top views. We are looking down on this doughnut. So the center of each of these circles is b away from the origin, or from the z-axis, what we're rotating around. It's always b away. So our x-coordinate, or our x- and y-coordinate, so if we go to the center of the circle here, we're going to be b away, and this is going to be b away, right over there. So let's think about where we are in the x-y plane, or how far the part of our, what we're, I guess you could imagine, if you were to project our point into the x-y plane, how far is that going to be from our origin? Well, it's always going to be, remember, let's go back to this drawing here. This might be the most instructive. This is just one particular circle on the z-y plane, but it could be any of them. If this is the z-axis, over here, this distance right here is always going to be b. We know that for sure. And so what is this distance going to be? We're at b to the center, and then we're going to have some angle s, and so depending on that angle s, this distance onto, I guess, the x-y plane, you know, if we're sitting on the x-y plane, how far are we from the z-axis, or the projection onto the x-y plane. Or you can, you know, the x or the y position. I'm saying it as many ways as possible. I think you're visualizing it. If z is a sine of theta, this distance right here, this little shorter distance right here, that's going to be a cosine if s. s is that angle right there. This distance right here is going to be a cosine of s. So if we talk about just straight distance from the origin, along the x-y plane, our distance is always going to be b plus a cosine of s. When s is out here, then it's actually going to become a negative number, and that makes sense, because our distance is going to be less than b. We're going to be at that point right there. So if you look at this top views over here, no matter where we are, that is b. And let's say we've rotated a little bit. That distance right here, if you're looking along the x-y plane, that is always going to be b plus a cosine of s. That's what that distance is to any given point. We're depending on our s's and t's. Now, as we rotate around, if we're at a point here, let's say we're at a point there, and that point, we already said, is b plus a cosine of s, away from the origin, on the xy plane. What are the x and y coordinates of that? Well, remember. This is a top-down. We're sitting on the z-axis looking straight down the x-y plane right now. We're looking down on the doughnut. So what are our x's and y's going to be? Well, you draw another right triangle right here. You have another right triangle. This angle right here is t. This distance right here is going to be this times the sine of our angle. So this right here, which is essentially our x, this is going to be our x-coordinate, x as a function of s and t, os going to be equal to the sine of t, t is our angle right there, times this radius. Times, we could write it either way, times b plus a cosine of s. Because remember, how far we are depends on how much around the circle we are, right? When we're over here, we're much further away. Here we're exactly b away, if you're looking only on the x-y plane. And then over here, we're b minus a away, if we're on the x-y plane. So that's x as a function of s and t. And actually, the way I defined it right here, our positive x-axis would actually go in this direction. So this is x positive, this is x in the negative direction. I could've flipped the signs, but hopefully, you know, this actually make sense that that would be the positive x, this is the negative x. Depends on whether using a right-handed or left-handed coordinate system, but hopefully that makes sense. We're just saying, OK, what is this distance right here that is b plus a cosine of s? We got that from this right here, when we're taking a view, just a cut of the torus. That's how far we are, in kind of the x-y direction at any point, or kind of radially out, without thinking about the height. And then if you want the x-coordinate, you multiply it times the sine of t, the way I've had it up here, and the y-coordinate is going to be this, right here, the way we've set up this triangle. So y as a function of s and t is going to be equal to the cosine of t times this radius. b plus a cosine of s. And so our parameterization, and you know, just play with this triangle, and hopefully it'll make sense. I mean, if you say that this is our y-coordinate right here, you just do SOCATOA, cosine of t, CA is equal to adjacent, which is y, right, this is the angle right here, over the hypotenuse. Over b plus a cosine of s. Multiply both sides of the equation times this, and you get y of s of t is equal to cosine of t times this thing, right there. Let me copy and paste all of our takeaways. And we're done with our parameterization. We could leave it just like this, but if we want to represent it as a position vector-valued function, we can define it like this. Find a nice color, maybe pink. So let's say our position vector-valued function is r. It's going to be a function of two parameters, s and t, and it's going to be equal to its x-value. Let me do that in the same color. So it's going to be, I'll do this part first. b plus a cosine of s times sine of t, and that's going to go in the x-direction, so we'll say that's times i. And this case, remember, the way I defined it, the positive x-direction is going to be here. So the i-unit vector will look like that. i will go in that direction, the way I've defined things. And then plus our y-value is going to be b plus a cosine of s times cosine of t in the y-unit vector direction. Remember, the j-unit vector will just go just like that. That's our j-unit vector. And then, finally, we'll throw in the z, which was actually the most straightforward. plus a sine of s times the k-unit vector, which is the unit vector in the z-direction. So times the k-unit vector. And so you give me, now, any s and t within this domain right here, and you put it into this position vector-valued function, it'll give you the exact position vector that specifies the appropriate point on the torus. So if you pick, let's just make sure we understand what we're doing. If you pick that point right there, where s and t are both equal to pi over 2, and you might even want to go through the exercise. Take pi over 2 in all of these. Actually, let's do it. So in that case, so when r of pi over 2, what do we get? It's going to be b plus a times cosine of pi over 2. Cosine of pi over 2 is 0, right? Cosine of 90 degrees. So it's going to be b, right, this whole thing is going to be 0, times sine of pi over 2. Sine of pi over 2 is just 1. So it's going to be b times i plus, once again, cosine of pi over 2 is 0, so this term right here is going to be b, and then cosine of pi over 2 is 0, so it's going to be 0 j. So it's going to be plus 0 j. And then finally, pi over 2, well, there's no t here, sine of pi over 2 is 1. So plus a times k. So there's actually no j-direction. So this is going to be equal to b times i plus a times k. So the point that it specifies, according to this parameterization, or the vector [UNINTELLIGIBLE], is b times i plus a times k. So b times i will get us right out there, and then a times k ill get us right over there. So the position of the vector being specified is right over there. Just as we predicted. That dot, that point right there, corresponds to that point, just like that. Of course, I picked points it was easy to calculate, but this whole, when you take every s and t in this domain right here, you're going to transform it to this surface. And this is the transformation, right here. And of course, we have to specify that s is between, we could write it multiple ways. s is between 2 pi and 0, and we could also say t is between 2 pi and 0. And you could, you know, we're kind of overlapping one extra time at 2 pi, so maybe we can get rid of one of these equal signs if you like, although that won't necessarily change the area any, if you're taking the surface area. But hopefully this gives you at least a gut sense, or more than a gut sense, of how to parameterize these things, and what we're even doing, because it's going to be really important when we start talking about surface integrals. And the hardest thing about doing all of this is just the visualization.