Parametric surfaces | Multivariable calculus | Khan Academy

So I have here a very complicated function. It's got a two-dimensional input—two different coordinates to its input—and then a three-dimensional output. Uh, specifically, it's a three-dimensional vector, and each one of these is some expression. It's a bunch of cosines and sines that depend on the two input coordinates.

In the last video, we talked about how to visualize functions that have a single input—a single parameter, um, like T—and then a two-dimensional vector output. So, some kind of expression of T and another expression of T, and this is sort of the three-dimensional analog of that.

What we're going to do, we're just going to visualize things in the output space, and we're going to try to think of all the possible points that could be outputs. So, for example, let's just start off simple; let's get a feel for this function by evaluating it at a simple, simple pair of points. So, let's say we evaluate this function f at T equals uh, zero. I think that would probably be pretty simple, and then S is equal to Pi.

So, let's think about what this would be. We go up and we say, okay, T of 0, cosine of 0 is 1, so this whole thing is going to be 1. Same with this one, and S of 0 is zero, so this over here is going to be zero, and this is also going to be zero. Now, cosine of Pi is negative 1, so this here is going to be negative 1. This one here is also going to be negative 1, and then S of Pi, just like S of 0, is zero. So, this whole thing actually ends up simplifying quite a bit.

So that the top is 3 * 1 + -1, 1 * 1 is 1, and we get 2. Then we have 3 * 0 plus 0, so the Y component is just zero, and then the Z component is also zero. So, what that would mean is that this output is going to be the point that's 2 along the x-axis. And there's nothing else to it; it's just 2 along the x-axis.

So I'll go ahead and move the graph about and add that point there. Um, so that's what would correspond to this one particular input, 0 and Pi. And you know, you could do this with a whole bunch, and you might add a couple other points based on other inputs that you find, uh, but this will take forever to start getting a feel for the function as a whole.

Another thing you can do is say, okay, maybe rather than thinking of evaluating at a particular point, imagine one of the inputs was constant. So let's imagine that S stayed constant at Pi, okay? Uh, but then we let T range freely. So that means we're going to have some kind of different output here, um, and we're going to let T just be some kind of variable while the output is Pi.

What that means is we keep all of these—this 1 and zero for what S of Pi is—but the output now is going to be 3 cosine of T cosine of T plus -1 * the cosine of T. So it's going to be minus cosine of T. Uh, the next part is going to still be Z. Uh, 3 S of T, this is no longer zero; I should probably erase those others, actually.

So, um, we're no longer evaluating I when T was zero. Um, so 3 * S of T, that's just still the function that we're dealing with—3 S of T and then minus 1 * S of T. So, minus S of T—keep drawing it in green just to be consistent—and then the bottom stays at zero.

This whole thing actually simplifies: 3 cosine T minus cosine T, that's just 2 cosine T, and then same deal for the other one; it's going to be 2 S of T. So this whole thing actually simplifies down to this. So this is, again, when we're letting S stay constant and T ranges freely.

When you do that, what you're going to end up getting is a circle that you draw. You can maybe see why it's a circle, because you have this cosine sine pattern. Um, it's a circle with radius 2, and it should make sense that it runs through that first point that we evaluated.

So that's what happens if you let just one of the variables run. But now let's do the same thing, but think instead about what happens as S varies and T stays constant. I encourage you to work it out for yourself. I'll go ahead and just kind of draw it because I kind of want to give the intuition here.

So, in that case, you're going to get a circle that looks like this. So again, I encourage you to try to think through, for the same reasons, imagine that you let S run freely, keep T constant at zero. Why is it that you would get a circle that looks like this?

In fact, if you let both T and S run freely, a very nice way to visualize that is to imagine that this circle, which represents S running freely, sweeps throughout space as you start to let T run freely. What you're going to end up getting when you do that is a shape that goes like this, and this is a donut.

We, uh, have a fancy word for this in mathematics: we call it a torus. Um, but it turns out the function here is a fancy way of drawing the torus. In another video, I'm going to go through in more detail if you were just given the torus, how you could find this function, how you can kind of get the intuitive feel for that.

It'll involve going through in a bit more detail why, when you sweep the circle out, it gets the torus just so. And what the relationship between this red circle and the blue circle is. But here, I just kind of want to give an intuition for what parametric surfaces are all about—how it's a way of visualizing something that has a two-dimensional input and a three-dimensional output.