Partial derivative of a parametric surface, part 1

So we've just computed a vector-valued partial derivative of a vector-valued function, but the question is, what does this mean? What does this jumble of symbols actually mean in a, you know, more intuitive geometric setting?

That has everything to do with how you visualize the function, and with this specific function, given that the input is two-dimensional but the output is three-dimensional. The output has more dimensions than the input, so it’s nice to visualize it as a parametric surface.

The way that I do that, maybe you could call this visualizing it as a transformation also. What I want to do is basically think of the ts plane, where all these input values live, and kind of think of how that’s going to map into three-dimensional space. But when I do that, I’m actually going to cheat a little bit. Rather than having a separate plane off there as the ts plane, I’m going to kind of overwrite onto the xy plane itself and plop the ts plane down like this.

This isn’t the full ts plane. This is actually supposed to represent just values of t that range from zero up to three. So each tick mark on the graph here corresponds with a half. So this is one, that’s two, and then up here is three. Same with s; s also ranges from zero to three.

The reason that I’m plopping it inside three-dimensional space to start with, kind of overwriting the xy plane with the ts plane, is just to make the animating a little bit easier. You could call it laziness, but the benefit here is that we can do is watch each point. Each one of these points, you’re thinking of is corresponding to some kind of ts pair, an input point which is just a pair of numbers. We’re going to watch each one of those points move to the corresponding output.

Right, the output is a three-dimensional value, a three-dimensional vector or point, however you want to think about it. What that looks like when we animate this, actually, is that each one of those points in our square of the ts plane moves to the corresponding output, and you end up with a certain surface. Just to make it a little more concrete, what’s actually going on here, let’s focus in on just one point.

We’ll focus in on this point, not just for the function visualization but for the partial derivative as well. The function, or the point rather, that I care about is going to be at (1, 1). So this point right here represents the pair of ts where each one of them is equal to (1, 1). You can start by predicting where you think this is going to output. To do that, you just plug it into the function.

This is kind of what the visualization means: we’re plugging this in. For t and s, we’re going to plug in just (1, 1). So that top part is going to look like (1 squared - 1 squared), which becomes 0. That middle part is going to be (1 times 1), which is 1, and then over here, we’re going to have (1 times 1 squared), which is 1 minus (1 times 1 squared), which is again 1.

You can probably see because of the symmetry there those also cancel out. You get 0, which means the output corresponding with this input should be the vector (0, 1, 0), a vector that’s of unit length pointing in the y direction. If we look here, you know, this is the x-axis; this here is the y-axis, so you would think it should be a vector that looks kind of like this unit vector in the y direction.

The point of the surface is what corresponds to the tip of that vector, right? This is how we visualize parametric things. You just think of the tip of the vector as kind of moving through space and drawing out the thing; in this case, the thing it’s drawing is a surface.

If we watch that animation again and let things play forward, that dot corresponding with the input (1, 1) does indeed land at the tip of that vector. So at least for that value, you can see that I’m not lying to you with the animation. In principle, you could do that for every single point, right? If any given input point you kind of plug it through the function and you draw the vector in three-dimensional space, as you watch this animation, it'll land at the tip of that vector.

Now, if we want to start thinking about what the partial derivative means, remember this little dt; this partial t is telling you to nudge it in the t direction. So what does movement, just not even nudges but just movement in general, look like in the t direction for our little snippet of the ts plane here?

Well, the t direction, I'm saying, is in this direction here, where, you know, this represents one, two, three of t values, and this line here represents the constant value for s. So this will be s constantly equaling one, which you can know because it’s passing through the point (1, 1), and then otherwise, you’re just letting t range freely.

If we watch how this gets transformed under the transformation, under the mapping to the parametric surface, you can get a feel for what varying the input t does in the output space. Right, so this whole pink line is basically telling you what happens if you let s constantly equal one, but you let the variable t, the input t, vary freely, and you get a certain curve in three-dimensional space. If you had a different constant for s, it would be another curve, and maybe you can kind of see on the grid lines what shape those other curves would have.

They’re all, in a sense, you know, parallelish to this curve corresponding to s equals one. So if instead of thinking about movement of t as a whole, you start thinking about nudges, this whole partial t is something where we’re just imagining a tiny, tiny little movement in the t direction—not really that much, just a tiny little move, and you’re recording its value as partial t.

So maybe if you’re being concrete, you’d say partial t would be something like 0.01. Really, it’s going to be a limiting variable that gets smaller and smaller, but I find it’s kind of nice to think about an actual value like (1/100). Then, if you let this whole thing undergo the transformation, we can kind of watch the input point, watch the line representing t.

That little nudge, that little nudge is going to get maybe stretched or squished, and it’s going to result in some kind of vector pointing along that curve, and it’ll be tangent to that curve, right? The vector that tells you how you move just a tiny, tiny little bit will be tangent in some way.

That vector, that output nudge, is what you’re thinking of as your tiny change to the output vector, that partial v. When you divide it by the tiny value, right? If your tiny value was 0.01 and you divide it by that, it’s going to become something bigger.

So the actual derivative isn’t going to be just some tiny little nudge that’s, you know, hardly visible, but it’s going to be that nudge vector scaled appropriately. In this case, it would be divided by (1/100) or multiplied by, you know, 100. It would be something that remains tangent to the curve but maybe it’s pointing big.

The larger it is, right? The longer it is, that’s telling you that as you let t vary and you’re kind of moving along this pink curve, tiny nudges in t correspond with larger movements, right? The ratio of the nudge sizes is bigger. If you were to have a very long partial derivative vector that’s still tangent but really goes out there, that would tell you that as you vary t, you’re zipping along super quickly.

If we look at this, you know, this particular one stemming off of one, you kind of get a feel for the curve around that point. You say, okay, okay. That curve, you’re moving positively in the x direction, right? You’re moving to the right, you’re moving positively in the y direction up there, and the z direction is actually negative, isn’t it? This curve kind of goes down as far as z is concerned.

Before even computing it, if I were to tell you that I’m going to plug in the value (1, 1) to this partial derivative that we computed in the last video, you would say, oh, well, just looking at the picture, you can kind of tell that the x value is going to be something positive, something greater than zero. The y value is also going to be something positive, and again, that’s because, you know, the movement is to the right. So positive x, it’s moving up, so positive y.

But the z value should actually be a little bit negative, right? Because as you look at this curve, it’s going down in a sense. With that being our prediction, if you start plugging in (1, 1) to t and s, what you’ll see is that, you know, (2 times 1) is 2, s equals 1, so that’s just 1. Then over here, this looks like (1 squared - 2 times 1 times 1), so this will be (1 - 2), and that’s negative 1.

So it is, in fact, that kind of positive, positive, negative pattern that you’re seeing. Maybe even from this curve, you can get a feel for why the movement in the x direction is twice as much as the movement in the y. It’s moving more to the right than it is up in the y direction.

Again, in principle, you can imagine doing this not just at the point (1, 1) but at any given point; maybe any given point along this curve or any given point along the surface, and the corresponding movement, the direction that, you know, nudges in the t direction, t will give you some vector in three-dimensional space.

That’s the interpretation; that is the meaning of the partial derivative of the vector-valued function here. Again, it’s not the tiny, the actual nudge vector itself, right? When you nudge the input and you get just a little smidgen in the output space here, but it’s that divided by the size of the nudge.

That’s why you’ll get kind of normal-sized vectors rather than tiny vectors. In the next video, I’ll do kind of the same thing for what happens when you nudge in the s direction, just to get a better feel for what’s going on in this example.