Computing a tangent plane

Hey guys! So, in the last video, I was talking about how you can define a function whose graph is a plane, and moreover, a plane that passes through a specified point and whose orientation you can somehow specify. We ended up seeing how specifying that orientation comes down to certain partial derivative information.

First, let me just kind of repeat what the conclusion was, but I'll put it in more abstract terms since I did it in a very specific example last time. Basically, if you want some kind of function which gives you a plane that passes through a certain point, well, first, let's say what that point is. Right, let's say the point was (x₀, y₀, z₀). So these are just constant values, and this is my way of abstractly describing a single point in space, using x₀ to represent a constant x value, y₀ to represent a constant y value, that kind of thing.

What it is, is it's going to be some sort of other constant a multiplied by (x - x₀). So this white x here is the variable, and then x₀ is just a constant. Let me go ahead and put that parenthesis there. Okay, so then we add to that b multiplied by (y - y₀), and then all of that you add z₀. Now, if you just present this as it is, it's kind of a lot, right? There are five different constants going on, but really what this is saying is you want something where the partial derivative with respect to x is just some constant, and you want to be able to specify what that constant is.

Similarly, the partial derivative with respect to y is another constant, and you just want to ensure that this passes through this point (x₀, y₀, z₀). If you imagine plugging in x (the variable) equals x₀ (the constant), this part goes to zero. Similarly, plugging in y₀ (the constant) makes this part go to zero. So this is a way of specifying that when you evaluate the function at (x₀, y₀), it equals z₀, and that's what makes sure that the graph passes through that point.

With that said, let's start thinking about how you can find the tangent plane to a graph. First of all, let's think about what that point is—how you specify such a point. Instead of specifying any three numbers in space, because you have to make sure the point is somewhere on the graph, you instead only specify two. You're basically going to say what's the x-coordinate, and in this case, let's say the x-coordinate was like 1. That would be like 1. Then the y-coordinate, which looks about like -2, and to make it easier, I'm just going to say, let's say it is -2. Then the z-coordinate is specified because this, as a graph, forces the z-coordinate to be whatever the output of the function is at (1, -2).

So this is going to be whatever the output of our function is at (1, -2). f here, f, is going to be whatever function gives us this graph. So maybe I should write down the actual function that I'm using for this graph. In this case, f, which is a function of x and y, is equal to 3 - (1/3)x² - y². Okay, so this is the function that we're using, and you evaluate it at that point, and this will give you your point in three-dimensional space that our linear function, that our tangent plane, has to pass through.

So we can start writing out our function. Right? We can say, okay, so our linear function has a function of x and y. It's got to make sure it goes through that (1, -2), so this is going to be some constant a, that we'll fill in in a moment, multiplied by (x - 1), plus and then b, also a constant that we'll specify in a moment, multiplied by (y + 2), so it's minus (-2). Then the whole thing that we add to it is f(1, -2).

Let's just go ahead and evaluate that. Let's say we plug in 1 and -2. So if we go up here and we plug in... so that would be 3 - x = 1, 3 - (1/3)(1)², so that's (1/3)(1) - and then (y - (-2)), so that will be -(-2)², so that's 3 - (1/3) - 4. So the whole thing is equal to—let's see—3 - 4 is -1 minus another (1/3), so okay, so that's what we add to this entire thing. We add -4/3.

Maybe I should just kind of make clear the separation here. So this is our function, but we don't know what A and B are; those are things that we need to plug in. Now, the whole idea of a tangent plane is that the partial derivative with respect to x should match that of the original function. If we go over to the graph here and start thinking about partial derivative information, if we want the partial derivative with respect to x, and you imagine moving purely in the x-direction here, this intersects the graph along some kind of curve.

What the partial derivative with respect to x at this point tells you is the slope of the tangent line. Well, I'm kind of bad at drawing. This is the slope of the tangent line in that direction of that point. So that's what the partial derivative with respect to x is telling you. What you want when you look at the tangent of the tangent plane is that the tangent plane also has that same slope. You kind of, if I line things up here, you'd want it also to have that same slope.

So you can specify over here and say a we want a to be equal to the partial derivative of the function with respect to x evaluated at this (1, -2), evaluated at that special point (1, -2). Similarly, B—for pretty much the same reasons. I'll draw it out here. So let's kind of, ah, let's erase this line. So instead of intersecting it with that slice, let's see what movement in the y-direction looks like. In this case, it looks like a very steep slope, right? Because in this case, the tangent line in that direction is a pretty steep slope.

Now when we bring in the tangent plane, it should intersect that constant x-value plane along that same slope. Made it kind of messy there, but you can see the line formed by intersecting these two planes should be that desired tangent. What that corresponds to in formulas is that this B, which represents the partial derivative of L—L is the tangent plane function—that should be the same as if we take the partial derivative of f with respect to y at that point, at this point (1, -2). This is stuff that we can compute and that we can figure out.

So let's just kind of start plugging that in. First, let me just copy this function, because we're going to need it. Copy. And now let's go on down here. I'm just gonna paste it down here, kind of in the bottom, because that's what we'll need. So let's compute the partial derivative of f with respect to x. When we look down here, the only place where x shows up is in this (1/3)x² context, so the partial derivative of f with respect to x is going to be just the derivative of this little guy, which is negative. We bring down the 2, -2/3 of x, so when we go ahead and plug in, you know, x = 1 to see what it looks like, when we evaluate at this point, that's just going to be equal to -2/3.

So that tells us that a is going to have to be -2/3. Now, for similar reasons, let's go ahead and compute the partial derivative with respect to y. We look down here. Well, the only place that y shows up in the entire expression is this (-1)y². So the partial derivative of f with respect to y is equal to just -2y. Now when we plug in y = -2, what we get is -2 multiplied, coincidentally, by -2. That didn't have to be the case that those were the same, and that whole thing equals 4.

So the partial derivative of f with respect to y evaluated at this point (1, -2) is equal to 4. So if we were to plug this information back up into our formula, we would replace a with -2/3; we would say -2/3, and we would replace B with 4. Replace B with 4, and that would give us the full formula, the full formula for the tangent plane.

This could be kind of a lot to look at at first, because we have to specify the input point, you know, (1, -2), and then we had to figure out where the function evaluates at that point, and then we had to figure out both of the partial derivatives with respect to x and with respect to y. But all in all, there's not actually a lot to remember for how you go about computing this. Looking at the graph actually makes things seem a lot more reasonable, because each of those terms has an actual meaning.

If we look at the (1, -2), that's just telling us the input, the kind of x and y coordinates of the input. And of course, we have to evaluate that because that tells us the z-coordinate that'll put us on the graph corresponding to that point. To get us a tangent plane, you just need to specify the two bits of partial differential information, and that'll tell you how this graph needs to be oriented. Once you start thinking things in that way—geometrically—even though there's a lot going on here, with five different numbers you have to put in, each one of them feels like, "Yeah, of course you need that number; otherwise, you couldn't specify a tangent plane."

There's kind of a lot of information required to put it in the appropriate spot. So with that, I will see you next video.