Second partial derivative test example, part 1

So one common type of problem that you see in a number of multivariable calculus classes will say something to the effect of the following: find and classify all of the critical points of, and then you'll insert some kind of multivariable function.

So first of all, this idea of a critical point basically means anywhere where the gradient equals zero. So you're looking for places where the gradient of your function at some kind of input, some specified input (x) and (y) that you're solving for, is equal to zero. As I've talked about in the last couple videos, the reason you might want to do this is because you're hoping to maximize the function or to maybe minimize the function.

Now, the second requirement of classifying those points, that's what the second derivative test is all about. Once you find something where the gradient equals zero, you want to be able to determine: is it a local maximum, is it a local minimum, or is it a saddle point?

So let's go ahead and work through this example. The first thing we're going to need to do, if we're solving for when the gradient equals 0, and remember when we say equal 0, we really mean the zero vector, but it's just a convenient way of putting it all on one line. We take both partial derivatives.

So the partial derivative with respect to (x) is, well, this first term, when we take the derivative of (3x^2 \cdot y) with respect to (x), that 2 hops down, so we have (6xy^3). Well, (y) looks like a constant, so (y^3) looks like a constant, minus (3x^2). So that 2 comes down, so we're subtracting off (6x). Again, this (3y^2) term, (y) looks like a constant, so everything here looks like a constant with zero derivative as far as the (x) direction is concerned.

Now, we do the partial of (f) with respect to (y). Then this first term looks like some sort of constant (3x^2); (x) looks like a constant, so some kind of constant times (y). So the whole thing looks like (3x^2). The second term, minus (x^3), minus (y^3), excuse me, looks like minus (3y^2) when we take the derivative, minus (3y^2).

Then this next term only has an (x), so it looks like a constant as far as (y) is concerned. Then this last term, we take down the 2 because we're differentiating (y^2), and you'll get (-6y), (-6 \cdot y).

So when we are finding the critical points, the first step is to set both of these guys equal to 0. So this first one, when we do set it equal to zero, we can simplify a bit by factoring out (6x). So this really looks like (6x \cdot (y - 1)) and then that's what we're setting equal to zero.

What this equation tells us is that either it's the (6x) term that equals 0, in which case that would mean (x) is equal to 0, or it's the case that (y - 1 = 0), in which case that would mean that (y = 1). So at least one of these things has to be true. That's kind of the first requirement that we've found.

Let me scroll down a little bit here. For the second equation, when we set it equal to 0, it's not immediately straightforward how you would solve for this in a nice way in terms of (x) and (y). But because we've already solved one, we can kind of plug them in and say, for example, if it was the case that (x = 0), and we kind of want to see what that turns our equation into, then we would have, well (3x^2) is nothing, that would be (0) and we'd just be left with (-3y^2 - 6y = 0).

We can factor out a bit, so I'm going to factor out a (-3y). So I'll factor out (-3y), which means that first term just has a (y) remaining, and then that second term has a (2), a positive (2) since I factored out (-3). So positive (2), and that equals (0). So what this whole situation would imply is that either (-3y = 0), which would mean (y = 0), or it would be the case that (y + 2 = 0), which would mean that (y = -2).

So that's the first situation where we plug in (x = 0). Now alternatively, there's the possibility that (y = 1). So we could say (y = 1) and what that gives us in the entire equation, we still have that (3x^2) because we're kind of solving for (x) now. (3x^2) and then the rest of it becomes, let's see, (-3 \cdot 1^2), so minus (3). We're plugging in (1) for (y), and we subtract off (6), plugging in that (1) for (y) again, and that whole thing is equal to (3x^2), then minus (3 - 6), so I'm subtracting off (9).

From here, I can factor out a little bit, and this will be (3 \cdot (x^2 - 3)). What that implies then, since this whole thing has to equal (0), what that implies is that (x^2 - 3 = 0). So we have (x = \pm \sqrt{3}).

Maybe I should kind of specify these are distinct things that we found. One of them was in the circumstance where (x = 0), and then the other was what we found in the circumstance where (y = 1). So this gives us a grand total of three different critical points because in the first situation where (x = 0), the critical points that we have, well both of them are going to have an (x) coordinate of (0) in them, an (x) coordinate of (0), and the two corresponding (y) coordinates are (0) or (-2). So you have (0) or (-2).

There's kind of two possibilities, and then there's another two possibilities here where if (y = 1), when (y = 1), we'll have (x) as positive or negative (\sqrt{3}). So we have positive (\sqrt{3}) and (y = 1), and then we have negative (\sqrt{3}) and (y = 1).

So these are the critical points, critical points which basically means all partial derivatives are equal to (0). In the next video, I will classify each of these critical points using the second partial derivative test.