Second partial derivative test example, part 2

In the last video, we were given a multivariable function and asked to find and classify all of its critical points. So, critical points just mean finding where the gradient is equal to zero, and we found four different points for that. I have them down here; they were (0, 0), (0, -2), (√3, 1), and (-√3, 1).

So then, the next step is to classify those, and that requires the second partial derivative test. What I'm going to go ahead and do is copy down the partial derivatives since we already computed those. I'll copy them and then just kind of paste them down here where we can start to use them for the second partial derivatives.

Let me clean things up a little bit, and we don't need the simplification of it. So, we've got our partial derivatives now. Since we know we want to apply the second partial derivative test, we've got to first just compute all of the different second partial derivatives of our function. That's just kind of the first thing to do.

So, let's go ahead and do it. The second partial derivative of the function with respect to X twice in a row will take this the partial derivative with respect to X and then do it with respect to X again. So, this first term looks like six times a variable times a constant, so it'll just be 6 times that constant, and then the second term, the derivative of -6x, is just -6.

Moving right along, when we do the second partial derivative with respect to Y twice in a row, we take the partial derivative with respect to Y and then do it again. So, this x² term looks like nothing; it looks like a constant as far as Y is concerned, so we ignore it. The derivative of -3y² is -6y, and then the derivative of -6y is just -6.

Then we can't forget that last crucially important mixed partial derivative term, which is the partial derivative of f, where first we do it with respect to X and then with respect to Y. The order doesn't really matter in this case since it's a perfectly ordinary polynomial function. So, we could do it either way; but I'm just going to choose to take a look at this guy and differentiate it with respect to Y. So, the derivative of the first term with respect to Y is 6x, and then that second term looks like a constant with respect to Y, so that's all we have.

Now what we're going to do is plug in each of the critical points to the special second partial derivative test expression. To remind you of what that is, that expression is we take the second partial derivative with respect to X twice. I'll just write it with a kind of shorter notation using subscripts, and we multiply that by the second partial derivative with respect to Y, and then we subtract off the mixed partial derivative term squared.

So, let's go ahead and do that for each of our points. When we do this at the point (0, 0), what we end up getting is, plugging that into the partial derivative with respect to X twice gives us 6 times 0, which is 0, so there just -6. This gives us -6 multiplied by, when we plug it into the partial derivative with respect to y² again, that y goes to zero, so we're left with just -6.

Then we subtract off the mixed partial derivative term, which in this case is zero because when we plug in x equals 0, we get zero. So we're subtracting off 0², and that entire thing equals -6 times -6, which is 36. We’ll get to analyzing what it means that that's positive in just a moment, but let's just kind of get all of them on the board so we can kind of start doing this with all of them.

If we do this with (0, -2), then once we plug in y = -2 to this expression, this time I'll write it out: 6 times -2 - 6, so that's -12 - 6, which gives us -18. Then, when we plug it into the partial derivative of f with respect to y², again I'll kind of write it out: -6 times -2 is equal to 12 - 6, so that will give us 6 that we plug in here.

Then for the mixed partial derivative, again x is equal to 0, so the mixed partial derivative is just going to look like zero when we do this. So, we’re subtracting off 0² and we get -18 times 6. Oh geez, what's -18 times 6? So that's going to be -108. The specific sign, the specific magnitude won't matter; it's going to be the sign that's important, and this is definitely negative.

Now, kind of moving right along, these examples can take quite a while. If we plug in (√3, 1), what we get now, instead of plugging in y = -2, we're plugging in y = 1. So, that'll be 6 times 1 - 6, so the whole thing is just 0.

Now, for the partial derivative with respect to y², instead of plugging in -2, now we're plugging in y = 1, so we have -6 times 1 - 6, which gives us -12. Now for the mixed partial derivative term which is 6x. X is equal to the √3, so now we're subtracting off √3².

What that equals is this first part is just entirely zero, and we're subtracting off three, so that's -3. Now we have (√3, 1), and this will be very similar because this first term just had a y, and we plugged in the Y, so it's also going to be zero for totally the same reasons.

The value of y didn't change, so that's also going to be 0, so it doesn't really matter because we're multiplying it by a zero. Right? And then, over here now we're plugging in -√3, and that's going to have the same square. So again, we're just subtracting off three.

So, what does this second partial derivative test tell us? Once we express this term, if it's greater than zero, we have a maximum or a minimum. That's what the test tells us, and then if it's less than zero, if it's less than zero, we have a saddle point.

So, in this case, the only term that's greater than zero is this first one. To analyze whether it's a maximum or a minimum, notice that the partial derivative with respect to X twice in a row or with respect to Y twice in a row was negative, which indicates a sort of negative concavity, meaning this corresponds to a maximum.

So this guy corresponds to a local maximum. Now all of the other three gave us negative numbers, so all of these other three give us saddle points. Saddle points!

So, the answer to the question—the original find and classify, you know, such and such points—is that we found four different critical points: (0, 0), (0, -2), (√3, 1), and (-√3, 1), and all of them are saddle points except for (0, 0), which is a local maximum.

All of that is something that we can tell without even looking at the graph of the function. And with that, I will see you in the next video!