WEBVTT
00:00:00.000 --> 00:00:07.430
JOEL LEWIS: Hi.
00:00:07.430 --> 00:00:08.920
Welcome back to recitation.
00:00:08.920 --> 00:00:11.170
In lecture, you've been
learning about critical points
00:00:11.170 --> 00:00:13.960
of functions, how to find them
using the first derivatives
00:00:13.960 --> 00:00:16.710
and how to classify them using
the second derivative test.
00:00:16.710 --> 00:00:19.340
So I have a question
here for you about that.
00:00:19.340 --> 00:00:21.060
So we have a function w.
00:00:21.060 --> 00:00:23.070
It's a function of two
variables, x and y,
00:00:23.070 --> 00:00:26.760
and it's given by this
polynomial function of them.
00:00:26.760 --> 00:00:31.330
So w equals x cubed
minus 3xy plus y cubed.
00:00:31.330 --> 00:00:34.090
So what I'd like you
to do is to first find
00:00:34.090 --> 00:00:36.860
the critical values
of this function
00:00:36.860 --> 00:00:38.710
and then classify
them-- are they
00:00:38.710 --> 00:00:42.230
minima or maxima
or saddle points--
00:00:42.230 --> 00:00:43.796
using the second
derivative test.
00:00:43.796 --> 00:00:45.170
So why don't you
pause the video,
00:00:45.170 --> 00:00:46.770
take some time to work that out.
00:00:46.770 --> 00:00:48.520
Come back and we can
work it out together.
00:00:57.050 --> 00:01:00.340
Hopefully, you had some luck
working out the solution
00:01:00.340 --> 00:01:01.270
to this question.
00:01:01.270 --> 00:01:02.440
Let's have a go at it.
00:01:02.440 --> 00:01:05.154
So in order to find
the critical points,
00:01:05.154 --> 00:01:06.820
we need to look at
the first derivative.
00:01:06.820 --> 00:01:08.500
So the critical
points are the points
00:01:08.500 --> 00:01:11.780
where both partial derivatives--
or all partial derivatives,
00:01:11.780 --> 00:01:14.470
if we had a function of more
variables-- are equal to zero.
00:01:14.470 --> 00:01:17.090
So we need to look at
the first partials.
00:01:17.090 --> 00:01:21.140
So the first partials here, w
sub x, the partial with respect
00:01:21.140 --> 00:01:22.900
to x-- well, it's
just a polynomial
00:01:22.900 --> 00:01:25.200
so it's easy to compute
those partial derivatives.
00:01:25.200 --> 00:01:30.580
It's going to be 3
x squared minus 3y,
00:01:30.580 --> 00:01:33.585
and then the last term gets
killed because we treat y
00:01:33.585 --> 00:01:36.780
as a constant, and so we want
that to be equal to zero.
00:01:36.780 --> 00:01:41.090
And similarly, we want the
first partial with respect to y,
00:01:41.090 --> 00:01:43.180
w sub y, to be equal to zero.
00:01:43.180 --> 00:01:50.750
And so that's minus 3x
plus 3 y squared equals 0.
00:01:50.750 --> 00:01:53.390
Now, luckily, these are
fairly simple equations,
00:01:53.390 --> 00:01:55.610
so to solve them, we
could, for example,
00:01:55.610 --> 00:01:58.200
take the first equation and we
could solve the first equation
00:01:58.200 --> 00:02:00.370
for y in terms of x.
00:02:00.370 --> 00:02:04.900
So that'll give us
y equals x squared.
00:02:04.900 --> 00:02:07.240
And now if we plug
y equals x squared
00:02:07.240 --> 00:02:12.430
into this second equation,
well, we get minus 3x plus 3 x
00:02:12.430 --> 00:02:15.550
squared squared-- so that's x
to the fourth-- is equal to 0,
00:02:15.550 --> 00:02:20.080
and we can divide out by
that 3, so that means minus x
00:02:20.080 --> 00:02:23.130
plus x to the fourth equals 0.
00:02:23.130 --> 00:02:27.000
Well, OK, so we could have x
equal to 0, or you can divide,
00:02:27.000 --> 00:02:29.510
and then you get
x cubed equals 1,
00:02:29.510 --> 00:02:31.480
and that has
solution x equals 1.
00:02:31.480 --> 00:02:34.480
So x equals 0 or 1.
00:02:34.480 --> 00:02:36.470
Those are the only
solutions to this equation.
00:02:36.470 --> 00:02:38.344
And then the corresponding
y-values, well, we
00:02:38.344 --> 00:02:47.630
know y is equal to x squared,
so this gives us critical points
00:02:47.630 --> 00:02:52.150
when x is 0, y is 0,
and when x is 1, y is 1.
00:02:52.150 --> 00:02:54.860
So this function has two
critical points: (0, 0)
00:02:54.860 --> 00:02:56.000
and (1, 1).
00:02:56.000 --> 00:02:58.400
Now we need to figure out
whether those critical points
00:02:58.400 --> 00:03:01.250
are minima, maxima,
saddle points, sum
00:03:01.250 --> 00:03:04.142
of several of those.
00:03:04.142 --> 00:03:05.600
So in order to do
that, we're going
00:03:05.600 --> 00:03:07.090
to use this nice
tool that we have:
00:03:07.090 --> 00:03:08.720
the second derivative test.
00:03:08.720 --> 00:03:10.860
So in order to apply the
second derivative test,
00:03:10.860 --> 00:03:14.300
the first thing I need is
the second derivatives.
00:03:14.300 --> 00:03:15.620
So let's compute them.
00:03:15.620 --> 00:03:20.780
So the second-- let's
do the xx first.
00:03:20.780 --> 00:03:24.940
So we take our first partial,
3 x squared minus 3y,
00:03:24.940 --> 00:03:28.350
and we take another partial
of it with respect to x.
00:03:28.350 --> 00:03:32.580
So that's, in this case,
that's just going to be 6x.
00:03:32.580 --> 00:03:40.850
And then we've got the other
pure second partial, yy,
00:03:40.850 --> 00:03:42.840
so we go back over
here, and we look
00:03:42.840 --> 00:03:45.720
at what our first
partial w_y was,
00:03:45.720 --> 00:03:48.350
and then we take another partial
of this with respect to y,
00:03:48.350 --> 00:03:51.620
so that's just going to be 6y.
00:03:51.620 --> 00:03:56.137
And then we have the mixed
partials w_xy and w_yx,
00:03:56.137 --> 00:03:58.220
which, of course, are equal
to each other whenever
00:03:58.220 --> 00:04:02.490
our function is nicely
behaved, like a polynomial.
00:04:02.490 --> 00:04:08.020
So w_xy, we just take
the two mixed partials
00:04:08.020 --> 00:04:10.770
and-- OK, so we take
the partial of w_x
00:04:10.770 --> 00:04:15.270
with respect to y, for example,
and that gives us minus 3.
00:04:15.270 --> 00:04:18.980
So these are our three partials,
and then often, we, you know,
00:04:18.980 --> 00:04:23.120
call this one A and this
one C and this one B. I
00:04:23.120 --> 00:04:27.410
guess I kind of mixed up the
order a little bit there.
00:04:27.410 --> 00:04:29.850
So we look at these
three expressions,
00:04:29.850 --> 00:04:32.386
and now we want to look
at what sometimes people
00:04:32.386 --> 00:04:34.760
call the discriminant, although
I don't know if Professor
00:04:34.760 --> 00:04:35.950
Auroux used that term.
00:04:35.950 --> 00:04:43.870
So we want to study what the
expression A*C minus B squared
00:04:43.870 --> 00:04:47.320
is, so we want to know is this
positive, is this negative?
00:04:47.320 --> 00:04:49.320
At the critical points.
00:04:49.320 --> 00:04:52.270
So at the critical
points, right?
00:04:52.270 --> 00:04:53.840
This is important.
00:04:53.840 --> 00:04:58.280
At the critical points.
00:04:58.280 --> 00:05:04.210
So let's do the
point (1, 1) first.
00:05:04.210 --> 00:05:09.010
So at (1, 1), we have A is equal
to-- well, we put in x is 1,
00:05:09.010 --> 00:05:13.580
y is 1 into the expression for
A here, and that just gives a 6.
00:05:13.580 --> 00:05:16.960
We put x 1, y 1 into
the expression for C,
00:05:16.960 --> 00:05:20.110
and that also gives
a 6, we put x 1, y 1
00:05:20.110 --> 00:05:22.750
into the expression for B,
and that gives us minus 3.
00:05:22.750 --> 00:05:33.960
So A is 6, B is minus 3, C is 6,
so A*C minus B squared is equal
00:05:33.960 --> 00:05:41.452
to, well it's equal to
36 minus 9, so that's 27.
00:05:41.452 --> 00:05:45.810
And, in particular,
it's positive.
00:05:45.810 --> 00:05:47.850
So when this is
positive, that means we
00:05:47.850 --> 00:05:50.780
either have a
maximum or a minimum.
00:05:50.780 --> 00:05:52.440
So in order to
figure out whether we
00:05:52.440 --> 00:05:55.924
have a maximum or a minimum,
we check the sign of A.
00:05:55.924 --> 00:05:57.715
So in this case, the
sign of A is positive.
00:05:57.715 --> 00:06:00.260
A is a positive number.
00:06:02.950 --> 00:06:06.630
So when you have that A*C minus
B squared is positive and A is
00:06:06.630 --> 00:06:08.400
positive, that means
you have a minimum.
00:06:13.850 --> 00:06:17.110
So the critical point
(1, 1) is a local minimum
00:06:17.110 --> 00:06:18.391
for this function.
00:06:18.391 --> 00:06:18.890
All right.
00:06:18.890 --> 00:06:26.170
Now we can do the same thing
for the critical point (0, 0).
00:06:26.170 --> 00:06:29.760
So recall A was
6x, so at (0, 0).
00:06:29.760 --> 00:06:34.010
A is equal to 0, B was equal
to negative 3 everywhere,
00:06:34.010 --> 00:06:37.150
and C was equal to
6y, so at (0, 0),
00:06:37.150 --> 00:06:41.540
that's 6 times 0
so that's also 0.
00:06:41.540 --> 00:06:45.520
So then our quantity that we
want to look at, A*C minus B
00:06:45.520 --> 00:06:49.640
squared, well, that's
0 times 0 minus 9,
00:06:49.640 --> 00:06:51.680
so that's equal to negative 9.
00:06:51.680 --> 00:06:56.200
And negative 9 is less than 0,
so when A*C minus B squared is
00:06:56.200 --> 00:06:58.899
less than 0, that means
we have a saddle point.
00:07:03.869 --> 00:07:05.660
So in this case, the
second derivative test
00:07:05.660 --> 00:07:07.730
was able successfully
to distinguish
00:07:07.730 --> 00:07:10.030
what kinds of critical
points we had,
00:07:10.030 --> 00:07:13.180
and it found that the first
critical point 1, 1 was
00:07:13.180 --> 00:07:15.120
a minimum and that the
second critical point
00:07:15.120 --> 00:07:17.270
0, 0 was a saddle point.
00:07:17.270 --> 00:07:21.650
So just to quickly rehash what
we did, we had a function.
00:07:21.650 --> 00:07:25.050
Back over here, we started
with this function w.
00:07:25.050 --> 00:07:27.190
We had a nice formula for it.
00:07:27.190 --> 00:07:29.650
We computed its
first derivatives.
00:07:29.650 --> 00:07:31.730
We set them both equal
to zero and we solved
00:07:31.730 --> 00:07:33.380
that system of equations.
00:07:33.380 --> 00:07:36.470
So we found two solutions
to that system of equations,
00:07:36.470 --> 00:07:38.640
and those two solutions
are the critical points,
00:07:38.640 --> 00:07:41.780
the points where both partial
derivatives are equal to zero.
00:07:41.780 --> 00:07:43.930
So when you have the
two critical points,
00:07:43.930 --> 00:07:46.040
then you want to apply
the second derivative test
00:07:46.040 --> 00:07:48.030
to figure out for
each critical point
00:07:48.030 --> 00:07:51.000
whether it's a saddle point,
a minimum or a maximum.
00:07:51.000 --> 00:07:58.710
So we took our two
critical points, (1, 1)
00:07:58.710 --> 00:08:03.920
and (0, 0), and at those
points, we evaluated
00:08:03.920 --> 00:08:05.280
the second derivative.
00:08:05.280 --> 00:08:08.240
So A is the xx
second derivative.
00:08:08.240 --> 00:08:14.170
B is the mixed partial w_xy, and
C is the yy second derivative.
00:08:14.170 --> 00:08:17.500
So we evaluate those expressions
at the points in question,
00:08:17.500 --> 00:08:20.040
and then we look at
A*C minus B squared.
00:08:20.040 --> 00:08:23.840
And then the sign of A*C minus
B squared, if it's negative,
00:08:23.840 --> 00:08:25.720
that gives us a saddle point.
00:08:25.720 --> 00:08:28.980
If it's positive, that gives us
either a maximum or a minimum,
00:08:28.980 --> 00:08:30.780
and we check which
one by looking
00:08:30.780 --> 00:08:33.130
at the sign of A. So
here A was positive,
00:08:33.130 --> 00:08:36.160
so we got a minimum at (1, 1).
00:08:36.160 --> 00:08:37.685
So I'll stop there.