WEBVTT

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JOEL LEWIS: Hi.

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Welcome back to recitation.

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In lecture, you've been learning
about critical points

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of functions, how to find them
using the first derivatives

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and how to classify them using
the second derivative test. So

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I have a question here
for you about that.

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So we have a function w.

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It's a function of two variables
x and y, and it's

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given by this polynomial
function of them.

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So w equals x cubed minus
3xy plus y cubed.

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So what I'd like you to do is
to first find the critical

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values of this function and
then classify them.

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Are they minima or maxima or
saddle points using the second

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derivative test?

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So why don't you pause the
video, take some time

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to work that out.

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Come back and we can work
it out together.

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Hopefully, you had some luck
working out the solution to

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this question.

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Let's have a go at it.

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So in order to find the critical
points, we need to

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look at the first derivative.

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So the critical points are the
points where both partial

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derivatives or all partial
derivatives, if we had a

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function of more variables,
are equal to zero.

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So we need to look at
the first partials.

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So the first partials here,
w sub x, the partial with

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respect to x, well, it's just
a polynomial so it's easy to

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compute those partial
derivatives.

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It's going to be 3x squared
minus 3y, and then the last

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term gets killed because we
treat y as a constant, and so

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we want that to be
equal to zero.

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And similarly, we want the first
partial with respect to

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y, w sub y, to be
equal to zero.

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And so that's minus 3x plus
3y squared equals 0.

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Now, luckily, these are fairly
simple equations, so to solve

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them, we could, for example,
take the first equation and we

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could solve the first equation
for y in terms of x.

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So that'll give us y
equals x squared.

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And now if we plug y equals
x squared into this second

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equation, well, we get minus
3x plus 3x squared squared.

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So that's x to the fourth is
equal 0, and we can divide out

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by that 3, so that means
minus x plus x to the

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fourth equals 0.

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Well, OK, so we could have x
equal to 0, or you can divide,

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and then you get x cubed
equals 1, and that has

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solution x equals 1.

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So x equals 0 or 1.

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Those are the only solutions
to this equation.

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And then the corresponding
y-values, well, we know y is

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equal to x squared, so this
gives us critical points when

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x is 0, y is 0 and when
x is 1, y is 1.

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So this function has
two critical

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points: 0, 0 and 1, 1.

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Now we need to figure out
whether those critical points

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are minima, maxima,
saddle points, sum

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of several of those.

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So in order to do that, we're
going to use this nice tool

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that we have: the second
derivative test. So in order

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to apply the second derivative
test, the first thing I need

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is the second derivatives.

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So let's compute them.

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So the second--

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let's do the xx first. So we
take our first partial, 3x

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squared minus 3y, and we take
another partial of it with

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respect to x.

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So that's, in this case, that's
just going to be 6x.

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And then we've got the other
pure second partial yy, so we

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go back over here, and we look
at what our first partial wy

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was, and then we take another
partial of this with respect

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to y, so that's just
going to be 6y.

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And then we have the mixed
partials wxy and wyx, which,

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of course, are equal to each
other whenever our function is

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nicely behaved like
a polynomial.

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So wxy, we just take the two
mixed partials and--

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OK, so we take the partial of
wx with respect to y, for

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example, and that gives
us minus 3.

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So these are our three partials,
and then often, we,

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you know, call this one A and
this one C and this one B. I

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guess I kind of mixed up the
order a little bit there.

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So we look at these three
expressions, and now we want

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to look at what sometimes people
call the discriminant,

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although I don't know
if Professor

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Auroux used that term.

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So we want to study what the
expression AC minus B squared

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is, so we want to know is this
positive, is this negative at

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the critical points.

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So at the critical
points, right?

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This is important.

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At the critical points.

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So let's do the point 1, 1
first. So at 1, 1, we have A

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is equal to what we put in--
x is 1, y is 1 into the

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expression for A here--

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and that just gives a 6.

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We put x 1, y 1 into the
expression for C, and that

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also gives a 6, we put x 1, y
1 into the expression for B,

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and that gives us minus 3.

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So A is 6, B is minus 3, C is
6, so AC minus B squared is

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equal to, well its equal to
36 minus 9, so that's 27.

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And, in particular,
it's positive.

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So when this is positive, that
means we either have a maximum

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or a minimum.

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So in order to figure out
whether we have a maximum or a

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minimum, we check the sign of
A. So in this case, the sign

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of A is positive.

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A is a positive number.

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So when you have that AC minus B
squared is positive and A is

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positive, that means
you have a minimum.

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So the critical point
1, 1 is a local

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minimum for this function.

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All right.

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Now we can do the same thing for
the critical point 0, 0.

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So recall A was 6x, so at 0,
0, A is equal to 0, B was

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equal to negative 3 everywhere,
and C was equal to

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6y, so at 0, 0, that's 6 times
0 so that's also 0.

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So then our quantity that we
want to look at, AC minus B

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squared, well, that's 0 times 0
minus 9, so that's equal to

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negative 9.

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And negative 9 is less than 0,
so when AC minus B squared is

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less than 0, that means we
have a saddle point.

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So in this case, the second
derivative test was able

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successfully to distinguish what
kinds of critical points

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we had, and it found that the
first critical point 1, 1 was

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a minimum and that the second
critical point 0, 0 was a

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saddle point.

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So just to quickly rehash what
we did, we had a function.

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Back over here, we started
with this function w.

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We had a nice formula for it.

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We computed its first
derivatives.

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We set them both equal
to zero and we solved

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that system of equations.

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So we found two solutions to
that system of equations, and

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those two solutions are the
critical points, the points

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where both partial derivatives
are equal to zero.

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So when you have the two
critical points, then you want

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to apply the second derivative
test to figure out for each

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critical point whether it's
a saddle point, a

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minimum or a maximum.

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So we took our two critical
points 1, 1 and 0, 0, and at

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those points, we evaluated
the second derivative.

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So A is the xx second
derivative.

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B is the mixed partial wxy,
and C is the yy second

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derivative.

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So we evaluate those expressions
at the points in

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question, and then we look
at AC minus B squared.

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And then the sign of AC minus
B squared, if it's negative,

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that gives us a saddle point.

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If it's positive, that gives
us either a maximum or a

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minimum, and we check which one
by looking at the sign of

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A. So here A was positive, so
we got a minimum at 1, 1.

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So I'll stop there.