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DAVID JORDAN: Hello, and welcome
back to recitation.

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So in this problem, we're
considering a function f of

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three variables f of x, y, z,
and it's differentiable, and

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we're not told a
formula for f.

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We just know that it's
differential at this point P,

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which is 1 minus 1, 1, and we're
told that the gradient

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of f at that point is this
particular vector 2i plus j

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minus 3k at that point P. So all
we understand about f is

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how it looks around
the point P.

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Now, we also have this
relation between the

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variables, so x, y and
z aren't unrelated.

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They're related by this
constraint that z is x squared

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plus y plus 1.

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So with this information, we
want to compute the gradient

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of a new function g, and the new
function g is a function

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of two variables, and this
function g is obtained from f

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by just plugging in our
relation for y.

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So we can use our constraint to
solve for y, and then this

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function g is just f with
that constraint applied.

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And what we really want to
do is we want to find the

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gradient of g at
the point 1, 1.

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So notice that when g is equal
to 1, 1, that means that--

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sorry, when the input of g is 1,
1, that means the input of

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f is P. OK?

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So why don't you pause
the video and work

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this out for yourself.

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Check back with me and we'll
work it out together.

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OK, welcome back.

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So let's get started.

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So this problem may not look
like it's a problem about

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partial derivatives with
constraints, but that's what

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it's really going to boil down
to, which is to say that when

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we want to compute--

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so when we want to answer this
question by computing the

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gradient, the first thing we're
going to want to do is

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compute the partial derivative
of g and its variable x.

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And from the way we set up the
problem, that's just the same

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as computing the partial
derivative of f with respect

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to x and keeping z constant.

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Now, remember, when we do
partial derivatives with

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constraints, what's important
about the

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notation is what's missing.

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The variable y is missing, and
that's because we are going to

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use the constraint to get rid of
it, and that's exactly how

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we define g, so this is
the key observation.

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So computing the gradient of g
is just going to be computing

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these partial derivatives
with constraints.

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So we'll do that in a moment,
and I'll also just write that

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the partial derivative of g in
the z-direction is partial f

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partial z, now keeping
x constrained.

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All right, so we need to
compute these partial

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derivatives with constraints.

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And so you remember
how this goes.

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The way that I prefer to do this
is to compute the total

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differentials.

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So let's compute over here.

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The total differential df is--

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the total differential of f is
just the partials of f in the

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x-direction, f in the
y-direction, f in the

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z-direction, and each of these
is multiplied by the

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corresponding differential.

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And we don't know f, so we
can't compute the partial

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derivatives of it in general,
but we do know these partial

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derivatives at this point.

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And so in the problem, we were
given that this is 2dx

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plus dy minus 3dz.

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So this is just using the fact
that the gradient of f we were

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given is 2, 1 minus 3, OK?

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So that's the total differential
of f, and now we

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have this constraint.

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And remember, when we do these
partial derivatives with

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constraints, the trick is to
take the differential of the

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constraint.

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So we had this equation z equals
x squared plus y plus

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1, and what we need to do is
take its differential.

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So we have dz is
2x dx plus dy.

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So that's our constraint.

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Now here we have this variable
x, but we're not varying x in

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this problem.

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We're only focused on the
point P, and at the

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point P, x is 1.

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So, in fact, dz is just
2dx plus dy, OK?

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So now what we need to do is
we need to combine the

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constraint equation and the
total differential for f into

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one equation, and so this is
just linear algebra now.

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So I'll just come over here.

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So we can rewrite our total
differential for the

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constraint as saying that dy is
equal to dz minus 2dx, and

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then we can plug that
back into our total

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differential for f.

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And so we get that df is
equal to 2dx plus--

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now I plug in dy here--

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so dz minus 2dx, and then
finally, minus 3dz.

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So altogether, I get a
minus 2dz, because

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this and this cancel.

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OK.

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We get a minus 2dz.

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So what does that tell us
about the gradient?

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So remember that the
differential now of g and its

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variables is partial
g partial x dx plus

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partial g partial z dz.

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And remember, the trick about
partial derivatives and their

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relation to total differentials
is that the

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partial derivative is just
this coefficient.

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So the fact that we computed
df and we found that it was

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minus 2dz, that tells
us that dg--

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so the thing that we need to
use is that g, remember, is

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just the specialization of f.

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So dg is equal to df in this
case, because g is just a

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special case of f with
constraints.

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So when we computed df here, we
were imposing exactly the

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constraints that we used to
define dg, and so what this

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tells us is that we can just
compare the coefficients here

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and so our gradient
is 0 minus 2.

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So the 0 is because there is no
dependence anymore on x at

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this point.

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There wasn't a dx term.

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There could have been
and there wasn't.

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And the minus 2 is because
that's the dependence on z.

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OK, so this is a bit
complicated, so why don't we

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review what we did.

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So going back over to the
problem statement, we first

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needed to realize that saying
that g was a special case of f

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where we use our constraints
to solve for y, that's what

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put us in the context of
problems with constraints.

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So the fact that the dependence
on y was so simple

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that we could just use
the constraint.

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OK.

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So then, what that allowed us to
do is it allowed us to say

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that the partial derivative of
g in the x-direction is just

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the partial derivative in the
x-direction subject to the

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constraint, and that's
what we did here.

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And so then, the next steps that
we did are the same steps

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that we would always do to
compute partial derivatives

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with constraints, and so we
finally got a relationship

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that said that, subject to these
constraints, df is equal

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to minus 2dz.

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And the point is that g is just
the function f with those

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constraints imposed, and so dg
and df are the same, and so,

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in particular, dg
is minus 2dz.

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And then, we remember that you
can always read off partial

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derivatives from the
total differential.

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They're just the coefficients.

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And so in the end, we got that
the gradient was 0 minus 2.

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And I think I'll leave
it at that.

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