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DAVID JORDAN: Hello, and welcome
back to recitation.

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So today, the problem I'd like
to work with you is about

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taking partial derivatives in
the presence of constraints.

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So this is a pretty
subtle business.

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So take your time when you
work these problems.

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So what we have is we have this
function w, and it's a

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function of four variables:
x, y, z, and t.

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OK?

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But it's not really a function
of these four variables

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because we have a constraint.

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So we want to study how w
changes as we vary the

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parameters, except that
we have imposed

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this constraint here.

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So that really we kind of only
have three variables, because

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we have four variables
and one constraint.

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So that's what partial
derivatives with constraints

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help us do.

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So let's explain first
the notation.

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OK?

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So it says partial w partial
z, and then we have the

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subscripts x and y.

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So what's important about this
notation is not what you see

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as much as what you don't see.

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What you don't see is
the variable t.

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OK?

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So what this notation means is,
as always, the denominator

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in our derivative expression--

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partial z here-- that means
that we want to vary z.

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And we want to see how w
changes as we vary z.

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And the x and y here mean that
we want to keep x and y fixed.

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So if we didn't have a
constraint, this x and y here

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would be superfluous.

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Because by partial derivative,
we always mean to keep the

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other unlisted variables
unchanged.

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However, the fact that t is
missing here, it means that--

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so if you think about it-- if we
vary z, and we keep x and y

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fixed, then t also is varying.

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Right?

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Because we have this
constraint here.

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And so it wouldn't make sense
for me to ask you to compute

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the partial derivative of w in z
varying x, y and t because--

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excuse me--

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keeping x, y, and t fixed,
because then there would be no

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room for z to vary.

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OK?

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So this notation means that z
is going to be allowed to

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vary, but it's going to vary
in a way that we're

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just going to ignore.

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So you will see how this works
out in the problem.

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So what we're really interested
in is making sure

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that x and y stay fixed
and that z varies.

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And then we're going to need to,
when we do some algebra,

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we're going to need to get
rid of any mention of

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the variable t.

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OK.

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So the first way that we're
going to work this out is

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using total differentials.

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And I like to use total
differentials when I'm on new

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ground because they're not
the most computationally

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effective, because they involve
computing all the

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derivatives that we might
possibly need in sight.

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So they're not the most
efficient computationally.

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But if you go ahead and
compute the total

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differentials, then all the
other computations that you

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have to do are just
substitution.

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So it really just becomes linear
algebra, and that's

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what I like about it.

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In part b, we'll see a shortcut
using implicit

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differentiation and
the chain rule.

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And this is going to be
a little bit tricky.

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So we have these two equations,
we need to turn

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them both into differential
equations.

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And so we'll do that using
a combination of implicit

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differentiation and
the chain rule.

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So I'll let you pause the video
and get started on these

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problems. And you can
check back and

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we'll work it out together.

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OK, welcome back.

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So let's start by doing a, let's
start with problem a.

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So we have total differentials
is the suggested

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way to attack this.

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So why don't we just start
computing the total

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differentials that we know.

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So we have two equations.

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w in relation to the
other variables and

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the constraint equation.

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And what we first want to do
is just take the total

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differential of both of those
equations to get started.

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So we can take the first
one and it tells us

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that dw is equal to--

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OK so we have--

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3x squared y dx, plus x cubed
dy, minus 2zt dz, minus z

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squared dt.

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OK.

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Now right away, we can simplify
this equation.

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So this is the total
differential, but we have to

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remember that in the setting
we're interested in, x and y

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are held fixed.

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And so holding x and y fixed
means that the differentials

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dx and dy are both set to 0.

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So that lets us rewrite this
first differential equation is

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just dw equals minus 2zt
dz minus z squared dt.

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So that's our first equation
that we get.

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Let me just check with my
notes to make sure.

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That's right.

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OK.

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And so now, we have the
constraint equation from the

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original statement
of the problem.

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And we need to take
its differential.

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So on the one hand, we
get x dy plus y dx.

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That's the total differential
of the left-hand side.

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And then on the right-hand side,
we have t dz plus z dt.

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OK?

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And now we notice that now the
left-hand side of this

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equation is just 0 for
the same reason.

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dy and dx are being
held fixed.

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So the relation that we end up
getting is we get that dt is

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equal to minus t over
z dz by just doing

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straightforward algebra.

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OK.

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So with that in hand now we
can-- so remember I mentioned

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in the beginning our goal
was-- so from the very

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beginning, we knew that if we
varied z, because of our

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constraint, we're going to be
forced to be varying t.

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And that's exactly what this
equation says, doesn't it?

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We got this by just taking
the differential of the

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constraint.

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And it says if you vary z,
you have to vary t in an

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appropriate way, and
that's what this

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coefficient tells us.

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So what we're really interested
in is how does w

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vary in terms of z here.

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And so we want to get
rid of this dt here.

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And in fact, we can by
using the constraint.

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So combining this equation with
this equation, we get

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that dw here is equal to--

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OK, so we have--

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minus 2zt dz.

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And then we have--

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minus--

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OK--

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z squared times another
minus times t over

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z, so this all becomes--

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plus zt dz.

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So all I did is I plugged in for
dt using our formula here.

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And so this altogether is equal
to just minus zt dz.

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And that tells us that the
partial derivative that we're

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after is just this coefficient,
right?

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The partial derivative is just
defined to be the coefficient

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of the differential once you
work everything out.

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And so this is minus zt.

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OK, so that's a.

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So now let's see if we can use
some tricks to make the

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computation a bit shorter.

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So the tricks that we're going
to use are implicit

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differentiation and
the chain rule.

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So at the end of the
day-- excuse me--

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we're interested in partial
w partial z.

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And what we're going to do is
use the chain rule to just

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take a straightforward partial
derivative of our original

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expression.

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So remember, w was x cubed
y minus zt squared.

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And so let's just take a partial
derivative of that in

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the z-direction.

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So the partial derivative
in the z-direction of x

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cubed y is just 0.

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So that will go away.

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And so we only have minus,
we have a 2zt component.

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That's just because the partial
derivative of z

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squared is 2z.

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And then we have another term
which is minus z squared, and

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now we need to take the partial
derivative of t in the

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z-direction.

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So, you know, often times when
we take partial derivatives of

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one variable in terms of the
other, it's common to think

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that the partial derivative of
one variable in terms of the

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other is just 0.

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Because usually our variables
are independent.

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They don't vary in terms
of one another.

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But this is exactly a situation
where t does vary

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depending on z, and so
we had to include

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that into our notation.

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OK.

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So now this is almost what we
want, except we have this

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mystery component here.

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And of course, there's only
one way we can solve this

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mystery, which is the same way
we solved it in part a.

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We have to use the constraint.

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So let's take partial z of
our constraint equation.

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And remember, our constraint
equation was xy equals zt.

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OK.

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So if we take the partial
derivative of this equation,

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so if I take the partial
derivative x and y in terms of

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z, then I do get 0, because
x and y are genuinely

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independent from z.

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It's only t that depends on z.

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So on this side we get 0.

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Now, on the other side I just
need to use the product rule.

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So I get t, plus z partial
t partial z.

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OK?

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So we can rewrite this as saying
that partial t partial

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z is minus t over z.

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OK?

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Now, you might notice that,
you know, this is formally

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very similar to what we did in
part a, and of course, that's

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no surprise.

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When we are manipulating using
implicit differentiation and

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the chain rule, it's just a
compact way of doing what we

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were doing with the total
differentials.

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I mean, to me, the chain rule
is a computation which you

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could prove by doing the
corresponding thing with total

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differentials.

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And so we get this same
coefficient negative t over z,

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which you recall that
we got in part a.

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OK.

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So now we have, once again we
have this, two equations and

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we just can do substitution.

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So we get that partial
w partial z is

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equal to minus 2zt.

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And now again, we get minus
another minus, and z here

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cancels the z squared,
so we get plus zt.

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And so we get minus zt.

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OK, and finally, if we remember
our assumptions, our

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assumptions were that x and
y were independent of z.

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That was our notation.

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And we use that assumption
at this step right here.

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So in fact, we don't just have
the partial derivative of w

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with respect to z.

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We need to specify that
we held x and y fixed.

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OK.

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So just to review again, if we
look now at what we did in

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part b, you know, the meat of
the argument was the exact

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same as what we did in part a.

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The meat of the argument
was right here.

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We took some derivative and
then this was an unknown.

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The definition of w doesn't
know how t and z

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depend on one another.

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That you can only find by
looking at the constraint.

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And so we just went through
the problem and we took

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00:14:27,690 --> 00:14:30,590
derivatives of the constraint,
and that gave us an equation

259
00:14:30,590 --> 00:14:31,840
that we were looking for.

260
00:14:31,840 --> 00:14:35,270

261
00:14:35,270 --> 00:14:39,140
Now if we go back now
to part a over here.

262
00:14:39,140 --> 00:14:46,540

263
00:14:46,540 --> 00:14:48,530
So as you can see, there's
a lot more work

264
00:14:48,530 --> 00:14:49,510
involved in part a.

265
00:14:49,510 --> 00:14:51,670
On the other hand, to me it
was more straightforward.

266
00:14:51,670 --> 00:14:55,240
We just had to compute the total
differentials and then

267
00:14:55,240 --> 00:14:58,630
do some linear algebra
with cancellations.

268
00:14:58,630 --> 00:15:00,980
And somehow, when you do total
differentials, you just

269
00:15:00,980 --> 00:15:03,340
compute everything that could
possibly come up, and then you

270
00:15:03,340 --> 00:15:04,860
just substitute it in.

271
00:15:04,860 --> 00:15:09,950
And indeed, we got the same
answer: partial w partial z as

272
00:15:09,950 --> 00:15:12,690
being minus zt.

273
00:15:12,690 --> 00:15:14,980
OK, and I think I'll
stop there.

274
00:15:14,980 --> 00:15:15,115