WEBVTT
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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
00:00:11.098 --> 00:00:13.620
is about taking
partial derivatives
00:00:13.620 --> 00:00:15.120
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.
00:00:23.440 --> 00:00:25.860
So what we have is we
have this function w,
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and it's a function of four
variables: x, y, z, and t.
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OK?
00:00:31.490 --> 00:00:34.230
But it's not really a function
of these four variables
00:00:34.230 --> 00:00:36.890
because we have a constraint.
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So we want to
study how w changes
00:00:41.630 --> 00:00:46.190
as we vary the parameters,
except that we have imposed
00:00:46.190 --> 00:00:47.260
this constraint here.
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So that really we kind of
only have three variables,
00:00:50.820 --> 00:00:53.960
because we have four
variables and one constraint.
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So that's what partial
derivatives with constraints
00:00:58.630 --> 00:01:00.240
help us do.
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So let's explain
first the notation.
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OK?
00:01:02.810 --> 00:01:07.260
So it says partial w
partial z, and then we
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have the subscripts x and y.
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So what's important
about this notation
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is not what you see 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?
00:01:16.660 --> 00:01:19.260
So what this notation
means is, as always,
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the denominator in our
derivative expression--
00:01:22.990 --> 00:01:27.045
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,
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this x and y here
would be superfluous.
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Because by partial
derivative, we always
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mean to keep the other
unlisted variables unchanged.
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However, the fact that
t is missing here,
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it means that-- so if you
think about it-- if we vary z,
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and we keep x and y 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
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to ask you to compute the
partial derivative of w
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in z varying x, y, and t
because-- excuse me-- keeping
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x, y, and t fixed,
because then there
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would be no 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 vary,
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but it's going to vary
in a way that we're just
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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
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is making sure 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
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of the variable t.
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OK.
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So the first way that we're
going to work this out
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is using total differentials.
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And I like to use
total differentials
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when I'm on new ground
because-- they're not
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the most computationally
effective,
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because they involve computing
all the derivatives that we
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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 differentials,
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then all the other
computations that you have
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to do are just substitution.
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So it really just
becomes linear algebra,
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and that's what I like about it.
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In part b, we'll see a shortcut
using implicit differentiation
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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
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need to turn them both into
differential equations.
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And so we'll do that
using a combination
00:03:29.430 --> 00:03:32.300
of implicit differentiation
and the chain rule.
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So I'll let you pause
the video and get
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started on these problems.
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And you can check back and
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 way
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to attack this.
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So why don't we
just start computing
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the total 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 the constraint
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equation.
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And what we first
want to do is just
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take the total differential
of both of those equations
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to get started.
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So we can take the first one
and it tells us that dw is equal
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to-- OK, so we have 3 x
squared y dx, plus x cubed dy,
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minus 2z*t dz,
minus z 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,
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but we have to remember
that in the setting we're
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interested in, x and
y are held fixed.
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And so holding x and y fixed
means that the differentials dx
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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 2z*t
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
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from the 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 equation
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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
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is we get that dt is equal
to minus t over z dz by just
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doing straightforward algebra.
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OK.
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So, with that in
hand now, we can-- so
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remember I mentioned
in the beginning
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that our goal was-- so
from the very beginning,
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we knew that if we varied z,
because of our constraint,
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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 constraint.
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And it says if you
vary z, you have
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to vary t in an
appropriate way, and that's
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what this coefficient tells us.
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So what we're
really interested in
00:07:13.760 --> 00:07:18.460
is how does w 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,
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we get that dw here
is equal to-- OK,
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so we have minus 2z*t dz.
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And then we have minus-- OK--
z squared times another minus
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times t over z, so this
all becomes a plus z*t 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 z*t dz.
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And that tells us that the
partial derivative that we're
00:08:22.920 --> 00:08:31.520
after is just this
coefficient, right?
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The partial derivative
is just defined
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to be the coefficient
of the differential
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once you work everything out.
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And so this is minus z*t.
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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
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the computation a bit shorter.
00:08:54.075 --> 00:08:55.616
So the tricks that
we're going to use
00:08:55.616 --> 00:08:58.880
are implicit differentiation
and the chain rule.
00:09:09.067 --> 00:09:10.650
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
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to just take a straightforward
partial derivative
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of our original expression.
00:09:22.590 --> 00:09:27.450
So remember, w was x
cubed y minus z t squared.
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And so let's just take
a partial derivative
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of that in the z-direction.
00:09:33.890 --> 00:09:37.870
So the partial derivative in
the z-direction of x cubed y
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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 2z*t component.
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That's just because the partial
derivative of z squared is 2z.
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And then we have another term
which is minus z squared,
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and now we need to take
the partial derivative of t
00:10:00.829 --> 00:10:01.620
in the z-direction.
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So, you know, often times when
we take partial derivatives
00:10:14.420 --> 00:10:26.350
of one variable in
terms of the other,
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it's common to think that
the partial derivative of one
00:10:28.922 --> 00:10:30.630
variable in terms of
the 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
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t does vary depending on z,
and so we had to include that
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into our notation.
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OK.
00:10:43.460 --> 00:10:45.460
So now this is
almost what we want,
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except we have this
mystery component here.
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And of course,
there's only one way
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we can solve this mystery, which
is the same way we solved it
00:10:51.370 --> 00:10:52.100
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 x*y equals z*t.
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OK.
00:11:13.880 --> 00:11:18.780
So if we take the partial
derivative of this equation--
00:11:18.780 --> 00:11:21.740
so if I take the partial
derivative of x and y
00:11:21.740 --> 00:11:27.750
in terms of z, then I do
get 0, because x and y are
00:11:27.750 --> 00:11:31.790
genuinely independent from z.
00:11:31.790 --> 00:11:34.230
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?
00:12:00.730 --> 00:12:09.590
So we can rewrite this as saying
that partial t partial z is
00:12:09.590 --> 00:12:11.790
minus t over z.
00:12:15.090 --> 00:12:15.670
OK?
00:12:15.670 --> 00:12:18.640
Now, you might notice that,
you know, this is formally
00:12:18.640 --> 00:12:21.600
very similar to what we did
in part a, and of course,
00:12:21.600 --> 00:12:23.290
that's no surprise.
00:12:23.290 --> 00:12:27.740
When we are manipulating
using implicit differentiation
00:12:27.740 --> 00:12:30.365
and the chain rule,
it's just a compact way
00:12:30.365 --> 00:12:32.740
of doing what we were doing
with the total differentials.
00:12:32.740 --> 00:12:37.620
I mean, to me, the chain
rule is a computation
00:12:37.620 --> 00:12:41.130
which you could prove by
doing the corresponding thing
00:12:41.130 --> 00:12:43.720
with total differentials.
00:12:43.720 --> 00:12:46.920
And so we get this same
coefficient negative t over z,
00:12:46.920 --> 00:12:50.880
which you recall that
we got in part a.
00:12:50.880 --> 00:12:51.390
OK.
00:12:51.390 --> 00:12:59.790
So now we have, once again
we have this, two equations,
00:12:59.790 --> 00:13:01.880
and we just can do substitution.
00:13:01.880 --> 00:13:15.592
So we get that partial w partial
z is equal to minus 2z*t.
00:13:15.592 --> 00:13:20.670
And now again, we get
minus another minus,
00:13:20.670 --> 00:13:26.490
and z here cancels the z
squared, so we get plus z*t.
00:13:26.490 --> 00:13:29.400
And so we get minus z*t.
00:13:32.540 --> 00:13:36.380
OK, and finally, if we
remember our assumptions,
00:13:36.380 --> 00:13:40.390
our assumptions were that x
and y were independent of z.
00:13:40.390 --> 00:13:41.810
That was our notation.
00:13:41.810 --> 00:13:45.070
And we use that assumption
at this step right here.
00:13:45.070 --> 00:13:47.870
So in fact, we don't just have
the partial derivative of w
00:13:47.870 --> 00:13:49.780
with respect to z.
00:13:49.780 --> 00:13:54.140
We need to specify that
we held x and y fixed.
00:13:54.140 --> 00:13:56.150
OK.
00:13:56.150 --> 00:13:59.480
So just to review
again, if we look now
00:13:59.480 --> 00:14:06.180
at what we did in part b, you
know, the meat of the argument
00:14:06.180 --> 00:14:08.880
was the exact same as
what we did in part a.
00:14:08.880 --> 00:14:11.300
The meat of the
argument was right here.
00:14:11.300 --> 00:14:15.420
We took some derivative and
then this was an unknown.
00:14:15.420 --> 00:14:17.550
The definition of
w doesn't know how
00:14:17.550 --> 00:14:18.800
t and z depend on one another.
00:14:18.800 --> 00:14:21.230
That you can only find by
looking at the constraint.
00:14:21.230 --> 00:14:24.980
And so we just went
through the problem
00:14:24.980 --> 00:14:29.010
and we took derivatives
of the constraint,
00:14:29.010 --> 00:14:31.520
and that gave us an equation
that we were looking for.
00:14:35.270 --> 00:14:39.140
Now if we go back now
to part a over here.
00:14:46.540 --> 00:14:49.462
So as you can see, there's a lot
more work involved in part a.
00:14:49.462 --> 00:14:51.670
On the other hand, to me it
was more straightforward.
00:14:51.670 --> 00:14:54.980
We just had to compute
the total differentials
00:14:54.980 --> 00:14:58.630
and then do some linear
algebra with cancellations.
00:14:58.630 --> 00:15:00.557
And somehow, when you
do total differentials,
00:15:00.557 --> 00:15:02.890
you just compute everything
that could possibly come up,
00:15:02.890 --> 00:15:04.860
and then you just
substitute it in.
00:15:04.860 --> 00:15:07.560
And indeed, we got
the same answer:
00:15:07.560 --> 00:15:12.690
partial w partial z
as being minus z*t.
00:15:12.690 --> 00:15:14.615
OK, and I think I'll stop there.