WEBVTT
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DAVID JORDAN: Hello, and
welcome back to recitation.
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Today the problem I'd
like to work with you
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is about computing
partial derivatives
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and the total differential.
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So we have a function z which
is x squared plus y squared.
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So it depends on the
two variables x and y.
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Now the variables
x and y themselves
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depend on two auxiliary
variables, u and v.
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So that's the
setup that we have.
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So in part a, we just want to
compute the total differential
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dz in terms of dx and dy.
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So u and v aren't going
to enter into the picture.
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And then in part b,
we're going to compute
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the partial derivative
partial z partial u
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in two different ways.
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First, we're going to compute
it using the chain rule.
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And then we're going to compute
it using total differentials.
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And so we'll substitute
in some of the work
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that we had in a
to solve that part.
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So why don't you pause the video
now and work on the problem.
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We'll check back and
we'll do it together.
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Hi, and welcome back.
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Let's get started.
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So first, computing
a is not so bad.
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So we just need
to first remember,
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what does it mean to compute
the total differential?
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So the total
differential dz is just
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the partial derivative of z
in the x-direction dx plus
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z in the y-direction dy.
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OK?
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So now, looking at our
formula here for z,
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we have-- so the
partial derivative of z
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in the x-direction is
2x, so this is 2x dx.
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And the partial derivative
of z in the y is 2y,
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so we have 2y dy.
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OK, and that's all
we have to do for a.
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Now for b, we want to compute
this partial derivative
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in two different ways.
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First, using the chain rule.
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So let's remember what
the chain rule says.
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So whenever I think
about the chain rule,
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I like to draw this
dependency graph.
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OK?
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And this is just a
way for me to organize
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how the different variables
depend on one another.
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So at the top, we have z.
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And z is a function of
x and y, but x is itself
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a function of both u
and v, and y is also
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a function of u and v.
So z depends on x and y,
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and x and y each jointly
depend on u and v.
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So it's a little
bit complicated,
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the relationships here.
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So now, what the
chain rule says is
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that if we take a partial
derivative-- partial z
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partial u-- we have to go
through our dependency graph.
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Every way that we
can get from z to u,
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we get a term in our summation
for each one of those.
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So for instance, z
goes to x goes to u.
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So that means that we
have partial z partial x,
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partial x partial u.
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And then we can also go
z goes to y goes to u.
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And that will give us partial z
partial y, partial y partial u.
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And now these partials
are ones that we can just
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compute from our formulas.
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So for instance,
partial z partial x
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is 2x, which we computed.
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Now partial x partial u, we have
to remember that x is defined
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as u squared minus v squared.
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And so partial x
partial u, that's 2u.
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Partial z partial y, again,
is this 2y that we computed.
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And partial y partial u is v.
This v is just because u was
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u*v, and we take a partial
in the u-direction.
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OK.
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So altogether this
is 4u*x plus 2v*y,
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and that's our
partial derivative.
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So notice that, you know,
x is a function of u and v.
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So if I really wanted
to, I could substitute
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for x its formula for
u and v, but that's not
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really necessary.
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You know, what's interesting
about these problems is
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how the differentials
depend on one another,
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and I'm perfectly happy
with an answer that
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has mixed variables like this.
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That's fine.
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So now, let's go
over here and let's
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see if we can get the
same answer by using
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total differentials.
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Now, I have to say
that the chain rule
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that we used on the
previous problem,
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it's the quickest way to
do these sorts of things.
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I like to do total differentials
if I have some time
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to actually explore the problem
and get comfortable with it.
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I prefer to use total
differentials because I think
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it's a little bit clearer.
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Somehow, this chain
rule it's just, to me,
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it's just a prescription,
it's not an explanation.
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So why don't we compute
some total differentials.
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So we already saw-- let
me just repeat over here.
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We already saw that dz
is 2x dx plus 2y dy.
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OK.
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Now, we want to
use the fact that x
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is itself a function of
u and v. So that's what
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we need to do now.
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So that tells us that dx is 2u
du minus 2v dv in the same way.
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And dy.
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So remember, y was u*v.
So taking d of u*v,
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we get v du plus u dv.
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OK?
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So now, so what we've
done is we've just
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listed out all of the
total differentials.
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And the nice thing about
this is once you've
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done these computations,
now it's just substitution.
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So what we really want
to know is how does z
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depend on u and v. And
so all we need to do
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is substitute in our
formulas for dx here.
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So this tells us
that dz is-- OK,
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so we have 2x-- instead of
dx, we just plug in here--
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so we have 2u du minus 2v dv.
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So that was this term.
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And now we have
plus 2y-- and now
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we just plug in this--
so v du plus u dv.
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You see?
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It's just substitution.
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So then now, we just
expand everything out.
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And so we get-- OK,
so let's collect
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all the things involving du.
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So if we collect all the
things involving du, we have--
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2 times 2 times x times
u-- 4x*u plus 2y*v.
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This whole quantity times du.
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And then if we collect the
terms in dv, we have 2y*u.
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So that's coming from here, and
then we have a minus 4x*v. OK?
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And now what that tells
us is that-- so let's
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just remember that
one definition
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of the partial derivative
partial z partial u
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is this coefficient.
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So if I go over here, if we
write the total differential
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dz, we can write that as
partial z partial u du
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plus partial z partial v dv.
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Right?
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Well, look.
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What we have here
on these two sides
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is essentially the
same expression.
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So that means if
we want to compute
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partial z partial u, then that's
just equal to this coefficient
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here.
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So we get that partial z
partial u is 4x*u plus 2--
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that should be v.
One of those is an x.
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Let's see.
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So where did this come from.
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Yeah, one of those
is an x, sorry--
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SPEAKER 1: It's a y.
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DAVID JORDAN: --is a y.
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2v*y, OK.
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Now just as a sanity
check, why don't we
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go back to the
middle of the board,
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and we'll see that we
got the same thing.
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So 4x*u plus 2v*y, that's what
we concluded for partial z
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partial u.
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And then going back to the
middle of the board, that's we
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found again.
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So let's just go over
the two different methods
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and compare them.
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So if I'm in a rush to
do a computation-- maybe
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I'm taking an
exam-- I definitely
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think it's the quickest
to just compute,
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to figure out what the
dependency of the variable is,
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and I use this dependency graph.
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And then I just trace
all the paths from z
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to the independent variable
u that I'm interested in.
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And then I multiply all
the partial derivatives
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that correspond to each edge
and I get an expression.
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Now if I have more
time, then I really
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prefer to use the method
of total differentials
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that we did on the third board.
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I like it, because once you
do some simple calculus,
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and then after that it's
just, it's basic algebra.
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I find that I'm
less likely to make
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a mistake doing that method.
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But as you saw, it
involves computing
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a lot more derivatives
that we didn't actually
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use in the final answer.
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For instance, when we
computed total differentials,
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we got an expression
for partial z partial v
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at the end of the
day, even though we
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weren't asked to do that.
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So it's lengthier, but I
think more conceptually
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straightforward.
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So I think I'll
leave it at that.