The chain rule with constraints | 18.02SC Multivariable Calculus, Fall 2010 Video Lecture - Engineering Mathematics

DAVID JORDAN: Hello, and welcome
back to recitation.
So today, the problem I'd like
to work with you is about
taking partial derivatives in
the presence of constraints.
So this is a pretty
subtle business.
So take your time when you
work these problems.
So what we have is we have this
function w, and it's a
function of four variables:
x, y, z, and t.
OK?
But it's not really a function
of these four variables
because we have a constraint.
So we want to study how w
changes as we vary the
parameters, except that
we have imposed
this constraint here.
So that really we kind of only
have three variables, because
we have four variables
and one constraint.
So that's what partial
derivatives with constraints
help us do.
So let's explain first
the notation.
OK?
So it says partial w partial
z, and then we have the
subscripts x and y.
So what's important about this
notation is not what you see
as much as what you don't see.
What you don't see is
the variable t.
OK?
So what this notation means is,
as always, the denominator
in our derivative expression--
partial z here-- that means
that we want to vary z.
And we want to see how w
changes as we vary z.
And the x and y here mean that
we want to keep x and y fixed.
So if we didn't have a
constraint, this x and y here
would be superfluous.
Because by partial derivative,
we always mean to keep the
other unlisted variables
unchanged.
However, the fact that t is
missing here, it means that--
so if you think about it-- if we
vary z, and we keep x and y
fixed, then t also is varying.
Right?
Because we have this
constraint here.
And so it wouldn't make sense
for me to ask you to compute
the partial derivative of w in z
varying x, y and t because--
excuse me--
keeping x, y, and t fixed,
because then there would be no
room for z to vary.
OK?
So this notation means that z
is going to be allowed to
vary, but it's going to vary
in a way that we're
just going to ignore.
So you will see how this works
out in the problem.
So what we're really interested
in is making sure
that x and y stay fixed
and that z varies.
And then we're going to need to,
when we do some algebra,
we're going to need to get
rid of any mention of
the variable t.
OK.
So the first way that we're
going to work this out is
using total differentials.
And I like to use total
differentials when I'm on new
ground because they're not
the most computationally
effective, because they involve
computing all the
derivatives that we might
possibly need in sight.
So they're not the most
efficient computationally.
But if you go ahead and
compute the total
differentials, then all the
other computations that you
have to do are just
substitution.
So it really just becomes linear
algebra, and that's
what I like about it.
In part b, we'll see a shortcut
using implicit
differentiation and
the chain rule.
And this is going to be
a little bit tricky.
So we have these two equations,
we need to turn
them both into differential
equations.
And so we'll do that using
a combination of implicit
differentiation and
the chain rule.
So I'll let you pause the video
and get started on these
problems. And you can
check back and
we'll work it out together.
OK, welcome back.
So let's start by doing a, let's
start with problem a.
So we have total differentials
is the suggested
way to attack this.
So why don't we just start
computing the total
differentials that we know.
So we have two equations.
w in relation to the
other variables and
the constraint equation.
And what we first want to do
is just take the total
differential of both of those
equations to get started.
So we can take the first
one and it tells us
that dw is equal to--
OK so we have--
3x squared y dx, plus x cubed
dy, minus 2zt dz, minus z
squared dt.
OK.
Now right away, we can simplify
this equation.
So this is the total
differential, but we have to
remember that in the setting
we're interested in, x and y
are held fixed.
And so holding x and y fixed
means that the differentials
dx and dy are both set to 0.
So that lets us rewrite this
first differential equation is
just dw equals minus 2zt
dz minus z squared dt.
So that's our first equation
that we get.
Let me just check with my
notes to make sure.
That's right.
OK.
And so now, we have the
constraint equation from the
original statement
of the problem.
And we need to take
its differential.
So on the one hand, we
get x dy plus y dx.
That's the total differential
of the left-hand side.
And then on the right-hand side,
we have t dz plus z dt.
OK?
And now we notice that now the
left-hand side of this
equation is just 0 for
the same reason.
dy and dx are being
held fixed.
So the relation that we end up
getting is we get that dt is
equal to minus t over
z dz by just doing
straightforward algebra.
OK.
So with that in hand now we
can-- so remember I mentioned
in the beginning our goal
was-- so from the very
beginning, we knew that if we
varied z, because of our
constraint, we're going to be
forced to be varying t.
And that's exactly what this
equation says, doesn't it?
We got this by just taking
the differential of the
constraint.
And it says if you vary z,
you have to vary t in an
appropriate way, and
that's what this
coefficient tells us.
So what we're really interested
in is how does w
vary in terms of z here.
And so we want to get
rid of this dt here.
And in fact, we can by
using the constraint.
So combining this equation with
this equation, we get
that dw here is equal to--
OK, so we have--
minus 2zt dz.
And then we have--
minus--
OK--
z squared times another
minus times t over
z, so this all becomes--
plus zt dz.
So all I did is I plugged in for
dt using our formula here.
And so this altogether is equal
to just minus zt dz.
And that tells us that the
partial derivative that we're
after is just this coefficient,
right?
The partial derivative is just
defined to be the coefficient
of the differential once you
work everything out.
And so this is minus zt.
OK, so that's a.
So now let's see if we can use
some tricks to make the
computation a bit shorter.
So the tricks that we're going
to use are implicit
differentiation and
the chain rule.
So at the end of the
day-- excuse me--
we're interested in partial
w partial z.
And what we're going to do is
use the chain rule to just
take a straightforward partial
derivative of our original
expression.
So remember, w was x cubed
y minus zt squared.
And so let's just take a partial
derivative of that in
the z-direction.
So the partial derivative
in the z-direction of x
cubed y is just 0.
So that will go away.
And so we only have minus,
we have a 2zt component.
That's just because the partial
derivative of z
squared is 2z.
And then we have another term
which is minus z squared, and
now we need to take the partial
derivative of t in the
z-direction.
So, you know, often times when
we take partial derivatives of
one variable in terms of the
other, it's common to think
that the partial derivative of
one variable in terms of the
other is just 0.
Because usually our variables
are independent.
They don't vary in terms
of one another.
But this is exactly a situation
where t does vary
depending on z, and so
we had to include
that into our notation.
OK.
So now this is almost what we
want, except we have this
mystery component here.
And of course, there's only
one way we can solve this
mystery, which is the same way
we solved it in part a.
We have to use the constraint.
So let's take partial z of
our constraint equation.
And remember, our constraint
equation was xy equals zt.
OK.
So if we take the partial
derivative of this equation,
so if I take the partial
derivative x and y in terms of
z, then I do get 0, because
x and y are genuinely
independent from z.
It's only t that depends on z.
So on this side we get 0.
Now, on the other side I just
need to use the product rule.
So I get t, plus z partial
t partial z.
OK?
So we can rewrite this as saying
that partial t partial
z is minus t over z.
OK?
Now, you might notice that,
you know, this is formally
very similar to what we did in
part a, and of course, that's
no surprise.
When we are manipulating using
implicit differentiation and
the chain rule, it's just a
compact way of doing what we
were doing with the total
differentials.
I mean, to me, the chain rule
is a computation which you
could prove by doing the
corresponding thing with total
differentials.
And so we get this same
coefficient negative t over z,
which you recall that
we got in part a.
OK.
So now we have, once again we
have this, two equations and
we just can do substitution.
So we get that partial
w partial z is
equal to minus 2zt.
And now again, we get minus
another minus, and z here
cancels the z squared,
so we get plus zt.
And so we get minus zt.
OK, and finally, if we remember
our assumptions, our
assumptions were that x and
y were independent of z.
That was our notation.
And we use that assumption
at this step right here.
So in fact, we don't just have
the partial derivative of w
with respect to z.
We need to specify that
we held x and y fixed.
OK.
So just to review again, if we
look now at what we did in
part b, you know, the meat of
the argument was the exact
same as what we did in part a.
The meat of the argument
was right here.
We took some derivative and
then this was an unknown.
The definition of w doesn't
know how t and z
depend on one another.
That you can only find by
looking at the constraint.
And so we just went through
the problem and we took
derivatives of the constraint,
and that gave us an equation
that we were looking for.
Now if we go back now
to part a over here.
So as you can see, there's
a lot more work
involved in part a.
On the other hand, to me it
was more straightforward.
We just had to compute the total
differentials and then
do some linear algebra
with cancellations.
And somehow, when you do total
differentials, you just
compute everything that could
possibly come up, and then you
just substitute it in.
And indeed, we got the same
answer: partial w partial z as
being minus zt.
OK, and I think I'll
stop there.