DAVID JORDAN: Hello, and welcome back to recitation.
So in this problem, we're considering a function f of
three variables f of x, y, z, and it's differentiable, and
we're not told a formula for f.
We just know that it's differential at this point P,
which is 1 minus 1, 1, and we're told that the gradient
of f at that point is this particular vector 2i plus j
minus 3k at that point P. So all we understand about f is
how it looks around the point P.
Now, we also have this relation between the
variables, so x, y and z aren't unrelated.
They're related by this constraint that z is x squared
plus y plus 1.
So with this information, we want to compute the gradient
of a new function g, and the new function g is a function
of two variables, and this function g is obtained from f
by just plugging in our relation for y.
So we can use our constraint to solve for y, and then this
function g is just f with that constraint applied.
And what we really want to do is we want to find the
gradient of g at the point 1, 1.
So notice that when g is equal to 1, 1, that means that--
sorry, when the input of g is 1, 1, that means the input of
f is P. OK?
So why don't you pause the video and work
this out for yourself.
Check back with me and we'll work it out together.
OK, welcome back.
So let's get started.
So this problem may not look like it's a problem about
partial derivatives with constraints, but that's what
it's really going to boil down to, which is to say that when
we want to compute--
so when we want to answer this question by computing the
gradient, the first thing we're going to want to do is
compute the partial derivative of g and its variable x.
And from the way we set up the problem, that's just the same
as computing the partial derivative of f with respect
to x and keeping z constant.
Now, remember, when we do partial derivatives with
constraints, what's important about the
notation is what's missing.
The variable y is missing, and that's because we are going to
use the constraint to get rid of it, and that's exactly how
we define g, so this is the key observation.
So computing the gradient of g is just going to be computing
these partial derivatives with constraints.
So we'll do that in a moment, and I'll also just write that
the partial derivative of g in the z-direction is partial f
partial z, now keeping x constrained.
All right, so we need to compute these partial
derivatives with constraints.
And so you remember how this goes.
The way that I prefer to do this is to compute the total
differentials.
So let's compute over here.
The total differential df is--
the total differential of f is just the partials of f in the
x-direction, f in the y-direction, f in the
z-direction, and each of these is multiplied by the
corresponding differential.
And we don't know f, so we can't compute the partial
derivatives of it in general, but we do know these partial
derivatives at this point.
And so in the problem, we were given that this is 2dx
plus dy minus 3dz.
So this is just using the fact that the gradient of f we were
given is 2, 1 minus 3, OK?
So that's the total differential of f, and now we
have this constraint.
And remember, when we do these partial derivatives with
constraints, the trick is to take the differential of the
constraint.
So we had this equation z equals x squared plus y plus
1, and what we need to do is take its differential.
So we have dz is 2x dx plus dy.
So that's our constraint.
Now here we have this variable x, but we're not varying x in
this problem.
We're only focused on the point P, and at the
point P, x is 1.
So, in fact, dz is just 2dx plus dy, OK?
So now what we need to do is we need to combine the
constraint equation and the total differential for f into
one equation, and so this is just linear algebra now.
So I'll just come over here.
So we can rewrite our total differential for the
constraint as saying that dy is equal to dz minus 2dx, and
then we can plug that back into our total
differential for f.
And so we get that df is equal to 2dx plus--
now I plug in dy here--
so dz minus 2dx, and then finally, minus 3dz.
So altogether, I get a minus 2dz, because
this and this cancel.
OK.
We get a minus 2dz.
So what does that tell us about the gradient?
So remember that the differential now of g and its
variables is partial g partial x dx plus
partial g partial z dz.
And remember, the trick about partial derivatives and their
relation to total differentials is that the
partial derivative is just this coefficient.
So the fact that we computed df and we found that it was
minus 2dz, that tells us that dg--
so the thing that we need to use is that g, remember, is
just the specialization of f.
So dg is equal to df in this case, because g is just a
special case of f with constraints.
So when we computed df here, we were imposing exactly the
constraints that we used to define dg, and so what this
tells us is that we can just compare the coefficients here
and so our gradient is 0 minus 2.
So the 0 is because there is no dependence anymore on x at
this point.
There wasn't a dx term.
There could have been and there wasn't.
And the minus 2 is because that's the dependence on z.
OK, so this is a bit complicated, so why don't we
review what we did.
So going back over to the problem statement, we first
needed to realize that saying that g was a special case of f
where we use our constraints to solve for y, that's what
put us in the context of problems with constraints.
So the fact that the dependence on y was so simple
that we could just use the constraint.
OK.
So then, what that allowed us to do is it allowed us to say
that the partial derivative of g in the x-direction is just
the partial derivative in the x-direction subject to the
constraint, and that's what we did here.
And so then, the next steps that we did are the same steps
that we would always do to compute partial derivatives
with constraints, and so we finally got a relationship
that said that, subject to these constraints, df is equal
to minus 2dz.
And the point is that g is just the function f with those
constraints imposed, and so dg and df are the same, and so,
in particular, dg is minus 2dz.
And then, we remember that you can always read off partial
derivatives from the total differential.
They're just the coefficients.
And so in the end, we got that the gradient was 0 minus 2.
And I think I'll leave it at that.