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 differentiable 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.