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.