The following content is
provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer
high quality educational resources for free.
To make a donation or to view additional materials from
hundreds of MIT courses, visit MIT OpenCourseWare at
ocw.mit.edu. OK, so we're going to continue
looking at what happens when we have non-independent variables.
So, I'm afraid we don't take deliveries during class time,
sorry. Please take a seat, thanks.
[LAUGHTER] [APPLAUSE]
OK, so Jason, you please claim your package
at the end of lecture. OK,
so last time we saw how to use Lagrange multipliers to find the
minimum or maximum of a function of several variables when the
variables are not independent. And, today we're going to try
to figure out more about relations between the variables,
and how to handle functions that depend on several variables
when they're related. So, just to give you an
example, in physics, very often,
you have functions that depend on pressure, volume,
and temperature where pressure, volume,
and temperature are actually not independent.
But they are related, say, by PV=nRT.
So, of course, then you can substitute and
expressed a function in terms of two of them only,
but very often it's convenient to keep all three.
But then we have to figure out, what are the rates of change
with respect to t, with respect to each other,
the rate of change of f with respect to these variables,
and so on. So, we have to figure out what
we mean by partial derivatives again.
So, OK, more generally,
let's say just for the sake of notation,
I'm going to think of a function of three variables,
x, y, z, where the variables are related
by some equation, but I will put in the form g of
x, y, z equals some constant. OK, so that's the same kind of
setup as we had last time, except now we are not just
looking for minima and maxima. We are trying to understand
partial derivatives. So, the first observation is
that if x, y, and z are related,
then that means, in principle,
we could solve for one of them, and express it as a function of
the two others. So, in particular,
can we understand even without solving?
Maybe we can not solve. Can we understand how the
variables are related to each other?
So, for example, z, you can think of z as a
function of x and y. So, we can ask ourselves,
what are the rates of change of z with respect to x,
keeping y constant, or with respect to y keeping x
constant? And, of course,
if we can solve, that we know the formula for
this. And then we can compute these
guys. But, what if we can't solve?
So, how do we find these things without solving?
Well, so let's do an example. Let's say that my relation is
x^2 yz z^3=8. And, let's say that I'm looking
near the point (x, y, z) equals (2,3,
1). So, let me check 2^2 plus three
times one plus 1^3 is indeed eight.
OK, but now, if I change x and y a little
bit, how does z change? Well, of course I could solve
for z in here. It's a cubic equation.
There is actually a formula. But that formula is quite
complicated. We actually don't want to do
that. There's an easier way.
So, how can we do it? Well, let's look at the
differential -- -- of this constraint quantity.
OK, so if we called this g, let's look at dg.
So, what's the differential of this?
So, the differential of x^2 is 2x dx plus, I think there's a
zdy. There's a ydz,
and there's also a 3z^2 dz. OK, you can get this either by
implicit differentiation and the product rule,
or you could get this just by putting here,
here, and here the partial
derivatives of this with respect to x, y, and z.
OK, any questions about how I got this?
No? OK.
So, now, what do I do with this? Well, this represents,
somehow, variations of g. But, well, I've set this thing
equal to eight. And, eight is a constant.
So, it doesn't change. So, in fact,
well, we can set this to zero because, well,
they call this g. Then, g equals eight is
constant. That means we set dg equal to
zero. OK, so, now let's just plug in
some values at this point. That tells us,
well, so if x equals two, that's 4dx plus z is one.
So, dy plus y 3z^2 should be 6dz equals zero.
And now, this equation, here, tells us about a relation
between the changes in x, y, and z near that point.
It tells us how you change x and y, well, how z will change.
Or, it tells you actually anything you might want to know
about the relations between these variables so,
for example, you can move dz to that side,
and then express dz in terms of dx and dy.
Or, you can move dy to that side and express dy in terms of
dx and dz, and so on. It tells you at the level of
the derivatives how each of the variables depends on the two
others. OK, so, just to clarify this:
if we want to view z as a function of x and y,
then what we will do is we will just move the dz's to the other
side, and it will tell us dz equals
minus one over six times 4dx plus dy.
And, so that should tell you that partial z over partial x is
minus four over six. Well, that's minus two thirds,
and partial z over partial y is going to be minus one sixth.
OK, another way to think about this: when we compute partial z
over partial x, that means that actually we
keep y constant. OK, let me actually add some
subtitles here. So, here that means we keep y
constant. And so, if we keep y constant,
another way to think about it is we set dy to zero.
We set dy equals zero. So if we do that,
we get dx equals negative four sixths dx.
That tells us the rate of change of z with respect to x.
Here, we set x constant. So, that means we set dx equal
to zero. And, if we set dx equal to
zero, then we have dz equals negative one sixth of dy.
That tells us the rate of change of z with respect to y.
OK, any questions about that? No?
What, yes? Yes, OK, let me explain that
again. So we found an expression for
dz in terms of dx and dy. That means that this thing,
the differential, is the total differential of z
viewed as a function of x and y. OK, and so the coefficients of
dx and dy are the partials. Or, another way to think about
it, if you want to know partial z partial x, it means you set y
to be constant. Setting y to be constant means
that you will put zero in the place of dy.
So, you will be left with dz equals minus four sixths dx.
And, that will give you the rate of change of z with respect
to x when you keep y constant, OK?
So, there are various ways to think about this,
but hopefully it makes sense. OK, so how do we think about
this in general? Well, if we know that g of x,
y, z equals a constant, then dg, which is gxdx gydy
gzdz should be set equal to zero.
OK, and now we can solve for whichever variable we want to
express in terms of the others. So, for example,
if we care about z as a function of x and y -- -- we'll
get that dz is negative gx over gz dx minus gy over gz dy.
And, so if we want partial z over partial x,
well, so one way is just to say that's going to be the
coefficient of dx in here, or just to write down the other
way. We are setting y equals
constant. So, that means we set dy equal
to zero. And then, we get dz equals
negative gx over gz dx. So, that means partial z over
partial x is minus gx over gz. And, see,
that's a very counterintuitive formula because you have this
minus sign that you somehow probably couldn't have seen come
if you hadn't actually derived things this way.
I mean, it's pretty surprising to see that minus sign come out
of nowhere the first time you see it.
OK, so now we know how to find the rate of change of
constrained variables with respect to each other.
You can apply the same to find, if you want partial x,
partial y, or any of them, you can do it.
Any questions so far? No?
OK, so, before we proceed further, I should probably
expose some problem with the notations that we have so far.
So, let me try to get you a bit confused, OK?
So, let's take a very simple example.
Let's say I have a function, f of x, y equals x y.
OK, so far it doesn't sound very confusing.
And then, I can write partial f over partial x.
And, I think you all know how to compute it.
It's going to be just one. OK, so far we are pretty happy.
Now let's do a change of variables.
Let's set x=u and y=u v. It's not very complicated
change of variables. But let's do it.
Then, f in terms of u and v, well, so f, remember f was x y
becomes u plus u plus v. That's twice u plus v.
What's partial f over partial u? It's two.
So, x and u are the same thing. Partial f over partial x,
and partial f over partial u, well, unless you believe that
one equals two, they are really not the same
thing, OK? So, that's an interesting,
slightly strange phenomenon. x equals u, but partial f
partial x is not the same as partial f partial u.
So, how do we get rid of this contradiction?
Well, we have to think a bit more about what these notations
mean, OK? So, when we write partial f
over partial x, it means that we are varying x,
keeping y constant. When we write partial f over
partial u, it means we are varying u, keeping v constant.
So, varying u or varying x is the same thing.
But, keeping v constant, or keeping y constant are not
the same thing. If I keep y constant,
then when I change x, so when I change u,
then v will also have to change so that their sum stays the
same. Or, if you prefer the other way
around, when I do this one I keep v constant.
If I keep v constant and I change u, then y will change.
It won't be constant. So, that means,
well, life looked quite nice and easy with these notations.
But, what's dangerous about them is they are not making
explicit what it is exactly that we are keeping constant.
OK, so just to write things, so here we change u and x that
are the same thing. But we keep y constant,
while here we change u, which is still the same thing
as x. But, what we keep constant is
v, or in terms of x and y, that's y minus x constant.
And, that's why they are not the same.
So, whenever there's any risk of confusion,
OK, so not in the cases that we had before because what we've
done until now, we didn't really have a problem.
But, in a situation like this, to clarify things,
we'll actually say explicitly what it is that we want to keep
constant. OK, so what's going to be our
new notation? Well, so it's not particularly
pleasant because it uses, now, a subscript not to
indicate what you are differentiating,
but rather what you were holding constant.
So, that's quite a conflict of notation with what we had
before. I think I can safely blame it
on physicists or chemists. OK, so this one means we keep y
constant, and partial f over partial u with v held constant,
similarly. OK, so now what happens is we
no longer have any contradiction.
We have partial f over partial x with y constant is different
from partial f over partial x with v constant,
which is the same as partial f over partial u with v constant.
OK, so this guy is one. And these guys are two.
So, now we can safely use the fact that x equals u if we are
keeping track of what is actually held constant,
OK? So now, that's going to be
particularly important when we have variables that are related
because, let's say now that I have a
function that depends on x, y, and z.
But, x, y, and z are related. Then, it means that I look at,
say, x and y as my independent variables, and z as a function
of x and y. Then, it means that when I do
partials, say, with respect to x,
I will hold y constant. But, I will let z vary as a
function of x and y. Or, I could do it the other way
around. I could vary x,
keep z constant, and let y be a function of x
and z. And so, I will need to use this
kind of notation to indicate which one I mean.
OK, any questions? No?
All right, so let's try to do an example where we have a
function that depends on variables that are related.
OK, so I don't want to do one with PV=nRT because probably,
I mean, if you've seen it, then you've seen too much of
it. And, if you haven't seen it,
then maybe it's not the best example.
So, let's do a geometric example.
So, let's look at the area of the triangle.
So, let's say I have a triangle, and my variables will
be the sides a and b. And the angle here, theta.
OK, so what's the area of this triangle?
Well, its base times height over two.
So, it's one half of the base is a, and the height is b sine
theta. OK, so that's a function of a,
b, and theta. Now, let's say,
actually, there is a relation between a, b,
and theta that I didn't tell you about,
namely, actually, I want to assume that it's a
right triangle, OK?
So, let's now assume it's a right triangle with,
let's say, the hypotenuse is b. So, we have the right angle
here, actually. So, a is here. b is here.
Theta is here. So, saying it's a right
triangle is the same thing as saying that b equals sine theta,
OK? So that's our constraint.
That's the relation between a, b, and theta.
And, this is a function of a, b, and theta.
And, let's say that we want to understand how the area depends
on theta. OK, what's the rate of change
of the area of this triangle with respect to theta?
So, I claim there's various answers.
I can think of at least three possible answers. So, what can we possibly mean
by the rate of change of A with respect to theta?
So, these are all things that we might want to call partial A
partial theta. But of course,
we'll have to actually use different notations to
distinguish them. So, the first way that we
actually already know about is if we just forget about the fact
that the variables are related, OK?
So, if we just think of little a, b,
and theta as independent variables,
and we just change theta, keeping a and b constant -- So,
that's exactly what we meant by partial A, partial theta,
right? I'm not putting any constraints.
So, just to use some new notation, that would be the rate
of change of A with respect to theta, keeping a and b fixed at
the same time. Of course, if we are keeping a
and b fixed, and we are changing theta, it means we completely
ignore this property of being a right triangle.
So, in fact, it corresponds to changing the
area by changing the angle, keeping these lengths fixed.
And, of course, we lose the right angle.
When we rotate this side here, but the angle doesn't stay at a
right angle. And that one,
we know how to compute, right, because it's the one
we've been computing all along. So, that means we keep a and b
fixed. And then, so let's see,
what's the derivatives of A with respect to theta?
It's one half ab cosine theta. OK, now that one we know.
Any questions? No?
OK, the two other guys will be more interesting.
So far, I'm not really doing anything with my constraint.
Let's say that actually I do want to keep the right angle.
Then, when I change theta, there's two options.
One is I keep a constant, and then of course b will have
to change because if this width stays the same,
then when I change theta, the height increases,
and then this side length increases.
The other option is to change the angle, keeping b constant.
So, actually, this side stays the same
length. But then, a has to become a bit
shorter. And, of course,
the area will change in different ways depending on what
I do. So, that's why I said we have
three different answers. So, the next one is keep,
I forgot which one I said first.
Let's say keep a constant. And, that means that b will
change. b is going to be some function
of a and theta. Well, in fact here,
we know what the function is because we can solve the
constraint, namely, b is a over cosine theta.
But we don't actually need to know that so that the triangle,
so that the right angle, so that we keep a right angle.
And, so the name we will have for this is partial a over
partial theta with a held constant, OK?
And, the fact that I'm not putting b in my subscript there
means that actually b will be a dependent variable.
It changes in whatever way it has to change so that when theta
changes, a stays the same while b changes so that we keep a
right triangle. And, the third guy is the one
where we actually keep b constant,
and now a, we think a as a function of b
and theta, and it changes so that we keep
the right angle. So actually as a function of b
and theta, it's given over there.
A equals b cosine theta. And so, this guy is called
partial a over partial theta with b held constant.
OK, so we've just defined them. We don't know yet how to
compute these things. That's what we're going to do
now. That is the definition,
and what these things mean. Is that clear to everyone?
Yes, OK. Yes?
OK, so the second answer, again, so one way to ask
ourselves, how does the area depend on
theta, is to say, well,
actually look at the area of the right triangle as a function
of a and theta only by solving for b.
And then, we'll change theta, keep a constant,
and ask, how does the area change?
So, when we do that, when we change theta and keep a
the same, then b has to change so that it
stays a right triangle, right, so that this relation
still holds. That requires us to change b.
So, when we write partial a over partial theta with a
constant, it means that, actually, b will be the
dependent variable. It depends on a and theta.
And so, the area depends on theta, not only because theta is
in the formula, but also because b changes,
and b is in the formula. Yes?
No, no, we don't keep theta constant.
We vary theta, right? The goal is to see how things
change when I change theta by a little bit.
OK, so if I change theta a little bit in this one,
if I change theta a little bit and I keep a the same,
then b has to change also in some way.
There's a right triangle. And then, because theta and b
change, that causes the area to change.
OK, so maybe I should re-explain that again.
So, theta changes. A is constant.
But, we have the constraint, a equals be plus sine theta.
That means that b changes. And then, the question is,
how does A change? Well, it will change in part
because theta changes, and in part because b changes.
But, we want to know how it depends on theta in this
situation. Yes?
Ah, that's a very good question. So, what about,
I don't keep a and b constant? Well, then there's too many
choices. So I have to decide actually
how I'm going to change things. See, if I just say I have this
relation, that means I have two independent variables left,
whichever two of the three I want.
But, I still have to specify two of them to say exactly which
triangle I mean. So, I cannot ask myself just
how will it change if I change theta and do random things with
a and b? It depends what I do with a and
b. Of course, I could choose to
change them simultaneously, but then I have to specify how
exactly I'm going to do that. Ah, yes, if you wanted to,
indeed, we could also change things in such a way that the
third side remains constant. And that would be,
yet, a different way to attack the problem.
I mean, we don't have good notation for this,
here, because we didn't give it a name.
But, yeah, I mean, we could. We could call this guy c,
and then we'd have a different formula, and so on.
So, I mean, I'm not looking at it for simplicity.
But, you could have many more. I mean, in general,
you will want, once you have a set of nice,
natural variables, you will want to look mostly at
situations where one of the variables changes.
Some of them are held fixed, and then some dependent
variable does whatever it must so that the constraint keeps
holding. OK, so let's try to compute one
of them. Let's say I decide that we will
compute this one. OK, let's see how we can
compute partial a, partial theta with a held
fixed. [APPLAUSE]
OK, so let's try to compute partial A, partial theta with a
held constant. So, let's see three different
ways of doing that. So, let me start with method
zero. OK, it's not a real method.
That's why I'm not getting a positive number.
So, that one is just, we solve for b,
and we remove b from the formulas.
OK, so here it works well because we know how to solve for
b. But I'm not considering this to
be a real method because in general we don't know how to do
that. I mean, in the beginning I had
this relation that was an equation of degree three.
You don't really want to solve your equation for the dependent
variable usually. Here, we can.
So, solve for b and substitute. So, how do we do that?
Well, the constraint is a=b cosine theta.
That means b is a over cosine theta.
Some of you know that as a secan theta.
That's the same. And now, if we express the area
in terms of a and theta only, A is one half of ab cosine,
sorry, ab sine theta is now one half of a^2 sine theta over
cosine theta. Or, if you prefer,
one half of a^2 tangent theta. Well, now that it's only a
function of a and theta, I know what it means to take
the partial derivative with respect to theta,
keeping a constant. I know how to do it.
So, partial A over partial theta,
a held constant, well, if a is a constant,
then I get this one half a^2 coming out times,
what's the derivative of tangent?
Secan squared, very good. If you're European and you've
never heard of secan, that's one over cosine.
And, if you know the derivative as one plus tangent squared,
that's the same thing. And, it's also correct.
OK, so, that's one way of doing it.
But, as I've already said, it doesn't get us very far if
we don't know how to solve for b.
We really used the fact that we could solve for b and get rid of
it. So, there's two systematic
methods, and let's say the basic rule is that you should give
both of them a chance. You should see which one you
prefer, and you should be able to use one or the other on the
exam. OK, most likely you'll actually
have a choice between one or the other.
It will be up to you to decide which one you want to use.
But, you cannot use solving in substitution.
That's not fair. OK, so the first one is to use
differentials. By the way, in the notes they
are called also method one and method two.
I'm not promising that I have the same one,
am I?
I mean, I might have one and
two switched. It doesn't really matter.
So, how do we do things using differentials?
Well, first, we know that we want to keep a
fixed, and that means that we'll set da equal to zero,
OK? The second thing that we want
to do is we want to look at the constraint.
The constraint is a equals b cosine theta.
And, we want to differentiate that.
Well, differentiate the left-hand side.
You get da. And, differentiate the
right-hand side as a function of b and theta.
You should get, well, how many db's?
Well, that's the rate of change with respect to b.
That's cosine theta db minus b sine theta d theta.
That's a product rule applied to b times cosine theta.
So -- Well, now, if we have a constraint that's
relating da, db, and d theta,
OK, so that's actually what we did, right,
that's the same sort of thing as what we did at the beginning
when we related dx, dy, and dz.
That's really the same thing, except now are variables are a,
b, and theta. Now, we know that also we are
keeping a fixed. So actually,
we set this equal to zero. So, we have zero equals da
equals cosine theta db minus b sine theta d theta.
That means that actually we know how to solve for db.
OK, so cosine theta db equals b sine theta d theta or db is b
tangent theta d theta. OK, so in fact,
what we found, if you want,
is the rate of change of b with respect to theta.
Why do we care? Well, we care because let's
look, now, at dA, the function that we want to
look at. OK, so the function is A equals
one half ab sine theta. Well, then, dA,
so we had to use the product rule carefully,
or we use the partials. So, the coefficient of d little
a will be partial with respect to little a.
That's one half b sine theta da plus coefficient of db will be
one half a sine theta db plus coefficient of d theta will be
one half ab cosine theta d theta.
But now, what do I do with that? Well, first I said a is
constant. So, da is zero.
Second, well, actually we don't like b at
all, right? We want to view a as a function
of theta. So, well, maybe we actually
want to use this formula for db that we found in here.
OK, and then we'll be left only with d thetas,
which is what we want. So, if we plug this one into
that one, we get da equals one half a sine theta times b
tangent theta d theta plus one half ab cosine theta d theta.
And, if we collect these things together, we get one half of ab
times sine theta times tangent theta plus cosine theta d theta.
And, if you know your trig, but you'll see that this is
sine squared over cosine plus cosine squared over cosine.
That's the same as secan theta. So, now you have expressed da
as something times d theta. Well, that coefficient is the
rate of change of A with respect to theta with the understanding
that we are keeping a fixed, and letting b vary as a
dependent variable. Not enough space: sorry. OK, in case it's clearer for
you, let's think about it backwards.
So, we wanted to find how A changes.
To find how A changes, we write da.
But now, this tells us how A depends on little a,
little b, and theta. Well, we know actually we want
to keep little a constant. So, we set this to be zero.
Theta, well, we are very happy because we
want to express things in terms of theta.
Db we want to get rid of. How do we get rid of db?
Well, we do that by figuring out how b depends on theta when
a is fixed. And, we do that by
differentiating the constraint equation, and setting da equal
to zero. OK, so -- I guess to summarize
the method, we wrote dA in terms of da, db, d theta.
Then, we say that a is constant means we set da equals zero.
And, the third thing is that because, well,
we differentiate the constraint.
And, we can solve for db in terms of d theta.
And then, we plug into dA, and we get the answer.
OK, oops. So, here's another method to do
the same thing differently is to use the chain rule.
So, we can use the chain rule with dependent variables,
OK? So, what does the chain rule
tell us? The chain rule tells us,
so we will want to differentiate -- -- the formula
for a with respect to theta holding a constant.
So, I claim, well, what does the chain rule
tell us? It tells us that,
well, when we change things, a changes because of the
changes in the variables. So, part of it is that A
depends on theta and theta changes.
How fast does theta change? Well, you could call that the
rate of change of theta with respect to theta with a
constant. But of course,
how fast does theta depend to itself?
The answer is one. So, that's pretty easy.
Plus, then we have the partial derivative, formal partial
derivative, of A with respect to little a times the rate of
change of a in our situation. Well, how does little a change
if a is constant? Well, it doesn't change.
And then, there is Ab, the formal partial derivative
times, sorry, the rate of change of b.
OK, and how do we find this one? Well, here we have to use the
constraint. OK, and we can find this one
from the constraint as we've seen at the beginning either by
differentiating the constraint, or by using the chain rule on
the constraint. So, of course the calculations
are exactly the same. See, this is the same formula
as the one over there, just dividing everything by
partial theta and with subscripts little a.
But, if it's easier to think about it this way,
then that's also valid. OK, so tomorrow we are going to
review for the test, so I'm going to tell you a bit
more about this also as we go over one practice problem on
that.
