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PROFESSOR: Hi.
Our lesson, today, hopefully will serve two purposes.
On the one hand, we will give a nice application of the
chain rule.
Now that we've had two units devoted to working with the
chain rule, I thought you might enjoy seeing how it's
used in places which aren't quite that obvious--
places that we wouldn't expect to be used, at least.
And secondly, I would like to pick as my application one
which comes up in many, many different contexts.
And without further ado, the topic I want to cover today is
called "Integrals Involving Parameters".
Now, that sounds like a big mouthful.
Let me motivate that for you, first of all, physically, and
then in term of a couple of geometric examples.
You all know from past experience my great ability
with physical applications, so I won't even try to find
anything profound here.
Let me just take a pseudo example, pointing out what
type of situation we're trying to deal with and leaving it to
your own backgrounds to see places where the same
principle could have been applied, but hopefully in a
more practical, meaningful way for you.
Imagine for example, that we have a platform that we're
looking along in the x direction.
And we've punched holes in this platform, say, and liquid
is trickling through these various holes.
The holes are all on a horizontal line this way.
What we're going to do is we're going to focus our
attention on a particular particle of the liquid, and
we're going to watch it as it falls.
And what we would like to do is find how far that particle
fell, say, during the first second of its flight.
Obviously, from a calculus point of view, the first thing
we have to do is know what the velocity function is, because
ultimately, we would like to integrate the velocity.
Now, notice that we would expect the velocity to be
depending on time.
If this were a freely falling situation, we'd expect the
usual gravitational type situation.
Or whatever the situation happened to be, we would
expect on the one hand that the velocity does
depend on of time.
On the other hand, because of how the streams are flowing--
in other words, we don't know what's happening above here
that's causing the water to shoot out-- we don't know what
the initial velocity is coming out of each of these holes in
the sense that a different opening may give rise to a
different velocity of stream coming out.
All I'm trying to bring out here is that as we try to
focus our attention on a particular opening, we find
that the velocity of the particle that will follow it
during that first second is a function both of its position
x-- in other words, x sub 0 in this case, because we're
focusing at x equals x0 and the time t as t
goes from 0 to 1.
And we then simply integrate along the
vertical direction here.
We find y as a function of x0 from 0 to 1-- v(x0, t) dt.
Once we're through integrating, you see, notice
that the integration is with respect to t.
So when we're through integrating, this being a
definite integral, t no longer appears.
We have a function of x0 alone, saying nothing more
than the distance that the particle falls during the
first second is a function of the position of the opening
along the line here.
Now, at any rate, the practical application is not
so much writing down this equation as the
inverse is the case.
Namely, in many practical applications, we are given
this particular integral.
And for some reason or other, want to determine
what v itself is.
In other words, we often want to find the derivative given
the integral.
All right.
Let's just let it go at that for the time being.
The important point is that I want you to see an example of
an integral.
Let me just write this here in more abstract form.
It appears to be a definite integral, a to b.
The function inside the integrand is apparently a
function of two variables, x and y--
say, in this case x0 and t.
One of the variables is a variable of integration--
in this case, y--
and the other variable is being treated as a constant.
And that's where the word parameter comes in. x is a
parameter, meaning a variable constant, in the sense that,
for this particular problem, x is chosen to be in some domain
and remains fixed.
In other words, this is some function of x when we're all
through here.
All right?
As a geometric example, imagine
the following situation.
We have a surface w = f(x,y).
We take the plane x = x0 and intersect this surface with
that particular plane.
We get a curve, you see.
Now, we look at that curve corresponding to two points, p
and q, where p and q are determined by the y
values a and b.
In other words, p corresponds to y = a.
q corresponds to y = b.
And now, a very natural question that might come up is
that we would like to find the area of this particular plane
region, in other words, the area of this slice
between p and q.
Now, you know, the first thing I hope that you'll notice is
that because this shape of the surface can be in many
different ways, the particular cross section that we get does
depend on the choice of x0.
Different slices--
different planes x = x0 will give us different curves of
intersection.
The point is that once we have the curve intersection x is
being treated as the parameter.
This particular curve is given by what equation?
f is a function of x0 and y.
See, x = x0 for every point on this curve.
And so the area of the region R is the integral from a to b,
f(x0, y) dy.
And the question that very often comes up is, how do you
find the derivative of A sub R with respect to x0, noticing,
you see, that A sub R is a function of x0 alone, the y
dropping out between the limits a and b when we perform
the operation of integration.
A third place that this type of situation occurs is in
solving certain differential equations.
For example, suppose we're given a particular curve and
that that curve determines a region R between the lines x =
a and x = b.
Suppose all we know about the curve is its
slope at any point.
We know that its slope at any point is given by dy/dx and
some function of x and y which we don't necessarily have to
go into right now.
And let's suppose, for the sake of argument, that we
solve this first order differential equation.
What we'll find, if we're lucky, is that y is some
function of x and an arbitrary constant.
Remember, once you have one solution to a differential
equation, you have an infinite family in terms of a one
parameter solution to a differential equation.
In other words, in finding the area of the region R in this
case, there are many curves that satisfy this particular
differential equation.
Until we know what specific points are being referred to
over here, the best we know for sure is what?
That these endpoints are a and b, that the integrand is
f(x,c), and we're integrating that with respect
to x from a to b.
I freudianly put a c in here, because I think what I was
trying to emphasize for you is that when you look at this
thing, observe that this integral is a
function of c alone.
Namely, when you integrate this thing, you integrate it
with respect to x.
The x drops out.
All you have left here is a c.
A sub R, then, is a function of c.
And in many cases, what we would like to do is see how
fast the area changes as a function of c.
In other words, how do we change the area as a function
of changing the arbitrary constant c?
And I have enough exercises in the assignment to give you
concrete drill on this.
All I'm trying to give you here is an overview of the
entire topic.
And the reason I want to give you this overview is that it's
hinted at in the textbook, but this topic
is not covered there.
In fact, the reason that I had you read that particular
section of the textbook before the lecture this time-- you
notice that usually we start with the lecture.
This time I had you read the textbook first, because the
way the textbook covered this topic is essentially nothing
more than the way we tackled a different
problem last semester.
And I want you to see that the problem done in the Thomas
text is not a new problem--
it's one that we've done before--
but that with the tools that we now have available, we
could've tackled a more significant problem.
And that's the one I'm electing to do here and what I
want to show you the key steps on, because
they're not in the text.
But I will leave the
reinforcement for the exercise.
At any rate, hopefully now, when you see an integral of
this type, it will not bother you too much.
In other words don't worry about how do you integrate a
function of two independent variables?
When you see something like this, it means that there is
some implicitly implied domain for x.
In other words, we have some function of x.
Let's say the domain of g might very well be, say, all
x's between two values, say c and d.
But who cares about that right now?
The important point is that g is defined on a certain
set of values x.
And what it says is to compute the output of the g machine.
For the given x, you fix that x and integrate f(x,y) dy
between a and b.
Notice, you see, during the integration, x is being
treated as a constant, so that for all intents and purposes,
this is an ordinary integral.
But because x isn't a bona fide constant, meaning what?
It's a constant only in the sense that once chosen, it
remains fixed for this particular integration.
Different values of x will give me
different integrals here.
And consequently a very natural question that comes up
is how does my function g-- which depends on x--
how does that vary as x varies?
In other words, the key question is simply this.
First of all, given g defined this way, one, does g prime
even exist?
Does dg/dx exist?
And two, if it does exist, what is it?
In other words, the question that we're raising is if we
can find g prime of x, how do we do it in terms of looking
at the right hand side?
And let me not try to guess the answer here.
The answer does turn out to be, in this particular case,
one that you might have guessed.
But I prefer to show you that we don't have to
guess, point one.
Point two, if you do guess, you won't
always be that lucky.
That's my finale for today's lecture.
But let me see if I can survive to get
to the finale first.
Let me see how we'll tackle a problem like this.
First of all, to see if g prime exists at some value x0,
what we have to do is--
way back to the very beginning of Part 1--
the same old definition for an ordinary derivative.
We have to compute the limit of g of x sub 0 plus h minus
g(x sub 0) over h, taking the limit as h approaches 0.
Notice what the g machine does...
What the g machine does is it feeds x into the integrand
here and integrates this with respect to y from a to b.
So if the input of my g machine is x0 plus h, that
means that the x is replaced by x0 plus h here. g of x0
plus h is simply integral from a to b f of x0
plus h comma y dy.
Similarly, g(x0) is integral from a to b f(x0, y) dy.
I now want to form this difference.
And noticing--
again, this is all calculus of a single variable--
that the difference of two definite integrals is the
definite integral of the difference, I can conclude
that g(x0 + h) minus g(x0) is simply this single
integral over here.
Now, my next step in determining g prime of x0 is I
must divide this by h.
Notice, by the way, that h is an arbitrary increment, but
once chosen, remains fixed.
Notice that h is a constant as far as this
integration is concerned.
Consequently, to divide by h, it is permissible to bring the
h inside the integrand.
In other words, technically speaking, the h should be
here, but since h is a constant with respect to y,
the integral can have the h brought in.
Why do I want to bring the h in here?
Let me again telegraph what I'm leading up to.
Obviously, when I'm going to compute g prime, my next step
is to take the limit of this as h approaches 0.
With h in here, I look at this and what I hope is that by
this amount of time at least the following minimum amount
of material has rubbed off on you in a second nature way--
that if I look at this expression in brackets as h
approaches 0, this is precisely the definition of
what we mean by the partial of f(x, y) with respect to x
evaluated at (x0, y).
See, this is what?
The change in f-- see, y is held constant.
We're taking this over to change from x0 to x0 plus h
and dividing by h.
This is a partial of f with respect to x.
That's why I want to bring the h inside.
So now, I say, OK, g prime of x0, by
definition, is this limit.
I now want to take the limit of this expression.
And by the way, notice what I'd love to do now is to jump
right in here and say, aha, this is just the partial of f
with respect to x evaluated at (x0, y0).
But the thing I would like you to notice--
and again, going back to Part 1 of course, one of the big
things that we talked about under the heading of uniform
convergence.
There is a very dangerous thing in general to
interchange the order of limit and integration.
This says what?
First perform the integration, and then take the limit.
What we would like to be able to do is first take the limit
and then integrate the result.
Now, we did see that, provided the integrand was continuous,
these operations were permissible.
But we'll talk about that a little later.
For the time being, let's simply summarize by saying if
the limit operation and the integration operation can be
interchanged, then the derivative--
see, this thing here is what?
This is my g(x).
The derivative of g(x) with respect to x has the very
delightful form that, essentially, all I have to do
is take the derivative operation, come inside the
integrand, and replace the derivative with respect to x
by the partial derivative with respect to x.
In other words, the derivative of the integral from a to b,
f(x, y) dy is the integral from a to b, the partial of f
with respect to x dy--
provided, of course, that the limit and the integration are
interchangeable.
In particular, this will be true if f and f sub x exist
and are continuous and see, straighten out the range and
the domain and what have you once and for all.
Notice that y is allowed to exist between a and b.
We've said that x is going to exist on some domain
between c and d.
Notice that saying that y is between a and b and x is
between c and d geometrically says that f is
defined on a rectangle.
See?
OK.
And then what we're saying is, under these conditions, to
integrate an integral with a parameter, with respect to
that parameter, all we have to do is come inside the integral
sign and differentiate--
take the partial derivative--
with respect to what is being used as the parameter.
In this case, it's x, which is the parameter.
Now, the only danger with this particular thing-- and by the
way, notice, not only is there a danger here that I'm going
to mention.
The danger is this looks so easy, you may be saying, why
did he do it the hard way?
Why didn't he just tell us this was the right
way of doing it?
And the point is it just happens to be one of those
coincidences where the rigorous way yields a logical
answer which is consistent with what is probably our
intuitive guess.
But it's not always going to happen that way.
And the example that I have in mind now goes back to what we
were talking about the beginning of the lecture.
Namely, I wanted to give you an application
of the chain rule.
And here's where that application comes in.
I now call this--
I don't know what to call it.
So let's just call it variable limits of integration.
Same problem as before--
it's going to cause the chain rule to come in now.
The only difference is going to be--
and that's just a forewarning.
You don't have to know that right now.
I'm going to have the same problem as before.
What did I have before?
I had that g(x) was integral f(x, y) dy between the two
constants a and b.
Now, I'm going to let my constants of integration also
depend on the parameter.
See, all the constants of integration have to be our
constants as far as y is concerned.
What I'm saying is what makes this problem differ from the
previous one is suppose that it happens that instead of
being given a nice rectangle to play around with, I'm given
a couple of curves like this in the xy plane.
See, this would be a of x.
This would be y equals b(x).
See, what I'm saying now is that not only does the
integrand depend on what value of x I pick, but the limits of
integration as I'm finding a cross-sectional area of a
surface, you see.
The limits of the integral themselves depend on the
choice of x--
constant, as far as y is concerned, but depend on x.
You see, now, what happens is that my parameter appears in
the limits as well as just in the integrand.
And now, you see, also what this means is if I try the
previous approach of computing g(x plus delta x), et cetera,
I'm in trouble, because the only way I can combine two
integrals and put them under the same integral sign is if
they're between the same limits of integration.
Notice here, for example, that if I replace x by x plus delta
x, I not only change the integrand, but notice that the
limits become what?
a of x plus delta x, b(x plus delta x)-- and those in
general, unless a and b happen to be constants,
will vary with x.
In fact, let's look at it this way.
This is the problem we should have started with in the sense
that constant limits of integration are a
special case of this.
At any rate, what I wanted to show you was that this
particular problem can be handled very nicely in terms
of the chain rule.
Namely, what we do here is we observe that, first of all, y
is not really a variable here.
It's integrated out.
So what we think of is let's think of x as being some
variable u.
Let's think of b(x) as being some variable v. Let's write
down the function of three independent
variables u, v, and x.
OK.
What will that function be?
Let u, v, and x be arbitrary, independent variables.
Look at the integral from u to v f(x, y) dy.
This is obviously dependent upon u.
It's dependent upon v. And it's dependent upon x.
The place that the chain rule comes in is that in our
particular problem, u and v cannot be arbitrary, but
rather u must be that particular function a(x), and
v must be the particular function b(x).
Consequently g(x) is simply what?
It's h(u, v, x), where u is a(x) and v is b(x).
Consequently, to find g prime of x, what we want is h prime
of x and to find h prime as a function of x.
See the chain rule here?
u can be expressed in terms of x.
v can be expressed in terms of x.
Obviously, x is already expressed in terms of x.
So this is really implicitly a function of x alone.
So by the chain rule, what I'd like to be able to do is to
combine these three pieces of information to
find h prime of x.
And remember how the chain rule works.
Now, I'm not going to beat that to death.
We've just had two units on that.
Let's just say it rather quickly.
g prime of x is the partial of h with respect to u times u
prime of x plus the partial of h with respectively to v times
v prime of x times the partial of h with respect to x times x
prime of x, which, of course, is just 1.
In other words, writing this thing out, g prime of x is
simply this.
What is the partial of h with respect to u?
Let's come back here for a second and remember what h is.
h is this integral.
I want to take the partial of that with respect to u.
That means I have to investigate this.
Now, here's the interesting point.
Whereas u, v, and x are independent variables, what
does it mean when you say you're taking the partial with
respect to u?
It means that you're treating v and x as constants.
Now, if v and x are being treated as constants, what I
have is simply what?
I'm taking a derivative with respect to a variable where
the only place the variable appears is as the lower limit
of the integrand.
In other words, I claim that that's nothing more
than minus f(x, u).
Then I go inside the integrand.
In other words, I differentiate the integral.
That leaves me just the integrand--
and replace the variable by the variable of
integration u here.
And because it's the lower limit, I put
in the minus sign.
Now, why did I go through that very fast?
That's why I had you read this assignment first.
Notice that the assignment in the textbook does not touch
what I'm talking about, but rather seems to review that
topic that we covered under Part 1--
that if you wanted to take an ordinary derivative with
respect to u, integral from u to a, g(y) dy, the answer
would just be minus g(u).
And that's exactly what I did in here.
I treated v and x as constants here.
In other words, the only variable in here was u.
That will be emphasized again in the exercises.
In a similar way, the partial of h with respect to v means
this thing.
Notice now that u is being treated as a constant.
x is being treated as a constant.
To differentiate this, my variable appears only as an
upper limit on the integrand.
That means I come inside the integral sign, replace the
integral by just the function itself, replacing what?
Replacing y--
that's the only variables of integration--
by the upper limit v. In other words, this is f(x, v).
All right?
And finally, the partial of h with respect to x is this
integral here.
Notice now that u and v are being treated as constants.
With u and v being treated as constants, that's the special
case that we started out lecture with, namely, to
differentiate with respect to a parameter when the parameter
appears only as part of the integrand.
So how do we do that?
We come inside.
This is the interval from u to v, the partial of f with
respect to x dy.
Putting the whole thing together--
recalling, among other things, that du/dx, since u is a(x),
is just a prime of x and that dv/dx is b prime of x-- what
this says--
and again, I want to see the beauty of the chain rule here,
because at least to my way of thinking, I don't see anything
at all intuitive about the result I'm going to show you.
And that is as soon as you make the limits of integration
variable, to differentiate an integral involving a
parameter--
you see again, what's the parameter here? x.
We integrate it with respect to y.
This is a constant as far as y is concerned.
See, intuitively, you might say, gee, all you've got to do
is take the derivative sign, bring it inside, and this
should be the answer.
See, it's the same as we did before.
The point is-- and this is where many serious mistakes
are made in problems involving integrals of this type--
is that the reason that our intuitive way happened to be
right in the simpler case was that these were constants with
respect to x.
Now, they're variables.
Well, it turns out, if you just wrote this thing down,
you would be wrong.
What is the correction factor?
Again, come back to here.
The correction factor is this here, which we've just started
to compute.
Again, just saying it--
if you wrote this term down to get the correct answer, you
would have to tack on what?
b prime of x times f(x, b(x)).
That means what?
You think of a y as being over here.
For the particular value of x, you replace y by b(x).
In other words, you look at f(x, y), and every place you
see a y, replace it by b(x).
And then subtract that from that a prime of
x times f(x, a(x)).
Now again, I suspect that, for many of you, it's the first
time that you've seen something like this, because I
say it's a topic which I believe was a natural one to
occur in the textbook, but for some reason it
doesn't appear there.
Because of the importance of the concept, the number of
times it appears in physical applications, the number of
times that one has to differentiate with respect to
an integral--
I don't know the physical applications well enough to
lecture on them, but it does occur in probability theory,
among other places, it appears in any subject involving
integral equations and the like--
that I wanted to give you the experience of seeing what the
concept means, to have you hear me say it.
And then I will spend the exercises trying to drive home
the computational know how so that you will be able to do
these things at least in a mechanical way, independently
of whether the theory made that much sense because of the
lack of physical example motivation other than what I
did at the beginning.
At any rate, keep in mind, though, that in terms of our
present topic, where we're talking about the chain rule,
this is a certainly noble application to show an
important place in the physical world where knowledge
of the chain rule plays a very important role.
And with that, I might just as well conclude today's lecture.
And until next time, good bye.
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