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PROFESSOR: Today we're going
to continue our discussion
of parametric curves.
I have to tell you
about arc length.
And let me remind me where
we left off last time.
This is parametric
curves, continued.
Last time, we talked about
the parametric representation
for the circle.
Or one of the parametric
representations for the circle.
Which was this one here.
And first we noted that
this does parameterize,
as we say, the circle.
That satisfies the
equation for the circle.
And it's traced
counterclockwise.
The picture looks like this.
Here's the circle.
And it starts out here
at t = 0 and it gets up
to here at time t = pi / 2.
So now I have to talk
to you about arc length.
In this parametric form.
And the results should
be the same as arc length
around this circle ordinarily.
And we start out with this
basic differential relationship.
ds^2 is dx^2 + dy^2.
And then I'm going to take
the square root, divide by dt,
so the rate of change
with respect to t of s
is going to be the square root.
Well, maybe I'll write
it without dividing.
Just write it as ds.
So this would be
(dx/dt)^2 + (dy/dt)^2, dt.
So this is what you get
formally from this equation.
If you take its
square roots and you
divide by dt squared
in the-- inside
the square root, and you
multiply by dt outside,
so that those cancel.
And this is the formal
connection between the two.
We'll be saying just
a few more words
in a few minutes about how to
make sense of that rigorously.
Alright so that's the set of
formulas for the infinitesimal,
the differential of arc length.
And so to figure it out, I have
to differentiate x with respect
to t.
And remember x is up here.
It's defined by a cos t, so
its derivative is -a sin t.
And similarly, dy/dt = a cos t.
And so I can plug this in.
And I get the arc
length element,
which is the square root of
(-a sin t)^2 + (a cos t)^2, dt.
Which just becomes the square
root of a^2, dt, or a dt.
Now, I was about to divide by t.
Let me do that now.
We can also write the rate
of change of arc length
with respect to t.
And that's a, in this case.
And this gets
interpreted as the speed
of the particle going around.
So not only, let me
trade these two guys,
not only do we have the
direction is counterclockwise,
but we also have that the speed
is, if you like, it's uniform.
It's constant speed.
And the rate is a.
So that's ds/dt.
Travelling around.
And that means that we can
play around with the speed.
And I just want to point out--
So the standard thing, what
you'll have to get
used to, and this
is a standard presentation,
you'll see this everywhere.
In your physics classes and
your other math classes,
if you want to change
the speed, so a new speed
going around this would be, if
I set up the equations this way.
Now I'm tracing around
the same circle.
But the speed is
going to turn out
to be, if you figure
it out, there'll
be an extra factor of k.
So it'll be ak.
That's what we'll work
out to be the speed.
Provided k is positive
and a is positive.
So we're making
these conventions.
The constants that we're
using are positive.
Now, that's the first
and most basic example.
The one that comes
up constantly.
Now, let me just make those
comments about notation
that I wanted to make.
And we've been treating these
squared differentials here
for a little while
and I just want
to pay attention
a little bit more
carefully to these
manipulations.
And what's allowed
and what's not.
And what's justified
and what's not.
So the basis for this was this
approximate calculation that we
had, that (delta s)^2 was
(delta x)^2 + (delta y)^2.
This is how we justified the
arc length formula before.
And let me just show you
that the formula that I
have up here, this
basic formula for arc
length in the
parametric form, follows
just as the other one did.
And now I'm going to do it
slightly more rigorously.
I do the division really
in disguise before I take
the limit of the infinitesimal.
So all I'm really doing
is I'm doing this.
Dividing through by this,
and sorry this is still
approximately equal.
So I'm not dividing by something
that's 0 or infinitesimal.
I'm dividing by
something nonzero.
And here I have ((delta
x)/(delta t))^2 + ((delta
y)/(delta t))^2 And
then in the limit,
I have ds/dt is equal to
the square root of this guy.
Or, if you like, the
square of it, so.
So it's legal to divide by
something that's almost 0
and then take the
limit as we go to 0.
This is really what
derivatives are all about.
That we get a limit here.
As the denominator goes to 0.
Because the numerator's
going to 0 too.
So that's the notation.
And now I want to warn you,
maybe just a little bit,
about misuses, if you
like, of the notation.
We don't do absolutely
everything this way.
This expression that
came up with the squares,
you should never
write it as this.
This, put it on the board
but very quickly, never.
OK.
Don't do that.
We use these square
differentials,
but we don't do it
with these ratios here.
But there was another place
which is slightly confusing.
It looks very
similar, where we did
use the square of the
differential in a denominator.
And I just want to point out
to you that it's different.
It's not the same.
And it is OK.
And that was this one.
This thing here.
This is a second derivative,
it's something else.
And it's got a dt^2
in the denominator.
So it looks rather similar.
But what this represents is
the quantity d/dt squared.
And you can see the
squares came in.
And squared the two expressions.
And then there's
also an x over here.
So that's legal.
Those are notations
that we do use.
And we can even calculate this.
It has a perfectly good meaning.
It's the same as the
derivative with respect
to t of the derivative of x,
which we already know was minus
sine-- sorry, a sin t, I guess.
Not this example,
but the previous one.
Up here.
So the derivative
is this and so I can
differentiate a second time.
And I get -a cos t.
So that's a perfectly
legal operation.
Everything in there makes sense.
Just don't use that.
There's another really
unfortunate thing,
right which is that the 2 creeps
in funny places with sines.
You have sine squared.
It would be out here,
it comes up here
for some strange reason.
This is just because
typographers are lazy
or somebody somewhere
in the history
of mathematical typography
decided to let the 2 migrate.
It would be like
putting the 2 over here.
There's inconsistency
in mathematics, right.
We're not perfect and people
just develop these notations.
So we have to live with them.
The ones that people
accept as conventions.
The next example that
I want to give you
is just slightly different.
It'll be a non-constant
speed parameterization.
Here x = 2 sin t.
And y is, say, cos t.
And let's keep track
of what this one does.
Now, this is a skill
which I'm going
to ask you about quite a bit.
And it's one of several skills.
You'll have to connect
this with some kind
of rectangular equation.
An equation for x and y.
And we'll be doing a certain
amount of this today.
In another context.
Right here, to see the pattern,
we know that the relationship
we're going to want to use
is that sin^2 + cos^2 = 1.
So in fact the right thing to do
here is to take 1/4 x^2 + y^2.
And that's going to turn
out to be sin^2 t + cos^2 t.
Which is 1.
So there's the equation.
Here's the rectangular equation
for this parametric curve.
And this describes an ellipse.
That's not the only information
that we can get here.
The other information
that we can get
is this qualitative
information of where
we start, where we're
going, the direction.
It starts out, I
claim, at t = 0.
That's when t = 0, this is
(2 sin 0, cos 0), right?
(2 sin 0, cos 0) is equal
to the point (0, 1).
So it starts up up here.
At (0, 1).
And then the next little
place, so this is one thing
that certainly you
want to do. t = pi/2
is maybe the next
easy point to plot.
And that's going to be
(2 sin(pi/2), cos(pi/2)).
And that's just (2, 0).
And so that's over
here somewhere.
This is (2, 0).
And we know it travels
along the ellipse.
And we know the minor axis is
1, and the major axis is 2,
so it's doing this.
So this is what
happens at t = 0.
This is where we
are at t = pi/2.
And it continues all
the way around, etc.
To the rest of the ellipse.
This is the direction.
So this one happens
to be clockwise.
Alright, now let's keep
track of its speed.
Let's keep track of the speed,
and also the arc length.
So the speed is the square
root of the derivatives here.
That would be (2 cos
t)^2 + (sin t)^2.
And the arc length is what?
Well, if we want to
go all the way around,
we need to know that that
takes a total of 2 pi.
So 0 to 2 pi.
And then we have to integrate
ds, which is this expression,
or ds/dt, dt.
So that's the square root
of 4 cos^2 t + sin^2 t, dt.
The bad news, if you
like, is that this is not
an elementary integral.
In other words,
no matter how long
you try to figure out
how to antidifferentiate
this expression, no matter how
many substitutions you try,
you will fail.
That's the bad news.
The good news is this is
not an elementary integral.
It's not an elementary integral.
Which means that this is
the answer to a question.
Not something that
you have to work on.
So if somebody asks you for
this arc length, you stop here.
That's the answer, so it's
actually better than it looks.
And we'll try to--
I mean, I don't
expect you to know already
what all of the integrals
are that are impossible.
And which ones are hard
and which ones are easy.
So we'll try to coach
you through when
you face these things.
It's not so easy to decide.
I'll give you a few clues, but.
OK.
So this is the arc length.
Now, I want to move on to
the last thing that we did.
Last type of thing
that we did last time.
Which is the surface area.
And yeah, question.
STUDENT: [INAUDIBLE]
PROFESSOR: The question,
this is a good question.
The question is, when
you draw the ellipse,
do you not take into
account what t is.
The answer is that
this is in disguise.
What's going on here
is we have a trouble
with plotting in the plane
what's really happening.
So in other words, it's
kind of in trouble.
So the point is that we have
two functions of t, not one.
x(t) and y(t).
So one thing that I can do if
I plot things in the plane.
In other words, the
main point to make here
is that we're not talking
about the situation
y is a function of x.
We're out of that realm now.
We're somewhere in
a different part
of the universe in our thought.
And you should drop
this point of view.
So this depiction is not
y as a function of x.
Well, that's obvious because
there are two values here,
as opposed to one.
So we're in trouble with that.
And we have that
background parameter,
and that's exactly
why we're using it.
This parameter t.
So that we can depict
the entire curve.
And deal with it as one thing.
So since I can't really draw
it, and since t is nowhere
on the map, you should
sort of imagine it as time,
and there's some kind of
trajectory which is travelling
around.
And then I just labelled
a couple of the places.
If somebody asked you to
draw a picture of this,
well, I'll tell you exactly
where you need the picture
in just one second, alright.
It's going to come up
right now in surface area.
But otherwise, if
nobody asks you to,
you don't even have to put
down t = 0 and t = pi / 2 here.
Because nobody
demanded it of you.
Another question.
STUDENT: [INAUDIBLE]
PROFESSOR: So, another
very good question
which is exactly
connected to this picture.
So how is it that we're
going to use the picture,
and how is it we're going
to use the notion of the t.
The question was, why is
this from t = 0 to t = 2 pi?
That does use the t
information on this diagram.
the point is, we do
know that t starts here.
This is pi / 2, this is pi, this
is 3 pi / 2, and this is 2 pi.
When you go all the
way around once,
it's going to come
back to itself.
These are periodic
functions of period 2 pi.
And they come back to
themselves exactly at 2 pi.
And so that's why we know
in order to get around once,
we need to go from 0 to 2 pi.
And the same thing is going
to come up with surface area
right now.
That's going to be the
issue, is what range of t
we're going to need when we
compute the surface area.
STUDENT: [INAUDIBLE]
PROFESSOR: In a question,
what you might be asked
is what's the
rectangular equation
for a parametric curve?
So that would be
1/4 x^2 + y^2 = 1.
And then you might
be asked, plot it.
Well, that would be a
picture of the ellipse.
OK, those are types of questions
that are legal questions.
STUDENT: [INAUDIBLE]
PROFESSOR: The
question is, do I need
to know any specific formulas?
Any formulas that you know
and remember will help you.
They may be of limited use.
I'm not going to ask
you to memorize anything
except, I guarantee you that
the circle is going to come up.
Not the ellipse, the circle
will come up everywhere
in your life.
So at least at MIT,
your life at MIT.
We're very round here.
Yeah, another question.
STUDENT: I'm just a tiny bit
confused back to the basics.
This is more a question
from yesterday, I guess.
But when you have your
original ds^2 = dx^2 + dy^2,
and then you integrate that to
get arc length, how are you,
the integral has dx's and dy's.
So how are you just
integrating with respect to dx?
PROFESSOR: OK, the question
is how are we just integrating
with respect to x?
So this is a question which
goes back to last time.
And what is it with arc length.
So.
I'm going to have to answer
that question in connection
with what we did today.
So this is a subtle question.
But I want you to realize
that this is actually
an important
conceptual step here.
So shhh, everybody, listen.
If you're representing
one-dimensional objects,
which are curves,
maybe, in space.
Or in two dimensions.
When you're keeping
track of arc length,
you're going to have to
have an integral which is
with respect to some variable.
But that variable,
you get to pick.
And we're launching now
into this variety of choices
of variables with
respect to which you
can represent something.
Now, there are
some disadvantages
on the circle to
representing things
with respect to the variable x.
Because there are two
points on the circle here.
On the other hand,
you actually can
succeed with half the circle.
So you can figure out
the arc length that way.
And then you can set it
up as an integral dx.
But you can also set it up
as an integral with respect
to any parameter you want.
And the uniform parameter
is perhaps the easiest one.
This one is perhaps
the easiest one.
And so now the thing that's
strange about this perspective
- and I'm going to make this
point later in the lecture
as well - is that the
letters x and y-- As I say,
you should drop this notion
that y is a function of x.
This is what we're throwing
away at this point.
What we're thinking
of is, you can
describe things in terms
of any coordinate you want.
You just have to say what each
one is in terms of the others.
And these x and y
over here are where
we are in the Cartesian
coordinate system.
They're not-- And
in this case they're
functions of some
other variable.
Some other variable.
So they're each functions.
So the letters x and
y just changed on you.
They mean something different.
x is no longer the variable.
It's the function.
Right?
You're going to have
to get used to that.
That's because we
run out of letters.
And we kind of want to use
all of them the way we want.
I'll say some more
about that later.
So now I want to do this
surface area example.
I'm going to just take the
surface area of the ellipsoid.
The surface of the
ellipsoid formed
by revolving this previous
example, which was Example 2.
Around the y-axis.
So we want to set up that
surface area integral here
for you.
Now, I remind you that the
area element looks like this.
If you're revolving
around the y-axis,
that means you're
going around this way
and you have some curve.
In this case it's this
piece of an ellipse.
If you sweep it
around you're going
to get what's
called an ellipsoid.
And there's a little chunk here,
that you're wrapping around.
And the important thing you
need besides this ds, this arc
length piece over here, is
the distance to the axis.
So that's this
horizontal distance here.
I'll draw it in another color.
And that horizontal
distance now has a name.
And this is, again, the virtue
of this coordinate system.
The t is something else.
This has a name.
This distance has a name.
This distance is called x.
And it even has a formula.
Its formula is 2 sin t.
In terms of t.
So the full formula
up for the integral
here is, I have to take
the circumference when
I spin this thing around.
And this little
arc length element.
So I have here 2
pi times 2 sin t.
That's the x variable here.
And then I have here ds,
which is kind of a mess.
So unfortunately I don't
quite have room for it.
Plan ahead.
Square root of 4 cos^2 t + sin^2
t, is that what it was, dt.
Alright, I guess I
squeezed it in there.
So that was the arc
length, which I re-copied
from this board above.
That was the ds piece.
It's this whole thing
including the dt.
That's the answer
except for one thing.
What else do we need?
We don't just need
the integrand,
this is half of
setting up an integral.
The other half of setting up
an integral is the limits.
We need specific limits here.
Otherwise we don't have a
number that we can get out.
So we now have to think
about what the limits are.
And maybe somebody can see.
It has something to
do with this diagram
of the ellipse over here.
Can somebody guess what it is?
0 to pi.
Well, that was quick.
That's it.
Because we go from
the top to the bottom,
but we don't want
to continue around.
We don't want to
go from 0 to 2 pi,
because that would be
duplicating what we're going
to get when we spin around.
And we know that we start at 0.
It's interesting
because it descends
when you change
variables to think
of it in terms of the y variable
it's going the opposite way.
But anyway, just one piece
of this is what we want.
So that's this setup.
And now I claim that this is
actually a doable integral.
However, it's long.
I'm going to spare
you, I'll just tell you
how you would get started.
You would use the
substitution u = cos t.
And then the du is
going to be -sin t dt.
But then, unfortunately,
there's a lot more.
There's another
trig substitution
with some other multiple
of the cosine and so forth.
So it goes on and on.
If you want to check
it yourself, you can.
There's an inverse
trig substitution which
isn't compatible with this one.
But it can be done.
Calculated.
In elementary terms.
Yeah, another question.
STUDENT: [INAUDIBLE]
PROFESSOR: So, if you
get this on an exam,
I'm going to have to
coach you through it.
Either I'm going to have to
tell you don't evaluate it
or, you're going to have
to work really hard.
Or here's the first step,
and then the next step
is, keep on going.
Or something.
I'll have to give you some cues.
Because it's quite long.
This is way too long for an
exam, this particular one.
OK.
It's not too long
for a problem set.
This is where I would leave
you off if I were giving it
to you on a problem set.
Just to give you an idea
of the order of magnitude.
Whereas one of the ones that I
did yesterday, I wouldn't even
give you on a problem
set, it was so long.
So now, our next job is to
move on to polar coordinates.
Now, polar coordinates involve
the geometry of circles.
As I said, we really
love circles here.
We're very round.
Just as I love 0, the rest of
the Institute loves circles.
So we're going to
do that right now.
What we're going to talk about
now is polar coordinates.
Which are set up in
the following way.
It's a way of describing
the points in the plane.
Here is a point in
a plane, and here's
what we think of as
the usual x-y axes.
And now this point is
going to be described
by a different pair of
coordinates, different pair
of numbers.
Namely, the distance
to the origin.
And the second parameter
here, second number here,
is this angle theta.
Which is the angle
of ray from origin
with the horizontal axis.
So that's what it
is in language.
And you should put this
in quotation marks,
because it's not
a perfect match.
This is geometrically what
you should always think of,
but the technical
details involve
dealing directly with formulas.
The first formula is
the formula for x.
And this is the
fundamental, these two
are the fundamental ones.
Namely, x = r cos theta.
The second formula
is the formula
for y, which is r sin theta.
So these are the
unambiguous definitions
of polar coordinates.
This is it.
And this is the thing from
which all other almost correct
statements almost follow.
But this is the one you
should trust always.
This is the
unambiguous statement.
So let me give you an
example something that's
close to being a good
formula and is certainly
useful in its way.
Namely, you can think of r as
being the square root of x^2 +
y^2.
That's easy enough
to derive, it's
the distance to the origin.
That's pretty obvious.
And the formula for theta,
which you can also derive,
which is that it's the
inverse tangent of y y/x.
However, let me just warn
you that these formulas are
slightly ambiguous.
So somewhat ambiguous.
In other words, you can't
just apply them blindly.
You actually have
to look at a picture
in order to get them right.
In particular, r could
be plus or minus here.
And when you take
the inverse tangent,
there's an ambiguity between,
it's the same as the inverse
tangent of (-y)/(-x).
So these minus signs are a
plague on your existence.
And you're not going to get a
completely unambiguous answer
out of these formulas
without paying attention
to the diagram.
On the other hand, the
formula up in the box
there always works.
So when people mean
polar coordinates,
they always mean that.
And then they have conventions,
which sometimes match things up
with the formulas over
on this next board.
Let me give you various
examples here first.
But maybe first I
should I should draw
the two coordinate systems.
So the coordinate system
that we're used to
is the rectangular
coordinate system.
And maybe I'll draw it
in orange and green here.
So these are the coordinate
lines y = 0, y = 1, y = 2.
That's how the
coordinate system works.
And over here we have the
rest of the coordinate system.
And this is the way we're
thinking of x and y now.
We're no longer thinking of
y as a function of x and x
as a function of
y, we're thinking
of x as a label of
a place in a plane.
And y as a label of
a place in a plane.
So here we have x =
0, x = 1, x = 2, etc.
Here's x = -1.
So forth.
So that's what the rectangular
coordinate system looks like.
And now I should draw the other
coordinate system that we have.
Which is this guy here.
Well, close enough.
And these guys here.
Kind of this bulls-eye
or target operation.
And this one is,
say, theta = pi/2.
This is theta = 0.
This is theta = -pi/4.
For instance, so I've
just labeled for you three
of the rays on this diagram.
It's kind of like
a radar screen.
And then in pink, this is
maybe r = 2, the radius 2.
And inside is r = 1.
So it's a different coordinate
system for the plane.
And again, the letter
r represents measuring
how far we are from the origin.
The theta represents
something about the angle,
which ray we're on.
And they're just two
different variables.
And this is a very different
kind of coordinate system.
OK so, our main job is
just to get used to this.
For now.
You will be using
this a lot in 18.02.
It's very useful in physics.
And our job is just to
get started with it.
And so, let's try a
few examples here.
Tons of examples.
We'll start out very slow.
If you have (x, y) = (1, -1),
that's a point in the plane.
I can draw that point.
It's down here, right?
This is -1 and this is 1,
and here's my point, (1, -1).
I can figure out what
the representative is
of this in polar coordinates.
So in polar coordinates,
there are actually
a bunch of choices here.
First of all, I'll
tell you one choice.
If I start with the
angle horizontally,
I wrap all the way
around, that would
be to this ray here--
Let's do it in green again.
Alright, I labeled
it actually as -pi/4,
but another way of looking at
it is that it's this angle here.
So that would be r
= square root of 2.
Theta = 7pi/4.
So that's one possibility of
the angle and the distance.
I know the distance is a square
root of 2, that's not hard.
Another way of looking
at it is the way
which was suggested
when I labeled this
with a negative angle.
And that would be r = square
root of 2, theta = -pi/4.
And these are both legal.
These are perfectly
legal representatives.
And that's what
I meant by saying
that these representations over
here are somewhat ambiguous.
There's more than one answer
to this question, of what
the polar representation is.
A third possibility, which is
even more dicey but also legal,
is r equals minus
square root of 2.
Theta = 3pi/4.
Now, what that corresponds to
doing is going around to here.
We're pointing out
3/4 pi direction.
But then going negative
square root of 2 distance.
We're going backwards.
So we're landing
in the same place.
So this is also legal.
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: The question
is, don't the radiuses
have to be positive
because they represent
a distance to the origin?
The answer is I
lied to you here.
All of these things that I said
are wrong, except for this.
Which is the rule for what
polar coordinates mean.
So it's maybe plus or minus the
distance, is what it is always.
I try not to lie to you
too much, but I do succeed.
Now, let's do a little
bit more practice here.
There are some easy
examples, which
I will run through
very quickly. r = a,
we already know
this is a circle.
And the 3 theta equals
a constant is a ray.
However, this involves an
implicit assumption, which
I want to point out to you.
So this is Example 3.
Theta's equal to a
constant is a ray.
But this implicitly
assumes 0 <= r < infinity.
If you really wanted to allow
minus infinity < r < infinity
in this example, you
would get a line.
Gives the whole line.
It gives everything behind.
So you go out on some ray,
you go backwards on that ray
and you get the whole line
through the origin, both ways.
If you allow r going to
minus infinity as well.
So the typical
conventions, so here
are the typical conventions.
And you will see people assume
this without even telling you.
So you need to watch out for it.
The typical conventions
are certainly this one,
which is a nice thing to do.
Pretty much all the time,
although not all the time.
Most of the time.
And then you might have
theta ranging from minus pi
to pi, so in other words
symmetric around 0.
Or, another very popular
choice is this one.
Theta's >= 0 and
strictly less than 2pi.
So these are the
two typical ranges
in which all of these
variables are chosen.
But not always.
You'll find that
it's not consistent.
As I said, our job is
to get used to this.
And I need to work up
to some slightly more
complicated examples.
Some of which I'll give
you on next Tuesday.
But let's do a few more.
So, I guess this is Example 4.
Example 4, I'm
going to take y = 1.
That's awfully simple in
rectangular coordinates.
But interestingly,
you might conceivably
want to deal with it
in polar coordinates.
If you do, so here's how
you make the translation.
But this translation
is not so terrible.
What you do is, you plug
in y = r sin(theta).
That's all you have to do.
And so that's going
to be equal to 1.
And that's going to give
us our polar equation.
The polar equation is
r = 1 / sin(theta).
There it is.
And let's draw a picture of it.
So here's a picture
of the line y = 1.
And now we see that if we take
our rays going out from here,
they collide with the
line at various lengths.
So if you take an angle,
theta, here there'll
be a distance r
corresponding to that
and you'll hit this
in exactly one spot.
For each theta you'll
have a different radius.
And it's a variable radius.
It's given by this formula here.
And so to trace this
line out, you actually
have to realize that there's
one more thing involved.
Which is the possible
range of theta.
Again, when you're
doing integrations
you're going to need to know
those limits of integration.
So you're going to
need to know this.
The range here goes
from theta = 0,
that's sort of when
it's out at infinity.
That's when the
denominator is 0 here.
And it goes all the way to pi.
Swing around just one half-turn.
So the range here
is 0 < theta < pi.
Yeah, question.
STUDENT: [INAUDIBLE]
PROFESSOR: The
question is, is it
typical to express r
as a function of theta,
or vice versa, or
does it matter?
The answer is that for the
purposes of this course,
we're almost always going to
be writing things in this form.
r as a function of theta.
And you can do
whatever you want.
This turns out to be what
we'll be doing in this course,
exclusively.
As you'll see when we
get to other examples,
it's the traditional
sort of thing
to do when you're thinking
about observing a planet
or something like that.
You see the angle, and then
you guess far away it is.
But it's not necessary.
The formulas are
often easier this way.
For the examples that we have.
Because it's usually a
trig function of theta.
Whereas the other way, it would
be an inverse trig function.
So it's an uglier expression.
As you can see.
The real reason is that we
choose this thing that's
easier to deal with.
So now let me give you a
slightly more complicated
example of the same type.
Where we use a shortcut.
This is a standard example.
And it comes up a lot.
And so this is an
off-center circle.
A circle is really easy
to describe, but not
necessarily if the center
is on the rim of the circle.
So that's a different problem.
And let's do this with
a circle of radius a.
So this is the point (a,
0) and this is (2a, 0).
And actually, if you
know these two numbers,
you'll be able to remember the
result of this calculation.
Which you'll do about five
or six times and then finally
you'll memorize it during 18.02
when you will need it a lot.
So this is a standard
calculation here.
So the starting place is
the rectangular equation.
And we're going to pass to
the polar representation.
The rectangular representation
is (x-a)^2 + y^2 = a^2.
So this is a circle centered
at (a, 0) of radius a.
And now, if you like, the
slow way of doing this
would be to plug in x = r
cos(theta), y = r sin(theta).
The way I did in
this first step.
And that works perfectly well.
But I'm going to do it
more quickly than that.
Because I can sort of see in
advance how it's going to work.
I'm just going to
expand this out.
And now I see the a^2's cancel.
And not only that,
but x^2 + y^2 = r^2.
So this becomes r^2.
That's x^2 + y^2 - 2ax = 0.
The r came from the fact
that r^2 = x^2 + y^2.
So I'm doing this the rapid way.
You can do it by
plugging in, as I said.
r equals-- So now that
I've simplified it,
I am going to use
that procedure.
I'm going to plug in.
So here I have r^2 -
2ar cos(theta) = 0.
I just plugged in for x.
As I said, I could have
done that at the beginning.
I just simplified first.
And now, this is the same
thing as r^2 = 2ar cos(theta).
And we're almost done.
There's a boring part of this
equation, which is r = 0.
And then there's,
if I divide by r,
there's the interesting
part of the equation.
Which is this.
So this is or r = 0.
Which is already included
in that equation anyway.
So I'm allowed to divide by r
because in the case of r = 0,
this is represented anyway.
Question.
STUDENT: [INAUDIBLE]
PROFESSOR: r = 0
is just one case.
That is, it's the
union of these two.
It's both.
Both are possible.
So r = 0 is one point on it.
And this is all of it.
So we can just ignore this.
So now I want to say one
more important thing.
You need to understand
the range of this.
So wait a second and we're going
to figure out the range here.
The range is very important,
because otherwise you'll
never be able to integrate
using this representation here.
So this is the representation.
But notice when theta =
0, we're out here at 2a.
That's consistent,
and that's actually
how you remember
this factor 2a here.
Because if you remember this
picture and where you land when
theta = 0.
So that's the theta = 0 part.
But now as I tip
up like this, you
see that when we get to
vertical, we're done.
With the circle.
It's gotten shorter and
shorter and shorter,
and at theta = pi/2,
we're down at 0.
Because that's cos(pi/2) = 0.
So it swings up like this.
And it gets up to pi/2.
Similarly, we swing
down like this.
And then we're done.
So the range is
-pi/2 < theta < pi/2.
Or, if you want to
throw in the r = 0 case,
you can throw in this,
this is repeating,
if you like, at the ends.
So this is the range
of this circle.
And let's see.
Next time we'll figure out
area in polar coordinates.
