WEBVTT

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Welcome back to recitation.

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In this video, I'd like us to
do the following problem,

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which is going to be relating
polar and Cartesian

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coordinates.

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So I want you to write each of
the following in Cartesian

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coordinates, and that means our
x y coordinates, and then

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describe the curve.

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So the first one is r squared
equals 4r cosine theta, and

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the second one is r equals 9
tangent theta secant theta.

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So again, what I'd like you to
do is convert each of these to

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something in the Cartesian
coordinates, in the x y

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coordinates, and then I want you
to describe what the curve

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actually looks like.

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So I'll give you a little while
to work on it, and then

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when I come back, I'll
show you how I do it.

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OK, welcome back.

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Well, hopefully you were able
to get pretty far in

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describing these two curves
in x y coordinates.

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And I will show you how I
attacked these problems. So

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we'll start with a.

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So for A--

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I'm going to rewrite the problem
up here, so we can

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just be focused on
what's up here.

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So we had r squared equals
4r cosine theta.

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Well, we know what
r squared is.

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That's nice in terms of
x and y coordinates.

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That's just x squared
plus y squared.

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So we know that, so we'll
replace that.

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And then we can actually replace
all the r's and thetas

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over here pretty easily,
as well.

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Because we know r cosine
theta describes x.

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So the cartesian coordinate x
is the polar coordinate--

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or described in polar
coordinates as r cosine theta.

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So we can just write
that as 4x.

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And the reason I asked you to
describe the curve is because

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from here, you could say, oh,
well I wrote it in the

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Cartesian coordinates.

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I wrote it in x y, and
so now I'm done.

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But the point is that you can
actually work on this equation

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right here and get into a form
that you can recognize.

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That it'll be a recognizable
curve.

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So let's see if we can sort of
play around with this, and

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come up with something
that looks familiar.

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And what you might think to do,
would be, say, you know,

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subtract off the x squared, or
subtract off the y squared.

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Try and solve for x
or solve for y.

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But that can be a little bit
dangerous in this situation,

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because in fact, y might
not be a function of x.

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So we might run into
some trouble there.

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But if you'll notice, there's
something kind of, a glaring

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way we should go.

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And that's because we have this
x squared plus y squared

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together-- this maybe could look
something like a circle

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or an ellipse or something like
that, if we could figure

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out a way to put this part
in with the x squared.

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So this is kind of--

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it's a good way to think
about what direction to

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head in this problem.

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In particular, it would be a bad
idea for this problem for

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you to subtract x squared
and take the square

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root of both sides.

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Because you would lose some
information about

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what this curve was.

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OK?

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Because when you take the square
root, you would have to

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say, well, do I want the
positive square root, or do I

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want the negative square root?

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We'd lose a little bit
of information.

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So we do not want
to solve for y.

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So let's do what I said.

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Let's try and figure out a way
to get this 4x into something

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to do with this x
squared term.

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So I'm going to subtract 4x and
rewrite the equation here.

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And so you might say, well,
Christine, this doesn't really

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seem that helpful.

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It's just the same thing
moved around.

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But we're going to use one of
our favorite techniques from

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integration, which is completing
the square.

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So we can actually complete the
square on this guy right

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here, and turn it into
a perfect square.

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We'll have to add an extra term,
but once we do that,

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we'll have a perfect
square, an extra

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term, and a y squared.

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And we're getting more into
the form of something that

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actually looks like a circle.

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So let's see.

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Completing the square on this,
it's going to be x squared

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minus 4x plus 4.

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How did I know that?

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Well, if I want to complete
the square on this, I need

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something that, multiplied by
2, gives me negative 4.

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That's 2.

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And then 2 squared is 4.

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So that's where the
4 comes in.

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To keep this equal, I'll add 4
to the other side, as well.

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So if I add 4 to both sides, I
haven't changed the equality,

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and I keep my y squared
along for the ride.

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So now I have a perfect
square.

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What does this give me?

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This gives me x minus 2 quantity
squared plus 4-- plus

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four squared-- plus y squared.

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So x minus 2 quantity squared.

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That came from these
three terms. Plus y

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squared equals four.

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And now it's a curve we
can describe, clearly.

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What curve is this?

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Well, it's obviously a circle.

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It's centered at the point 2
comma 0, and it has radius 2.

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We've talked, or you've seen
this in the lecture videos.

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I believe what the form for a
circle is, x minus a quality

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squared plus y minus b quantity

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squared equals r squared.

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So this is, a is 2, b is 0, and
r is 2. so it's a circle

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of radius 2, centered
at 2 comma 0.

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So we have a good way to
describe what started off in

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polar coordinates.

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We can now describe it
in x y coordinates.

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OK.

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So now let's move on to B. And
I'm going to rewrite B over

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here as well, so we don't
have to worry

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about it, looking back.

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r equals 9 tan theta
secant theta.

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OK.

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So let's look at this.

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Now, there's some information
buried in here, in terms of x

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y coordinates.

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And one thing that should stand
out to you is, what is

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secant theta?

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Secant theta is 1 over
cosine theta.

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Right?

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And if we have 1 over cosine
theta over here, we can

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multiply both sides by cosine
theta, and we get an r cosine

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theta over here.

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So I'm going to write
that down.

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That this actually
is in the same--

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this is the same as r cosine
theta equals 9 tan theta.

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Right?

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I mean, you could get mad at me
about where this is defined

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in terms of theta, but I'm not
worrying about that in this

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situation, just right now.

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We're just trying to figure
out how we could write

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this in x and y.

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We know what our cosine
theta is.

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Again, it's x, as
it was before.

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What about tan theta?

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Tangent theta, remember, if you
recall, this tangent theta

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is opposite over adjacent,
right?

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And in this case, opposite is
the y, and adjacent is the x.

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This is something you saw a
picture of, you can see a

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picture of pretty easily.

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So this is x is equal
to 9 times y over x.

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Right?

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Which is x squared
is equal 9y.

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So this is in fact how you could
write this expression

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that's in r and theta
in terms of x and y.

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And so this, if you look at
it, is actually a parabola

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that goes through the point 0,
0, and is stretched by a

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factor of 9, or 1/9.

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Well, I guess you can say,
there's a vertical stretch or

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horizontal stretch, you can
pick which one it is.

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And in one case, it's going
to be by 3 or 1/3.

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I always mix those up.

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I'd have to check.

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Or by 9 or 1/9.

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So essentially, it's going to
be a parabola with some

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stretching on it.

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Now, the problem is that you
might say, well, it's not

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really all of that.

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Because secant theta is not
going to be defined for all

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theta the way cosine is.

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So you do potentially run into
some problems. You might have

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to worry about what part of
the domain makes sense for

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theta, so that this
is well-defined.

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And so that this is
well-defined, what part of the

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curve is carved out.

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That's a little more technical
than I want

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to go in this video.

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But some of you might look
at it and say, oh,

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she's missing something.

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Yeah, you caught something
that I'm

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intentionally ignoring.

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So the main point of this was
just so that you could see how

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you can take these functions of
r and theta and turn them

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into functions of x and y, and
then figure out kind of what

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the curves might look like.

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So I'm going to stop there.

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Hopefully this was a good
exercise to get you

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understanding how these
different coordinates relate

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to one another.

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And yeah, that's where
we'll leave it.