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, which
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is going to be relating polar
and Cartesian 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,
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and then describe the curve.
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So the first one is r squared
equals 4r cosine theta,
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and 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
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to something in the Cartesian
coordinates, in the (x, y)
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coordinates, and then I
want you to describe what
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the curve actually looks like.
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So I'll give you a little
while to work on it,
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and then 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
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in describing these two
curves in (x, y) coordinates.
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And I will show you how I
attacked these problems.
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So we'll start with (a).
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So for (a)-- I'm going to
rewrite the problem up here,
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so we can 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 over
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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
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in the 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
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work on this equation right
here and get into a form
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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,
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and 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,
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a glaring way we should go.
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And that's because we
have this x squared
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plus y squared together-- this
maybe could look something
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like a circle or an ellipse
or something like that,
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if we could figure out
a way to put this part
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in with the x squared.
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So this is kind
of-- it's a good way
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to think about what direction
to head in this problem.
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In particular, it would be
a bad idea for this problem
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for you to subtract
x squared and take
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the square root of both sides.
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Because you would lose
some information about what
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this curve was.
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OK?
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Because when you
take the square root,
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you would have to
say, well, do I
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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,
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this doesn't really
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
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from 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
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do that, we'll have a perfect
square, an extra term,
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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
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going to be x squared
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,
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I need something
that, multiplied by 2,
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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 4
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squared-- plus y squared.
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So x minus 2 quantity
squared-- that came
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from these three terms--
plus y 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.
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x minus a quality squared plus
y minus b quantity squared
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equals r squared.
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So this is, a is 2,
b is 0, and r is 2.
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so it's a circle of radius
2, centered at 2 comma 0.
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So we have a good way to
describe what started off
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in 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).
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And I'm going to rewrite
(b) over here as well,
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so we don't have to worry
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,
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in terms of (x, y) coordinates.
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And one thing that
should stand out to you
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is, what is 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,
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we can multiply both
sides by cosine theta,
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and we get an r cosine
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
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this is defined
in terms of theta,
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but I'm not worrying about
that in this situation,
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just right now.
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We're just trying
to figure out how
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we could write 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,
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this tangent theta 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,
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you can see a 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
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this expression 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
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a parabola that goes
through the point (0, 0),
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and is stretched by a
factor of 9, or 1/9.
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Well, I guess you
can say, there's
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a vertical stretch or
horizontal stretch,
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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 stretching
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on it.
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Now, the problem is that
you might say, well,
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it's not really all of that.
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Because secant
theta is not going
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to be defined for all
theta the way cosine is.
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So you do potentially
run into some problems.
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You might have to worry
about what part of the domain
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makes sense for theta, so
that this is well-defined.
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And so that this
is well-defined,
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what part of the
curve is carved out.
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That's a little more
technical than I
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want to go in this video.
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But some of you might look
at it and say, oh, she's
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missing something.
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Yeah, you caught something that
I'm intentionally ignoring.
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So the main point
of this was just so
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that you could see how you
can take these functions of r
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and theta and turn them
into functions of x and y,
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and then figure out kind of
what 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
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to get you understanding how
these different coordinates
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relate to one another.
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And yeah, that's
where we'll leave it.