WEBVTT mathematics/trigonometry/murray
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Hi these are the trigonometry lectures for educator.com, and today we're going to learn about polar coordinates which is really a kind of a new way of looking at the coordinate plane.
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It's going to be a lot of fun to check this out.
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I want to remind you how you have been keeping track of points in the plane.
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You've been using ...
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That's a really ugly access.
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You've been using x and y coordinates which are known as rectangular coordinates or Cartesian coordinates.
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The way that works is you graph a point based on its distance from the x-axis and its rectangular distance from the y-axis.
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You look at that distance, that's the y-coordinate, and that distance is the x-coordinate.
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That's all based on rectangles, it's also called Cartesian coordinates because it comes from Descartes, and you've done that before.
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Polar coordinates are the new idea here, and it's really quite different.
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If you have a point somewhere in the plane, instead of trying to orient that point based on rectangular distances from the x and y axis, what you do is you draw a line straight out from the origin to that point.
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You describe that point in terms of how long that line is, r, it's the distance from the origin, in terms of the angle that line makes with the positive x-axis.
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There's the positive x-axis.
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You measure what that angle is from the positive x-axis, and you measure the distance along that line, then give the coordinates of that point in terms of r and θ.
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That's really a new idea.
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We're going to figure out how.
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If you know what the point is in terms of x and y, how do you figure out what the r and θ are and vice versa.
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Let's check that out now.
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If you know the x and y coordinates of a point, the rectangular coordinates of a point, then you can figure out r and θ based on the Pythagorean theorem.
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There's x and there's y, there's θ, and there's r.
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r, just based on the Pythagorean theorem, is the square root of x²+y².
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That's the same as the magnitude of a vector that we learned in the lecture last time about vectors.
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You don't really have to remember any new formulas here, it's the same formula r=x²+y².
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Same with θ, we know that the tan(θ)=y/x, that's based on SOH CAH TOA, which means that θ is arctan(y/x).
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Here is where it gets tricky.
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Remember that arctan(y/x) will always give you an angle in the fourth quadrant or in the first quadrant.
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What you do if you have a point that's in the second or third quadrant, or if you have a point in the second or third quadrant, we said before that you add 180 degrees to θ.
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In polar coordinates, you often use radians instead of degrees.
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I've changed that formula slightly.
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It's the same formula except in terms of radians, it's π+arctan(y/x), instead of 180+arctan(y/x).
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You use this formula when your point is in quadrant 2 or 3, is when you have to add on a π to the arctan(y/x).
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In other words, that occurs when the x-coordinate is negative, when x is less than 0.
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This formula, the basic formula arctan(y/x), you use that for quadrants 1 and 4, or when your x-coordinate is positive.
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That's how if you know the x and the y, you can find the r and the θ.
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If you know the r and the θ, you can find the x and the y, by taking rcos(θ) and rsin(θ).
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Those are the same formulas we had when we started with the magnitude and angle of a vector, and we convert it back to find the components of the vector.
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You don't really have to remember any new formulas here.
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It's the same formulas as before, x=rcos(θ), y=rsin(θ).
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Those come from SOH CAH TOA, so it's not very hard to derive those formulas based on the SOH CAH TOA relationships in a right angle.
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That's θ, that's r, that's x and that's y.
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You can use SOH CAH TOA to find x and y equals rcos(θ) and rsin(θ).
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There's some conventions that we tried to follow but it's not absolutely essential.
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The r is usually assumed to be positive.
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Certainly, if you used this formula to find r, you get a square root, so it's always positive.
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But you can also talk about -r's.
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You think about a -r, if you have an angle θ, what would a -r represent?
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Well that just represents going r units in the other direction.
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If r is less than 0, then that just represents going r units on the other side of the origin.
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That's what a -r would represent, but we try to use positive r's if we can.
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Similarly, θ, we try to keep in between the range 0 to 2π.
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Here's 0, π/2, π, 3π/2, and 2π.
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We try to keep our θ in between 0 and 2π.
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If it's outside of that range, then what you can do is you could add or subtract multiples of 2π to try to get it back into that range.
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That's how we try to restrict the values of r and θ.
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You'll see some examples of this as we practice converting points.
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It's probably best just to move on to some examples and you'll see how we generally find r's that are positive and θ's between 0 and 2π.
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But if someone gives us values that are not in those ranges, we can modify them to find different sets of coordinates that do have r's and θ's in those ranges.
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Let's move on to some examples.
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First example is to convert the following points from rectangular coordinates to polar coordinates.
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Remember, we're given here the x and the y, and we want to find the r's and the θ's, because we're given rectangular coordinates that's x and y.
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We want to find polar coordinates, that's r and θ.
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I'm going to draw some little graphs of these points to help me keep track of where they are.
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You can also figure them out just using the formulas that we learned on the previous slide.
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For the first point here, we've got 3 square root of 2 and -3 square root of 2.
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That's positive in the x direction, and negative in the y direction.
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That's a point down there.
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I want to figure out where that is.
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I'm going to use the formulas to figure that out, r is equal to the square root of x²+y².
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Let me remind you of those formulas up here because we're going to use them quite a bit, x²+y².
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θ=arctan(y/x), that's the case if x is greater than 0, or π+arctan(y/x) if the case x is less than 0.
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In this case, our r, our magnitude is the square root of 3 root 2 squared, plus negative 3 root 2 squared.
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3 root 2 squared is 9 times 2, that's the square root of 18, plus another 18, which is the square root of 36, which is 6.
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θ is arctan of negative 3 root 2 over 3 root 2, because it's y/x, which is arctan(-1).
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Arctan(-1), that's a common value that I remember.
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That's -π/4, but I would like to get an answer in between 0 and 2π, -π/4 doesn't qualify.
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I'm going to add +2π, and that will give me, 2π is 8π/4, 7π/4.
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My polar coordinates for that point (r,θ) are (6,7π/4).
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That really corresponds with what I can check visually.
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That angle is -π/4 if you go down south from the x-axis, but if you go all the way around the long way, it's 7π/4.
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That's our first set of polar coordinates.
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Moving on to the next one, (-4,-3), that's -4 in the x direction, -3 in the y direction.
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That's somewhere down there.
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So, r is the square root of 4²+3².
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Since we're squaring them, I'm not going to worry about the negative signs.
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16+9=25, square root of that is 5.
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θ=arctan(-4/-3).
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Now, we know that this angle is in the third quadrant, its x-coordinate is negative, that means we have to add a π to the θ.
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This is arctan(4/3)+π.
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If you plug in arctan(4/3), your calculator would give you an angle up here.
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We have to add π radians to whatever the calculator's answer is.
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Now, arctan(4/3), one thing that's very important here is that I'm going to switch my calculator over to radian mode, that's because I'm looking for answers in terms of radians now.
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I don't want to mix up my radians and my degrees.
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The π is certainly in radians, I didn't say 180 degrees, I said π, I need to make sure that my calculator's in radian mode when I do arctan(4/3).
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I do arctan(4/3), negative's cancelled, that gives me 0.93+π.
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That simplifies down to 4.07.
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My (r,θ) for that second one is (5,4.07).
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Just a reminder that that 4.07 is a radian measure.
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It's not so obvious when you don't have the π in there, when you don't have a multiple of π, but it is a radian measure there.
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That's the radius and the reference angle for that point.
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The third one here, negative square root of 3 and 1.
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Let me graph that one.
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Negative square root of 3 and positive 1, that's the point over there.
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My r is, root 3 squared is 3, plus 1, so that's 2, θ is arctan(y/x), so 1 over negative root 3.
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The x is negative, I have to add a π there.
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That's arctan, if we rationalize that, that's negative root 3/3, plus π.
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Now, negative root 3 over 3, that's a common value.
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I recognize that as something that I know the arctangent of, that's -π/6+π, that's 5π/6.
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My polar coordinates for that point, our (r,θ), is (2,5π/6).
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Finally, we have -2π.
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That's -2 in the x direction, π in the y direction.
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That's a point somewhere up there in the second quadrant, so r is the square root of 2²+5², which is the square root of 29, nothing very much I can do with that.
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θ, arctan(5/-2), my x is negative and I'm in the second quadrant so I'm using the other formula for the θ, I have to add on a π there.
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My calculator's set to radian mode, I'm going to do arctan(-5/2), and I get -1.19+π, which is 1.95.
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As long as I'm giving decimal approximations, I may as well give a decimal approximation for the r.
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The square root of 29 is approximately equal to 5.39.
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My (r,θ) is (5.39,1.95) radians.
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That's the answer for that one.
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All of these, it really helped me to draw a picture even though I didn't really need that for the calculations, but it was really useful to kind of check that I was in the right place.
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In all four, I drew a picture of where the point was, I figured out which quadrant it was in.
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I worked out the r using the square root of x²+y².
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I worked out θ using arctan(y/x), but then I had to check which quadrant the point was in.
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If it was in the second or third quadrant, then I had to add π to the answer that the calculator gave me for arctan(y/x).
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In this fourth one, for example, the calculator gave me an answer of -1.19.
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The calculator's answer was down there, that's why I had to add π to it inorder to get the answer.
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I came back to these formulas for r and θ every time, but it's still very helpful to draw your coordinate axis and to show where the point is on those axis.
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For our second example, we have to convert the following points in polar coordinates to standard form.
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That means that the r should be positive and the θ should be between 0 and 2π, then we'll convert these points to rectangular coordinates.
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We're going to start by graphing each one of these points.
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First one, we have -3π/4, that means we're going down 3π/4 in the negative direction.
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That's the same as going around 5π/4 in the positive direction.
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We can write that as (8,5π/4).
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That is legit because it's between 0 and 2π.
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If you didn't like doing that graphically, you could do -3π/4+2π, and that gives you immediately the 5π/4.
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That's how you could figure it out using equations instead of doing it graphically.
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That's the standard form of that point.
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Now, I have to find the rectangular coordinates, and I'm going to use rcos(θ) for x, and rsin(θ) for y.
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My x is 8cos(5π/4).
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The cos(5π/4), that's a common value that I've got it memorized, that's 8 times root 2 over 2, but it's negative because the x is negative there, negative root 2 over 2.
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y is sin(5π/4), the y is also negative there down on the third quadrant, negative root 2 over 2.
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This is why it's so useful to draw a graph where the point is.
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It helps you find the sine and cosine and remember which one is positive or negative.
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(x,y) together simplify down to (-4 root 2,-4 root 2).
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That was my first point.
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The second one, I'm going around 11π/6, that's just short of 2π, but it's in the negative direction, so I go around just short of 2π in the negative direction.
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That's actually the same as going π/6 in the positive direction.
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If you don't like doing that graphically, you can add 2π to 11π/6, and you'll get π/6.
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But then I'm going -6 units in that direction.
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That really takes me 6 units in the opposite direction.
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Really, my reference angle is down here in the third quadrant.
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I can write that as 6, that's π/6, that's an angle of π/6.
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This is π/6 beyond π.
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I have here (6,7π/6) would be the standard version of that point in polar coordinates.
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Now I've got a positive radius and a positive angle between 0 and 2π.
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Now I've got to find the rectangular coordinates for that point.
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I'm going to use rcos(θ) and rsin(θ).
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x=6cos(7π/6), which is 6×cos(7π/6), is negative root 3 over 2.
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y=6sin(7π/6), these are common values that I've gotten memorized which is 6×(-1/2) down in the third quadrant.
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The (x,y) collectively, the coordinates, simplify down to -3 root 3, and -3.
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Okay, next one, -2 and 11π/3.
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Let's figure out where that one is.
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11π/3 is pretty big, I'm going to subtract 2π from that right away.
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11π/3-2π, 2π=6π/3, that's 5π/3.
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That's up there in the second quadrant, 5π/3.
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We'll have to go -2 units in that direction, which takes me down in the negative direction.
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Sorry, I graphed that in the wrong place.
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I was graphing 5π/6 instead of 5π/3.
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Let me modify that slightly.
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5π/3, that's in the fourth quadrant, that's down there.
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5π/3 is down there, not 5π/6.
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I want to go -2 units in that direction.
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That's up there, back in the second quadrant.
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That's really our reference angle.
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We want to go 2 units in that direction.
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Where is that direction?
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That is 2π/3.
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Our final answer for the standard polar coordinates there is (2,2π/3).
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The way we got that just to remind you is we took 11π/3.
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That was really big so I subtracted 2π right away, I got 5π/3.
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Then I graphed 5π/3, but because I had to go -2 units in that direction, I subtracted π off from 5π/3, that's how I got the 2π/3.
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Now, my x and y, rcos(θ) and rsin(θ).
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2cos(2π/3), cos(2π/3)=-1/2, because the x-coordinate is negative, 2×-1/2.
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y is 2sin(2π/3) which is 2 times positive root 3/2, because the y-coordinate's positive in the third quadrant.
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x and y together are, 2×-1/2=-1, 2×root 3/2=root 3.
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I've got x and y for that third point.
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Finally, the fourth point here is (3,-2π/3).
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Let me graph that one.
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-2π/3, we're going down in the opposite direction from the x-axis, that's 2π/3.
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If we make that something in the positive direction, that's 4π/3 in the positive direction.
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If you don't like doing that graphically, just add 2π to -2π/3, and you'll get 4π/3.
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This is the same as (3,4π/3).
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3 is already positive, so we don't have to do anything else clever there.
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That's our standard polar coordinate form.
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The x and y, x=3cos(4π/3).
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Cosine is negative there, -1/2, so 3×-1/2.
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The sine, the y-coordinate, is also negative there.
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3×sin(4π/3), common value, I remember that one, is 3 times negative root 3 over 2.
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The x and y collectively are -3/2 and negative 3 root 3/2.
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This one was a little bit tricky because we've been given r's and θ's that are not in the standard ranges.
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Some of the r's were less than 0, and some of the θ's were either less than 0, or bigger than 2π.
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The trick on all four of these points, is first of all, graph the point.
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Once you graph the point, if it's not in the standard range of 0 to 2π, add or subtract multiples of 2π until you get it into that standard range.
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That's what we did at first, we added or subtracted multiples of 2π to each of those to get it into the standard range.
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After we did that, we looked at the r's.
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If the r's were negative, then we had to go in the opposite direction from the direction we expected.
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That's what happened in this second and third problem with -r's.
00:27:23.000 --> 00:27:28.000
We had to go in the opposite direction from what we expected.
00:27:28.000 --> 00:27:38.000
What we did was we added or subtracted π to get the correct reference angle, and to get a positive value of r.
00:27:38.000 --> 00:27:42.000
That's how we got the r's and θ's into the standard range then.
00:27:42.000 --> 00:27:57.000
After that, we used x=rcos(θ), y=rsin(θ), in each case to give the x and y the rectangular coordinates of the point.
00:27:57.000 --> 00:27:59.000
That's how we did those conversions.
00:27:59.000 --> 00:28:01.000
We'll try some more examples.
00:28:01.000 --> 00:28:13.000
This one we have to graph the polar equation r=2sin(θ), then we're going to check our answers by converting the equation to rectangular coordinates and solving it algebraically.
00:28:13.000 --> 00:28:23.000
What I'm going to do here is make a little chart of all my reference angles.
00:28:23.000 --> 00:28:28.000
I'm going to make a list of all the possible values of θ that I can easily figure out 2sin(θ).
00:28:28.000 --> 00:28:33.000
Then I'll fill in all my 2sin(θ)'s.
00:28:33.000 --> 00:28:37.000
I'll try to graph them and see what happens.
00:28:37.000 --> 00:28:40.000
First of all, let me remind you which angles we know the common values for.
00:28:40.000 --> 00:28:47.000
I'm going to draw a unit circle here.
00:28:47.000 --> 00:29:00.000
The angles that I know, first, the easy ones, 0, π/2, π, 3π/2, and then 2π.
00:29:00.000 --> 00:29:07.000
I also know all the multiples of π/4 and I also know all the multiples of π/3.
00:29:07.000 --> 00:29:24.000
There's π/6, π/3, 2π/3, 5π/6, 7π/6, 4π/3, 5π/3 and 11π/6.
00:29:24.000 --> 00:29:27.000
I know every multiple of π/6 and π/4 between 0 and 2π.
00:29:27.000 --> 00:29:37.000
I'm going to make a chart showing all of these and then we'll try to graph those and see what it turns out to be.
00:29:37.000 --> 00:29:47.000
2sin(θ), first one is 0, sin(0)=0, that's just 0.
00:29:47.000 --> 00:29:59.000
π/6, 30-degree angle, the sin is 1/2, and 2 times that is 1.
00:29:59.000 --> 00:30:02.000
Next one is π/4, 45-degree angle.
00:30:02.000 --> 00:30:14.000
The sine of square root of 2 over 2, this stuff I have memorized, that's square root of 2, which for future reference is about 1.4.
00:30:14.000 --> 00:30:17.000
π/3 is the next one.
00:30:17.000 --> 00:30:27.000
The sine of that is square root of 3 over 2, 2sin(θ) square root of 3, which is about 1.7.
00:30:27.000 --> 00:30:35.000
Next one is π/2, sine of that is 1, 2 sine of that is 2.
00:30:35.000 --> 00:30:45.000
Next one is 2π/3, sine of the square root of 3 again, that's 1.7.
00:30:45.000 --> 00:30:57.000
3π/4, sine is root 2 over 2, twice that is root 2, 1.4.
00:30:57.000 --> 00:31:06.000
5π/6, sine of that is 1/2, 2 sine of that is 1.
00:31:06.000 --> 00:31:11.000
Finally, π, sine of that is just 0.
00:31:11.000 --> 00:31:20.000
Let me go through and figure out the values on the southern half of the unit circle.
00:31:20.000 --> 00:31:27.000
Then we'll get to put all these together and see what kind of graph we get.
00:31:27.000 --> 00:31:34.000
Below the unit circle, the first value after π is 7π/6.
00:31:34.000 --> 00:31:38.000
The sine of that is negative now, it's -1/2, 2 sine of that is -1.
00:31:38.000 --> 00:31:46.000
Then we get to 5π/4, negative root 2 over 2.
00:31:46.000 --> 00:31:53.000
We just multiply by 2 and get negative root 2, which is -1.4.
00:31:53.000 --> 00:32:07.000
Then we get 4π/3, sine of that is negative root 3 over 2, we get -1.7.
00:32:07.000 --> 00:32:16.000
3π/2, sine of that is -1, because we're down here at the bottom of the unit circle.
00:32:16.000 --> 00:32:19.000
This is -2.
00:32:19.000 --> 00:32:34.000
5π/3, the sine of that is negative root 3 over 2, so it's negative root 3 because we're multiplying by 2, about 1.7.
00:32:34.000 --> 00:32:45.000
7π/4, sine of that is negative root 2 over 2, so I multiply it by 2, you get 1.4.
00:32:45.000 --> 00:32:48.000
I forgot my negative sign up above.
00:32:48.000 --> 00:32:55.000
Finally, 11π/6, sine of that is -1/2.
00:32:55.000 --> 00:33:00.000
All of these are values that you should have memorized to be able to figure out very quickly.
00:33:00.000 --> 00:33:01.000
2 times the sine of that is -1.
00:33:01.000 --> 00:33:06.000
Finally, 2π, we're back to 0 again, sin(0).
00:33:06.000 --> 00:33:26.000
We've made this big chart, now I want to try and graph this thing.
00:33:26.000 --> 00:33:39.000
Those are my values of multiples of 0, π/2, π, 3π/2, and 2π.
00:33:39.000 --> 00:33:54.000
Now I'm filling in π/4, 3π/4, 5π/4, and 7π/4.
00:33:54.000 --> 00:34:10.000
This one is π/6 and 7π/6.
00:34:10.000 --> 00:34:27.000
That is π/3 and 4π/3.
00:34:27.000 --> 00:34:41.000
That's 2π/3 and 5π/3.
00:34:41.000 --> 00:34:46.000
Finally, 5π/6 and 11π/6.
00:34:46.000 --> 00:34:54.000
I want to graph each one of this radii on the spokes of this wheel.
00:34:54.000 --> 00:34:56.000
Starting at 0 ...
00:34:56.000 --> 00:35:01.000
I'm at 0, I'll just put a big ...
00:35:01.000 --> 00:35:03.000
I think I better do this in another color or it's going to get obscured.
00:35:03.000 --> 00:35:10.000
I'll do my graph in red.
00:35:10.000 --> 00:35:14.000
I'll put a big 0 on the 0 axis there.
00:35:14.000 --> 00:35:16.000
At π/6, I'm 1 unit out.
00:35:16.000 --> 00:35:22.000
Let's say, that's about 1 unit at π/6.
00:35:22.000 --> 00:35:26.000
At π/4, I'm 1.4 units out.
00:35:26.000 --> 00:35:30.000
It's a little bit farther out, about there.
00:35:30.000 --> 00:35:38.000
At π/3, I'm 1.7 units out, a little farther out.
00:35:38.000 --> 00:35:48.000
At π/2, I'm 2 units out in the π/2 direction.
00:35:48.000 --> 00:35:55.000
2π/3 is 1.7.
00:35:55.000 --> 00:35:57.000
1.7 in the 2π/3 direction.
00:35:57.000 --> 00:36:03.000
1.4 in the 3π/4 direction.
00:36:03.000 --> 00:36:09.000
1 in the 5π/6 direction, it's shrinking back down.
00:36:09.000 --> 00:36:13.000
When we get to π, it's 0.
00:36:13.000 --> 00:36:30.000
If I just connect up what I've got so far, I've got something that looks fairly circular.
00:36:30.000 --> 00:36:37.000
It's a little hard to tell, my graph isn't perfect.
00:36:37.000 --> 00:36:41.000
It's also kind of an optical illusion here because of the spokes on the wheel here, but it looks kind of circular.
00:36:41.000 --> 00:36:45.000
Remember that that was two units out.
00:36:45.000 --> 00:36:47.000
This is one unit out.
00:36:47.000 --> 00:36:51.000
I want to see what happens when I fill in the values from π to 2π.
00:36:51.000 --> 00:36:55.000
I'm going to fill those in in blue.
00:36:55.000 --> 00:37:00.000
It's 7π/6, I'm in the -1 direction.
00:37:00.000 --> 00:37:05.000
I look at the 7π/6 spoke, but then I go -1 in that direction.
00:37:05.000 --> 00:37:08.000
That means I go in the opposite direction for 1 unit.
00:37:08.000 --> 00:37:17.000
I actually end up back in the opposite direction, I end up back here.
00:37:17.000 --> 00:37:27.000
At 5π/4, that's down here, but I go -1.4 in that direction which puts me back there.
00:37:27.000 --> 00:37:34.000
At 4π/3, I go -1.7 which ends me back there.
00:37:34.000 --> 00:37:39.000
3π/2, -2 puts me up there.
00:37:39.000 --> 00:37:47.000
I start out in the 3π/2 direction but I go -2 units, that takes me back up here.
00:37:47.000 --> 00:37:52.000
5π/3 puts me at -1.7.
00:37:52.000 --> 00:37:57.000
7π/4 puts me at -1.4.
00:37:57.000 --> 00:38:00.000
11π/6 puts me at -1.
00:38:00.000 --> 00:38:06.000
At 2π, I'm back at 0 again, and it starts to repeat.
00:38:06.000 --> 00:38:15.000
If you connect up those dots, what you see is another copy of the same graph.
00:38:15.000 --> 00:38:24.000
I'm drawing them slightly separated so that you can see both of them, but it really is a graph just retracing itself again.
00:38:24.000 --> 00:38:35.000
You might have thought that you might get some action down here in the south side of the coordinate plane, but you don't.
00:38:35.000 --> 00:38:47.000
You just get a graph that traces itself over the top side of the coordinate plane, and then retraces itself even when the angles are down on the south side of the coordinate plane.
00:38:47.000 --> 00:38:51.000
Let me recap what happened there.
00:38:51.000 --> 00:38:54.000
We started out with an equation r=2sin(θ).
00:38:54.000 --> 00:39:00.000
First thing I did was I made a big chart of all the θ's that I know the common values of.
00:39:00.000 --> 00:39:03.000
I made a big chart of all my possible θ's.
00:39:03.000 --> 00:39:07.000
Then I filled in what 2sin(θ) is for each one.
00:39:07.000 --> 00:39:12.000
I drew my axis and I drew my spokes with all the angles listed here.
00:39:12.000 --> 00:39:18.000
I plotted the points for each one of those spokes.
00:39:18.000 --> 00:39:23.000
I went out the correct direction for each one of those spokes.
00:39:23.000 --> 00:39:38.000
Except that when I got down to these angles below the horizontal axis, all my signs were negative, which means I'm going in the opposite direction, which actually landed me up on the top side of the axis.
00:39:38.000 --> 00:39:46.000
That's why you have this blue graph that retraces the original red graph.
00:39:46.000 --> 00:39:47.000
That's how we did that one.
00:39:47.000 --> 00:40:00.000
That's really all we need to do to graph that equation, but the problem also asked us to check our answers by converting the equation to rectangular coordinates and solving it algebraically.
00:40:00.000 --> 00:40:07.000
Let's just remember here what this graph look like because I'm going to have to go to a new slide to check it.
00:40:07.000 --> 00:40:12.000
We have what looks like a circle sitting on top of the x-axis.
00:40:12.000 --> 00:40:19.000
It peaks at y=2, and it looks like its center is y=1.
00:40:19.000 --> 00:40:26.000
In terms of x and y coordinates, it looks like we have that circle peaking at y=2 and centered at y=1.
00:40:26.000 --> 00:40:34.000
We're going to check that by working with the equation algebraically.
00:40:34.000 --> 00:40:39.000
Now we're going to work with this equation algebraically, r=2sin(θ).
00:40:39.000 --> 00:40:44.000
What I'm going to do there is I'm going to multiply both sides by r.
00:40:44.000 --> 00:40:52.000
The reason I'm going to do that is because that will give me r²=2rsin(θ).
00:40:52.000 --> 00:41:00.000
I recognized because I remember my transformations of coordinates from rectangular to polar.
00:41:00.000 --> 00:41:08.000
I remember that x=rsin(θ), and y=rcos(θ).
00:41:08.000 --> 00:41:20.000
I also remember that r is equal to the square root of x²+y², r²=x²+y².
00:41:20.000 --> 00:41:30.000
With this nice new version of the equation r²=2rsin(θ), I can write that as x²+y².
00:41:30.000 --> 00:41:34.000
rsin(θ), that's exactly my y.
00:41:34.000 --> 00:41:36.000
I've converted that equation into rectangular coordinates.
00:41:36.000 --> 00:41:43.000
I'm going to solve that algebraically and see if the graph checks out to what I found on the previous slide.
00:41:43.000 --> 00:41:51.000
What I'm going to do is I'm going to write this as x²+y²-2y=0.
00:41:51.000 --> 00:41:55.000
I like to simplify that y²-2y.
00:41:55.000 --> 00:42:01.000
I'm going to complete the square there.
00:42:01.000 --> 00:42:09.000
We're going to complete the square.
00:42:09.000 --> 00:42:15.000
That's an old algebraic technique, hopefully, you'll learn about that in the algebra lectures on educator.com.
00:42:15.000 --> 00:42:18.000
This is y²-2².
00:42:18.000 --> 00:42:25.000
You take the middle number and you divide it in half and square it.
00:42:25.000 --> 00:42:30.000
Minus 2 divide that by 2, and you square it.
00:42:30.000 --> 00:42:35.000
That's -1²=1.
00:42:35.000 --> 00:42:38.000
Add that to both sides.
00:42:38.000 --> 00:42:42.000
The point of that is that now we have a perfect square.
00:42:42.000 --> 00:42:51.000
We have y²-2y+1, that's (y-1)²=1.
00:42:51.000 --> 00:42:55.000
This is actually an equation I recognize.
00:42:55.000 --> 00:43:21.000
Remember that if you have x²+y² equals a number, let's say 1, that's a circle of radius 1 centered at the origin.
00:43:21.000 --> 00:43:24.000
That's the unit circle.
00:43:24.000 --> 00:43:34.000
This y-1, what that does is, it just takes the original graph of the unit circle and raises it up in the y direction by 1 unit.
00:43:34.000 --> 00:44:04.000
This is a circle of radius 1, that's because of that 1 right there, centered at x=0, that's because we think of that as (x-0)², and y=1, that's because of the y-1.
00:44:04.000 --> 00:44:21.000
If we graph that, there's 1, there's 2, I'm going to graph a circle of radius 1 centered at the point (1,0).
00:44:21.000 --> 00:44:30.000
There's (1,0).
00:44:30.000 --> 00:44:40.000
There's my circle of radius 1 centered at the point (0,1).
00:44:40.000 --> 00:44:47.000
If you flip back to the previous slide, you'll see that that is exactly the same graph that we got on the previous slide.
00:44:47.000 --> 00:44:58.000
We got it by using polar coordinates and a lot of trigonometry, but the graph that we drew actually really resembled a circle of radius 1 centered at (0,1).
00:44:58.000 --> 00:45:00.000
Let's recap what we did there.
00:45:00.000 --> 00:45:04.000
We started by graphing this thing in polar coordinates.
00:45:04.000 --> 00:45:15.000
I made a big chart of θ's and 2sin(θ)'s, then I graph the r's, their radius, at each one of those values including when the radius turn out to be negative.
00:45:15.000 --> 00:45:18.000
I ended up graphing this.
00:45:18.000 --> 00:45:20.000
I connected them up into a circle.
00:45:20.000 --> 00:45:22.000
That's one way to solve that problem.
00:45:22.000 --> 00:45:30.000
The second way to check our answer was to start with the equation, I see r=2sin(θ).
00:45:30.000 --> 00:45:45.000
Remembering my conversion formulas x=rcos(θ) and y=rsin(θ).
00:45:45.000 --> 00:45:48.000
That's very important not to mix up.
00:45:48.000 --> 00:45:55.000
I multiplied both sides of the equation by r.
00:45:55.000 --> 00:46:00.000
On the left, we get r², on the right, we get 2rsin(θ).
00:46:00.000 --> 00:46:10.000
The point of that is that the rsin(θ) converts into y, and the x²+y² is what you get from r².
00:46:10.000 --> 00:46:15.000
That of course comes from this, r²=x²+y².
00:46:15.000 --> 00:46:23.000
Then it was a matter of doing some algebra to realize that that's the equation of a circle that involve completing a square.
00:46:23.000 --> 00:46:35.000
We figured out that it was equation of a circle, that we can rid of the radius of the circle is 1, the coordinates of the center of the circle are (0,1), so we can graph that equation algebraically.
00:46:35.000 --> 00:46:38.000
It checks out with what we got from the polar coordinates.
00:46:38.000 --> 00:46:40.000
We'll try some more examples later.
00:46:40.000 --> 01:07:35.000
Try them out yourself and then we'll work them out together.