Sketch the Curve in Polar Coordinates

1. Jun 28, 2012

pumike11

1. The problem statement, all variables and given/known data

Sketch the curve r = 1 + 2cosθ in polar coordinates.

2. Relevant equations

None that I can think of, it's graphing.

3. The attempt at a solution

What I was trying to was use the method of finding cartesian coordinates and plugging different values of θ into the equation to the the 'r' values. I came up with the graph that you can see on in the snapshot, but it looks different than the graph on my TI-89. Does it look like I am doing this correctly?

Edit: So I found figured out the graph was correct, it was just a little small to scale so I increased its size so it was easier to read. I guess the only concern I have now is that I am not exactly sure what is meant by label the coordinates of three points not on either of the axes. Can someone explain to me what that part of the question is asking?

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2. Jun 28, 2012

algebrat

You've found a good question, that is hard (for me) to explain. You have graphed the curve in the (r,θ) plane. You are to change the coordinates to find the curve in the (x,y) plane.

I'm not sure how to phrase it. You've graphed the equation, "rectangularly". You need to interpret it as an equation in the polar coordinates for the plane. For instance, the set {(r,θ):0≤r≤1,0≤θ≤2π} is a rectangle in the rθ-plane, but it is a disk in the xy-plane (after transforming?).

Try multiplying both sides by r, then using the transformation between polar and rectangular coordinates. Actually, that might not be to easy to figure out the shape, though that method is pretty standard, so you should try it just to see how that goes. After that, you might try another technique to get a qualitative (rough sketch) idea of what the graph looks like. You can see that the graph (in the rθ-plane) fluctuates from -1 to 3. So that's the radius in the xy-plane. Then notice when it gets to zero. This will happen at a standard angle, so try to find that. You can see from the solution that one place it happens is a little after π/2. I'll let you run with the rest, have fun.

Last edited: Jun 28, 2012
3. Jun 28, 2012

pumike11

Attached is graph a little larger, I used increments of .5 instead of 1 and it looks better now. The only part of the question I don't understand is the part after the graphing that says:

Carefully label at least three points not on either of the axes. I did not realize there were axes on a polar graph. Can you clarify what that is asking?

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4. Jun 28, 2012

algebrat

You may still refer to the x and y axes when talking about the polar representation of the xy-plane. Try to notice the difference I described above between the rθ-plane, the xy-plane, and the polar representation of the latter.

5. Jun 28, 2012

algebrat

You can enter it into your calculator as polar coordinate and get the picture on the right

r=1+2cos(θ)

Or you can enter it into your caclulator as rectangular coordinate and get the picture on the left

y=1+2cos(x)

6. Jun 28, 2012

pumike11

Correct, so that part of the problem I understand.. I am just digging right now trying to figure out how I can find the rectangular points. I know the formulas:

r^2 = x^2 + y^2
x = r cos(θ)
y = r sin(θ)

...and while typing this I think I figured it out. So I could pick random θ's that satisfy the equation and solve for r, then plug them in the bottom two formulas above to get the rectangular coordinates. Correct?

7. Jun 28, 2012

algebrat

You might try thinking about your questions with the equation r=cosθ. It's a little easier to manipulate. Let us know if you get stuck, and I do encourage others to try to explain it, I think it's tricky and I'd like to know how others describe these different coordinates for the same manifold.

8. Jun 28, 2012

algebrat

Okay, yeah I hadn't thought of that part yet, looks good. However, I'm not entirely clear on which equation you want θ to satisfy. Every θ is in the domain. But I might try convenient values related to the standard angles, like 30,45,60,120,135 or 150 degrees. Also, what you might mean to be heading toward is (x,y)=(rcosθ,rsinθ), where r=1+2cosθ. So for instance x(r,θ)=x(r(θ),θ)=X(θ).

(I switched to caps because it is more reliable to call it a different function at that moment. Though many books and undergraduate courses give little mention to this fallback. It is okay (and often convenient and practical and efficient) to call it x(θ), but it can easily cause problems. For instance, doesn't it look a little strange to write x(3,0)=x(0)?)