# Graphs of Polar Equations

## Homework Statement

Hello!
I will be grateful for your help in deciphering the meaning of a paragraph from the book. I honestly don't understand how they got the semi-circle on the xy graph by transferring it from rθ graph.

## Homework Equations

I attach the screen shot from the book.

## The Attempt at a Solution

(1) if on xy graph values of r are on the x-axis, and values of θ are on the y axis, then when r = 6, we have to plot θ = 0, it's clear and corresponds to the point (6,0);

(2) but then when r = 3√2, θ = π/4, and when r = 0, θ = π/2, so I don't understand how could they "finish" the curve at the origin, namely at (0,0)? How that xy curve was formed by transforming values from the rθ curve?

Thank you very much!

#### Attachments

• Screen Shot 2017-05-07 at 15.17.37.png
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• Screen Shot 2017-05-07 at 15.17.45.png
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## Answers and Replies

scottdave
Homework Helper
Polar coordinates plot by starting at the origin, pointing in a direction (theta, measured counterclockwise from the positive x axis), and then moving in that direction a distance r, to find the point to plot. They showed you the cosine 'waveform' so you could understand how the distance r is behaving. Anytime that r is equal to zero, you will be at the origin.
Also note that negative values of r, mean it will point in the opposite direction. Imagine pointing your car North, and putting it in reverse, then driving for 1 mile. You will have moved 1 mile South.

Polar coordinates plot by starting at the origin, pointing in a direction (theta, measured counterclockwise from the positive x axis), and then moving in that direction a distance r, to find the point to plot. They showed you the cosine 'waveform' so you could understand how the distance r is behaving. Anytime that r is equal to zero, you will be at the origin.
Also note that negative values of r, mean it will point in the opposite direction. Imagine pointing your car North, and putting it in reverse, then driving for 1 mile. You will have moved 1 mile South.
Yes, this is all clear, and understand that procedure. I just don't understand how physically that xy curve is formed.

Ray Vickson
Homework Helper
Dearly Missed
Yes, this is all clear, and understand that procedure. I just don't understand how physically that xy curve is formed.

Use ##x = r \cos(\theta)=6 \cos^2 (\theta), y = r \sin(\theta) = 6 \sin(\theta) \cos(\theta)##.
Now apply the double-angle trigonometric formulas.

Last edited:
Use ##x = r \cos(\theta)=6 \cos^2 (\theta), y = r \sin(\theta) = 6 \sin(\theta) \cos(\theta)##.
Now apply the double-angle trigonometric formulas.
Thank you. But, if I understood correctly, they say that they "simply" transformed (without such additional computations) values from rθ to xy graph, and thus got that semi-circle. Have misunderstood what they stated in the book?

Ray Vickson
Homework Helper
Dearly Missed
Thank you. But, if I understood correctly, they say that they "simply" transformed (without such additional computations) values from rθ to xy graph, and thus got that semi-circle. Have misunderstood what they stated in the book?

What were their exact words? Did they really say "without such additional computations", or are those your own words?

Anyway, they are not "additional computations": the very definition of polar coordinates is that ##x = r \cos(\theta)## and ##y = r \sin(\theta)##, so those conditions are merely part of what is meant when we say "polar".

What were their exact words? Did they really say "without such additional computations", or are those your own words?

Anyway, they are not "additional computations": the very definition of polar coordinates is that ##x = r \cos(\theta)## and ##y = r \sin(\theta)##, so those conditions are merely part of what is meant when we say "polar".
Oh, so you haven't actually read the paragraph I attached. That's often confusing when people try to help by answering a different question, not the one being asked. I know the definition, and I understand it. I have a very specific question not about the definition of the polar graph and its coordinates. I will be grateful if you take a look at my question, and at the paragraph I attached in my original message.

Mark44
Mentor
Oh, so you haven't actually read the paragraph I attached.
Many helpers won't make the effort to open attachments, especially screen shots, which are often posted sideways or upside down, or blurry, or so dark you can't read them.

Many helpers won't make the effort to open attachments, especially screen shots, which are often posted sideways or upside down, or blurry, or so dark you can't read them.
I always check pictures before posting them. Can you, please, suggest another way to show the drawing from a book, or any other copied material, to make it easier for people to see?

Mark44
Mentor
I always check pictures before posting them. Can you, please, suggest another way to show the drawing from a book, or any other copied material, to make it easier for people to see?
Instead of posting thumbnails, post the pictures so that they appear full size.

Regarding you question in post #1
Vital said:
(2) but then when r = 3√2, θ = π/4, and when r = 0, θ = π/2, so I don't understand how could they "finish" the curve at the origin, namely at (0,0)?
They finish the curve at the origin, because r = 0. Unlike the Cartesian coordinate system points in polar form aren't unique. The "pole" is any point of the form ##(0, \theta)##, where ##\theta## is completely arbitrary.

## Homework Statement

Hello!
I will be grateful for your help in deciphering the meaning of a paragraph from the book. I honestly don't understand how they got the semi-circle on the xy graph by transferring it from rθ graph.

## Homework Equations

I attach the screen shot from the book.

## The Attempt at a Solution

(1) if on xy graph values of r are on the x-axis, and values of θ are on the y axis, then when r = 6, we have to plot θ = 0, it's clear and corresponds to the point (6,0);

(2) but then when r = 3√2, θ = π/4, and when r = 0, θ = π/2, so I don't understand how could they "finish" the curve at the origin, namely at (0,0)? How that xy curve was formed by transforming values from the rθ curve?
Thank you very much!
I think the problem may be that you're trying to directly associate the axes in the $\theta r$ graph with the axes in the $xy$ graph. This is not what the book is trying to say. Refer to the arrows that the book gives in its explanation. Consider a point $P$ located somewhere on the semicircle in the $xy$ plane. The $r$ coordinate tells you the distance $P$ is from the origin in the $xy$ plane. The $\theta$ coordinate tells you the counter-clockwise angle $P$ is from the positive $x$ axis. So to make the semicircle given in the text, you start with $\theta=0$ and $r=2\pi$. Your $r$ then gets slighly smaller since you get closer to the origin in the $xy$ plane and your $\theta$ gets slightly larger since the angle made with the positive $x$ axis is slightly increased. Proceed like this for every point on the semicircle. My advice is to start with the points on the semicircle and find the corresponding radius and angle represented by a point in the $\theta r$ plane. After you understand this, you will more easilly be able to go the other way around.

Instead of posting thumbnails, post the pictures so that they appear full size.
Sorry for asking further: I didn't find the way to post pictures. I see the Image icon, but when I click on it, it asks me to upload a url to the picture. So the only way I managed to find is to use the Upload button which is next to the Post reply and Preview ones.

Ray Vickson
Homework Helper
Dearly Missed
Oh, so you haven't actually read the paragraph I attached. That's often confusing when people try to help by answering a different question, not the one being asked. I know the definition, and I understand it. I have a very specific question not about the definition of the polar graph and its coordinates. I will be grateful if you take a look at my question, and at the paragraph I attached in my original message.

I answered the question you asked: you said "I just don't understand how physically that xy curve is formed." Well, OK, that is not a question as such, but that is what I addressed: I told you how to form the xy curve. I was not addressing anything the book did, or did not say; I was dealing directly with your problem. If you then do not like the answer, I cannot help that, but I can leave this thread now and avoid helping in the future.

I think the problem may be that you're trying to directly associate the axes in the $\theta r$ graph with the axes in the $xy$ graph. This is not what the book is trying to say. Refer to the arrows that the book gives in its explanation. Consider a point $P$ located somewhere on the semicircle in the $xy$ plane. The $r$ coordinate tells you the distance $P$ is from the origin in the $xy$ plane. The $\theta$ coordinate tells you the counter-clockwise angle $P$ is from the positive $x$ axis. So to make the semicircle given in the text, you start with $\theta=0$ and $r=2\pi$. Your $r$ then gets slighly smaller since you get closer to the origin in the $xy$ plane and your $\theta$ gets slightly larger since the angle made with the positive $x$ axis is slightly increased. Proceed like this for every point on the semicircle. My advice is to start with the points on the semicircle and find the corresponding radius and angle represented by a point in the $\theta r$ plane. After you understand this, you will more easilly be able to go the other way around.
Ah, I might start getting sense out of this; and yes, indeed, I was trying to associate rθ with xy, and it seems I see my mistake now. Thank you very much. I will have to think about it. I will be back in case I still have questions about it.

I answered the question you asked: you said "I just don't understand how physically that xy curve is formed." Well, OK, that is not a question as such, but that is what I addressed: I told you how to form the xy curve. I was not addressing anything the book did, or did not say; I was dealing directly with your problem. If you then do not like the answer, I cannot help that, but I can leave this thread now and avoid helping in the future.
Seems you misunderstood me. I am sorry if I somehow sounded ungrateful, which I would never intend to do. I am very grateful for the help I get here, including your one.