Graph r = 6 cos() issues with plotting on xy-plane

[Vital]

Homework Statement

Hello!
Last week I have came here for the help related to this problem. I am creating a new thread to describe the issue more precisely. I will be grateful for your help and explanation.

I post the explanation for the book first accompanied by attached pictures, and below I post my questions.

Homework Equations

Example on how to graph the polar equation
r = 6 cos(θ)

Quote part 1:
We graph one cycle of r = 6 cos(θ) on the θr-plane and use it to help graph the equation on the xy-plane. We see that as θ ranges from 0 to π/2 , r ranges from 6 to 0. In the xy-plane, this means that the curve starts 6 units from the origin on the positive x-axis (θ = 0) and gradually returns to the origin by the time the curve reaches the y-axis (θ = π/2 ). The arrows drawn in the figure below are meant to help you visualize this process. In the θr-plane, the arrows are drawn from the θ-axis to the curve r = 6 cos(). In the xy-plane, each of these arrows starts at the origin and is rotated through the corresponding angle , in accordance with how we plot polar coordinates.
End of the quote part 1.

Picture attached.

The Attempt at a Solution

Quote part 2:
Next, we repeat the process as θ ranges from π/2 to π. Here, the r values are all negative. This means that in the xy-plane, instead of graphing in Quadrant II, we graph in Quadrant IV, with all of the angle rotations starting from the negative x-axis.
End of the quote part 2.

So, if θ = 3π/4, then r = -3√2
θ = π , then r = -6

In the first part we started at the angle θ = 0 and thus r = 6, which we plotted as x = 6; then rotating counter-clockwise as all values of r become smaller as θ approaches π/2. This is clear to me.

And now I am confused by the second part. It is said that r values are negative, so I don't understand why we plot these values along the positive x-axis and rotate clockwise. How did they come up with this rotation, what is the reason that I fail to understand? The phrase on the picture saying "r < 0 so we plot here" gives a sense that this is obvious, but not to me. Please, help me to understand it.

It seems they are plotting absolute values of r along x-axis, so all x values are positive. But how's this justified mathematically?

Here is also the next, even more confusing, quote:
As θ ranges from π to 3π/2, the r values are still negative, which means the graph is traced out in Quadrant I instead of Quadrant III.
End of quote.

Interesting. The second part stated that as values of r are negative, we have to plot in QIV; and the third quote says that as values are still negative, we obviously have to plot in QI. I am utterly confused. :) Please, help.
Thank you very much!

 

[FactChecker]

It seems they are plotting absolute values of r along x-axis, so all x values are positive. But how's this justified mathematically?

I wouldn't say that. It looks like they are plotting points on the line through the origin at angle θ, where positive r is in direction θ and negative r is in the opposite direction. The arrow lines show the direction that the point is in from the origin.

 

[Vital]

I wouldn't say that. It looks like they are plotting points on the line through the origin at angle θ, where positive r is in direction θ and negative r is in the opposite direction. The arrow lines show the direction that the point is in from the origin.

But:
(1) when the interval is [0, π/2] and r = 6, they start at y = x = 6 and go all the way "up" (counter-clockwise) till angle reaches π/2; it's clear;
(2) when the interval is [π/2 , π] and the value of r = -6 at angle π, they most likely start at the angle -π/2, move counter-clockwise "up" till 2π, and surprisingly end at y = x = 6, not x = -6;
(3) and then, when even more puzzling thing happens - even though values are still negative and interval is [π, 3π/2], they get back to the interval [0, π/2] in QI;
(4) and for the last interval of [3π/2, 2π], they get back to QIV.
I genuinely don't get it. It seems there is something very easy-peasy in all this, some very basic notion that I miss.

 

[FactChecker]

But:
(1) when the interval is [0, π/2] and r = 6, they start at y = x = 6 and go all the way "up" (counter-clockwise) till angle reaches π/2; it's clear;
(2) when the interval is [π/2 , π] and the value of r = -6 at angle π, they most likely start at the angle -π/2, move counter-clockwise "up" till 2π, and surprisingly end at y = x = 6, not x = -6;

That is not what it says in part 2. They say they are measuring the angle (still counter-clockwise, I assume) starting on the negative x-axis. That puts them in quadrant IV. I prefer to think of it as measuring the angle as always (so still in quadrant II) and negative r points in the opposite direction through the origin, putting you in quadrant IV.

(3) and then, when even more puzzling thing happens - even though values are still negative and interval is [π, 3π/2], they get back to the interval [0, π/2] in QI;

Same as in (2). Since r is negative, they are measuring their angle counter-clockwise starting on the negative x axis. That pits them in quadrant I. As before, I prefer to think of it as measuring the angle as always (so still in quadrant III) and negative r points in the opposite direction through the origin, putting you in quadrant I.

(4) and for the last interval of [3π/2, 2π], they get back to QIV.

Their approach for positive r measures the angle normally (counter-clockwise starting from the positive x axis), which puts them in quadrant IV.

 

[Mark44]

From another perspective, it's possible to convert the equation ##r = 6\cos(\theta)## to Cartesian form.
First step: Multiply both sides by r, to get ##r^2 = 6r\cos(\theta)##.
Doing this adds a solution r = 0, but in the original equation, when ##\theta = \pi/2##, r = 0, so we're not adding a solution that wasn't already there.

Now convert ##r^2 = 6r\cos(\theta)## to Cartesian (or rectangular) form, resulting in ##x^2 + y^2 = 6x##. By completing the square in the x terms, it's fairly easy to show that the equation represents a circle.

 

[Vital]

From another perspective, it's possible to convert the equation ##r = 6\cos(\theta)## to Cartesian form.
First step: Multiply both sides by r, to get ##r^2 = 6r\cos(\theta)##.
Doing this adds a solution r = 0, but in the original equation, when ##\theta = \pi/2##, r = 0, so we're not adding a solution that wasn't already there.

Now convert ##r^2 = 6r\cos(\theta)## to Cartesian (or rectangular) form, resulting in ##x^2 + y^2 = 6x##. By completing the square in the x terms, it's fairly easy to show that the equation represents a circle.

Sorry, but you took a wrong direction. I am happy because now I have an explanation. In fact it's much easier: when r is negative, we sort of flip all values 180 degrees.

 

[Mark44]

Sorry, but you took a wrong direction.

How so?

In post 1, you have the following in your Relevant Equations section:

Example on how to graph the polar equation
r = 6 cos(θ)

My explanation gave an alternate way to get the graph directly.

I am happy because now I have an explanation. In fact it's much easier: when r is negative, we sort of flip all values 180 degrees.

That's not what FactChecker was saying in post 4.

 

[FactChecker]

Sorry, but you took a wrong direction. I am happy because now I have an explanation. In fact it's much easier: when r is negative, we sort of flip all values 180 degrees.

Good. But you should understand their approach because they might use it again later (maybe for other things).

 

[Vital]

Good. But you should understand their approach because they might use it again later (maybe for other things).

Of course. :-) It is a normal approach for such polar equations: whenever r<0, we first go r values on the usual θ ray, and then add π, namely go 180 on the opposite direction. That is why when r < 0 and θ is, say, in QII, that means we plot that part in QIV; if θ is in QIII and r < 0, we plot in QI. This is what I missed, and hence didn't understand all those steps, described in quotes in the original message. And, actually, this is exactly what you have said in your post above. :-)

 

[FactChecker]

Of course. :-) It is a normal approach for such polar equations: whenever r<0, we first go r values on the usual θ ray, and then add π, namely go 180 on the opposite direction. That is why when r < 0 and θ is, say, in QII, that means we plot that part in QIV; if θ is in QIII and r < 0, we plot in QI. This is what I missed, and hence didn't understand all those steps, described in quotes in the original message. And, actually, this is exactly what you have said in your post above. :-)

Oh! Of course! They are just adding the 180 at the beginning rather than at the end. I didn't see that.