I will be grateful for your help and explanation on how to decipher what authors of the Precalculus book did when plotting the equation from $r,\theta$ -plane into xy-plane. Below I quote 3 parts of their explanations, and attach screen shots of graphs
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 polar 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.
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.
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.
Summarizing a bit:
(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.
Thank you very much!