# Specifying domain range of polar coordinates

I apologize if this is easy to you, I haven't done math in a long time (3 years)

Question :A straight line is defined by equation $$y = 2x + 3$$ in Cartesian coordinate system $$XY$$.

(i) Define this straight line in polar coordinates $$r$$, a as an explicit function $$r = f (\alpha).$$

(ii) Specify the domain range for the polar coordinate $$\alpha$$ which is valid for this straight line

What I have done : for i), I have subbed in $$x = r \cos(\alpha)$$ and $$y = r \sin(\alpha)$$ into the equation, with the result of $$r = \frac{3}{\sin(\alpha)-2\cos(\alpha)},$$ after manipulation.

(ii) is where I am having issues, I have googled the term domain range to see how I could implement it, but unfortunately I do not see the link between my result, am I supposed to be trying to find the min and max possible values from that definition?

For example, I am thinking $$\alpha = 0$$ being invalid for that particular result, is that correct?

Once again I apologize if this is a simple problem, and I would appreciate it if I was linked to terms to read up upon in order to learn what I should do

As you've said: $$x= r\cos(\alpha),$$

$$y= r\sin(\alpha).$$

(you also have $$r^2 = x^2 +y^2$$)

So if you plug this in, you have

$$r\sin(\alpha) = 2r\cos(\alpha) +3,$$ $$r(\sin(\alpha) - 2\cos(\alpha)) = 3 \Longrightarrow r(\alpha) = \frac{3}{\sin(\alpha) - 2\cos(\alpha)}.$$ Now, for the range of $$\alpha$$, if you draw the line it looks like (using google):

$$\hspace{3cm}$$

You can see that when you consider a very distant point on the line (the black line on the plot below) the angle of this line (in red) approaches some quantity $$\alpha_0$$ degrees. If you consider distant point in the other direction (not drawn), the angle would approach $$\alpha_1 = 180 + \alpha_0$$ degrees:

$$\hspace{3cm}$$

Overall, the range is $$r \in [r_0, \infty)$$ and $$\alpha \in (\alpha_0, \alpha_1)$$ for $$r_0$$ you need to find the closest point on the line to the origin.

If you do the calculations: you get $$\alpha_0 = \mathrm{arctang}(2)$$ (the ratio of $$y$$ to $$x$$ approaches $$2$$ as we look at a distant points) and $$r_0 = \sqrt{3^2 + 1.5^2}$$ (Pythagorean theorem).

To reiterate my comment:

The range of $$r$$ and $$\alpha$$ are the values of $$r$$ and $$\alpha$$ that you would have to plug into the equation of the line $$r(\alpha) = f(\alpha)$$ in order to trace out the line. For example $$𝑟 = 0$$ is never used so it is not in the range, same is for $$𝛼= 250^\circ$$ or $$\alpha=0$$, both are not in the range.

• First of all, thank you very much for such a detailed answer, I can see this alot more clearly now. I understand why the range of r is infinity, but I am not very sure about the definition of range of a specifically in this instance, does this simply mean the opposite direction, because of my vector plane? Aug 17, 2019 at 11:24
• The range is the values of $r$ and $\alpha$ that you would have to plug in to $r(\alpha)=f(\alpha)$ in order to trace the line. For example $r=0$ is never used so it is not in the range, same is for $\alpha = 250^\circ$ also not in rage,
– them
Aug 17, 2019 at 11:29
• @arc2test please see, I fixed a mistake $\alpha$ is not $45^\circ$
– them
Aug 17, 2019 at 11:40
• thank you, α1 in this case would still be referring to the maximum angle in use right? If you don't mind, may I have the terms to search in order to find resources for further reading on this particular subject? Thank you very much for all the help you have done Aug 17, 2019 at 11:47
• I'm not sure there is a fixed terminology, the word "range" can be used in different ways so it should be put into context. Maybe check here en.wikipedia.org/wiki/Range_(mathematics)
– them
Aug 17, 2019 at 11:53