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


1 Answer 1


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):

$\hspace{3cm}$ enter image description here

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:

$\hspace{3cm}$ enter image description here

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.

  • $\begingroup$ 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? $\endgroup$
    – arc2test
    Aug 17, 2019 at 11:24
  • $\begingroup$ 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, $\endgroup$
    – them
    Aug 17, 2019 at 11:29
  • $\begingroup$ @arc2test please see, I fixed a mistake $\alpha$ is not $45^\circ$ $\endgroup$
    – them
    Aug 17, 2019 at 11:40
  • $\begingroup$ 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 $\endgroup$
    – arc2test
    Aug 17, 2019 at 11:47
  • $\begingroup$ 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) $\endgroup$
    – them
    Aug 17, 2019 at 11:53

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