# Graphing θ=π/4 on a Polar Coordinate System

## Homework Statement:

Graph $θ=\frac{π}{4}$ on a Polar Coordinate System.

## Homework Equations:

Why does the line go into the opposite quadrant as well?
When you graph something like $θ=\frac{π}{4}$ on a Polar Coordinate System:
Why does the line go into the opposite quadrant as well?
I can intuitively understand why it is in the first quadrant: $θ = 45°$ there and so all possible values of $r$ would apply there, giving you a straight line headed outwards at an angle of $45°$.
So, why does the line go into the opposite quadrant as well?
Isn't $r$ always positive? Or, is that something we define beforehand as being either positive or negative?

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QuantumQuest
Gold Member
In polar coordinates, the angle that corresponds to a certain point is not unique. For example, let's say that we have a point, at a distance of 2 units from the origin on the radius with $θ=30^0$. It has polar coordinates $r=2,θ=30^0$. It has also coordinates $r=2,θ=−330^0$ and $r=2,θ=390^0$. So, it depends on what direction we traverse the circle - as in Trigonometry an angle $θ$ is positive when it runs counter-clockwise, and how many times we traverse the circle.

EDIT: Also, the radius $r$, can be negative. So, we talk about an oriented distance. For instance, the radius with $θ=30^0$ and the one with $θ=210^0$ form a straight line which passes through the origin. The point, say, $P(2,210^0)$ which is at a distance of $2$ units from the origin on the radius with $θ=210^0$, has polar coordinates $r=2,θ=210^0$. We can reach this point if we go in the positive direction $210^0$ and walk $2$ units forward. We can also reach the same point if we turn $30^0$ in the positive direction and walk $2$ units backwards. The turns we take are with reference to the initial radius ($\theta = 0$), given that we are at the origin ($O$). So, the previous point we talked about, has also polar coordinates $r = -2, \theta = 30^0$.

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In polar coordinates, the angle that corresponds to a certain point is not unique. For example, let's say that we have a point, at a distance of 2 units from the origin on the radius with $\theta = 30^0$. It has polar coordinates $r = 2, \theta = 30^0$. It has also coordinates $r = 2, \theta = -330^0$ and $r = 2, \theta = 390^0$. So, it depends on what direction we traverse the circle - as in Trigonometry an angle $\theta$ is positive when it runs counter-clockwise, and how many times we traverse the circle.
But, in my question above, isn't the opposite angle $225°$ and not some other angle equivalent to $45°$ (like $-315°$ or $405°$)?
Am I stuck thinking in terms of the unit circle definition?

Math_QED
Homework Helper
2019 Award
If you don't allow negative radius, I think you are right. Then it should be a half line in the first quadrant. Changing the angle to an equivalent doesn't change this.

QuantumQuest
Gold Member
@lightlightsup see my edit in post #2.

EDIT: Also, the radius rr, can be negative. So, we talk about an oriented distance. For instance, the radius with θ=300θ=30^0 and θ=2100θ=210^0 form a straight line which passes through the origin. The point P(2,2100)P(2,210^0) which is at a distance, say, of 2 units from the origin on the radius with θ=2100θ=210^0, has polar coordinates r=2,θ=2100r=2,θ=210^0. We can reach this point if we go in the positive direction 2100210^0 and walk 2 units forward. We can also reach the same point if we turn 30030^0 in the positive direction and walk 2 units backwards. The turns we take are with reference to the initial radius (θ=0\theta = 0), given that we are at the origin (OO). So, the previous point we talked about has also polar coordinates r=−2,θ=300r = -2, \theta = 30^0.
So, it's kind of like $-\hat{r}$?
But, in the formal/mathematical case, we would explicitly have to state what $r$ is allowed to be (like $-r$)?

So, the way I'm understanding the sign on $θ$ is that it is defined independently in each quadrant based on whether it is counterclockwise (+) or clockwise (-):
For example: $+30°$ would occur in Quadrants I and III because it would be $30°$ counterclockwise from the Polar x-Axis there. $-30°$ would occur in Quadrants II and IV because it would be clockwise from the Polar x-Axis there.
So, the mystery remains: how is $r$'s sign defined? Left or Right from the Polar y-Axis?

QuantumQuest
Gold Member
So, the mystery remains: how is $r$'s sign defined?
There is no mystery. It's all a matter of convention. In math, conventions are there in order to talk about the same things in the same way. For the angle $\theta$ we use the positive sign ($+$) when we traverse the circle counter-clockwise and negative sign ($-$) when we traverse it clockwise. For the radius $r$ we use positive sign when we move from the origin to the outer side of the circle(s) on a certain radius and negative sign in the opposite case.

So, the convention is that $r$ is $+$ when it is in the $+$ direction of the Polar x-Axis.
And, $θ$'s signage is determined by clockwise/counterclockwise angling relative to the Polar x-Axis.
Does that seem like a correct interpretation?
I can't think of any other explanation for why, by convention, $(−r,−θ)$ is in Quadrant II and $(r,−θ)$ is in Quadrant IV.
Another convention I've seen is that $r$ is always pointing outwards based on how the $θ (+ or -)$ guides it. But, it can point it in the opposite direction with a $-$ before the $r$. This is the correct answer, I think. Determine the angle first, then figure out which direction you're heading in.

This is why when you graph $θ=\frac{π}{4}$, you include it in Quadrants I and III: you are considering all values of $r$ for which $θ=\frac{π}{4}$?

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QuantumQuest
Gold Member
So, the convention is that $r$ is $+$ when it is in the $+$ direction of the Polar x-Axis.
And, $θ$'s signage is determined by clockwise/counterclockwise angling relative to the Polar x-Axis.
Does that seem like a correct interpretation?
I can't think of any other explanation for why, by convention, $(−r,−θ)$ is in Quadrant II and $(r,−θ)$ is in Quadrant IV.
Another convention I've seen is that $r$ is always pointing outwards based on how the $θ (+ or -)$ guides it. But, it can point it in the opposite direction with a $-$ before the $r$. This is the correct answer, I think. Determine the angle first, then figure out which direction you're heading in.
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This is why when you graph $θ=\frac{π}{4}$, you include it in Quadrants I and III: you are considering all values of $r$ for which $θ=\frac{π}{4}$?
The whole point is the reference you use. In my example in post #2 I talked about the different polar coordinates of the same point. In the diagram you give, there is the same angle $\theta$ with different sign per case. So, in the first quadrant as per your diagram, we have obviously $(r,\theta)$. In the second quadrant the angle is $-\theta$ as it is seen clockwise beginning from the negative $-x$ axis. Then, the radius is $-r$ with respect to the direction we move. In the same manner we find the signs of $r$ and $\theta$ in the rest two quadrants.

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Think about the equation y = 2 in a rectangular coordinate system. Why is this a line that goes both to the left and the right? Because x can be any real value and the statement is still true. It is the same for polar coordinates. If I say θ=π/4, then r can be any value, positive or negative.