# Polar coordinates

1. Mar 18, 2013

### stunner5000pt

1. The problem statement, all variables and given/known data

Find two other pairs of polar coordinates of the given polar coordinate, one with r > 0 and one with r < 0. Then plot the point.

(2, 5π/3)

2. Relevant equations

I don't there are any.

3. The attempt at a solution

I'm not completely sure of how to do this actually.

I know that we can add 2 pi to the answer to get the coordinates for r > 0 this gives an answer of $(2,\frac{11\pi}{3}$

But I'm not sure about the r<0 option. Would I simply have to subtract 2 pi from 5 pi/3? In addition, would the r value change to the negative sign?

Last edited by a moderator: Mar 21, 2013
2. Mar 18, 2013

### tiny-tim

hi stunner5000pt!
are you sure that isn't θ > 0 and θ < 0 ?

3. Mar 20, 2013

### stunner5000pt

I chekced again and yes it is R and not theta ...

4. Mar 20, 2013

### eumyang

r can be negative. Think of yourself being in the origin of the coordinate plane, facing the positive y direction. If you walk in the positive y direction 5 units, then where you end up will correspond to
$\left( 5, \frac{\pi}{2} \right)$
in polar coordinates.

Think of yourself back in the origin, facing the positive y direction. Now walk backwards 5 units. By "walking backwards" your r is negative. Since you are still facing the positive y direction, this point will correspond to
$\left( -5, \frac{\pi}{2} \right)$
in polar coordinates. But this particular point is the same as facing the negative y direction and walking forwards 5 units, or
$\left( 5, \frac{3\pi}{2} \right)$

5. Mar 20, 2013

### stunner5000pt

based on what you have said, in the case of (2, 5pi/3), the moment arm for the 'backward' motion is in the second quadrant

If we 'walked' backward that would be $(-2, \frac{5\pi}{3})$
for the 'positive' motion it would be $(2, \frac{2\pi}{3})$

How does this sound?

6. Mar 21, 2013

### eumyang

Not quite. The three points you state are not the same. The three points I used in my example were not the same either - the last two formed a separate example from the first.

The question is this: which direction would you have to "face", if, when walking "backwards" 2 units, you end up at the same point as (2, 5π/3)? In other words,
$\left( 2, \frac{5\pi}{3} \right) = \left( -2, ? \right)$
Then you also need to consider coterminal angles.

7. Mar 21, 2013

### stunner5000pt

if we walked backward, then the angle would be pi/3

As for the answer with r > 0, the answer wuld be (2, -pi/3)

Is that correct?

8. Mar 21, 2013

### SammyS

Staff Emeritus

9. Mar 21, 2013

### eumyang

Sorry, that's still not correct. It may help if you looked at a polar coordinate grid like the one on this site.

Last edited: Mar 21, 2013
10. Mar 21, 2013

### stunner5000pt

Hmm... ok (2, -pi/3) is correct. This is because if we have a positive moment arm, the angle would be measured negative as we rotate clockwise

for the negative angle, wouldn't the angle be the same? But since the r is negative, would that imply that if we rotate clockwise, we get a positive angle?
would that mean that for r = -2, the angle should be + pi/3?

11. Mar 21, 2013

### eumyang

No. See the attached image. Point A is the point you want, $\left( 2, \frac{5\pi}{3} \right)$. Point B is $\left( -2, \frac{\pi}{3} \right)$, the point that you are saying that is the same as point A. They are not the same.

#### Attached Files:

• ###### polar_coord_ques.png
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12. Mar 21, 2013

### Staff: Mentor

In polar coordinates, the radial coordinate r is always positive.

13. Mar 21, 2013

### stunner5000pt

I am so confused

SO lets start from the top

if we consider a point $\left( 2,\frac{5\pi}{3} \right)$
then for r> 0 $\left( 2,\frac{11 \pi}{3} \right)$
and for r < 0 $\left( -2,\frac{2\pi}{3} \right)$

obtained r > 0 by simply rotating once (+2 pi)
obtained r < 0 by rotating half circle (- pi)

14. Mar 21, 2013

### skiller

Who says?
Bingo!

15. Mar 21, 2013

### Staff: Mentor

Every math book that I've ever seen. I challenge you to site one single reference in which r in polar coordinates is considered anything but positive (or zero).

Last edited: Mar 21, 2013
16. Mar 21, 2013

### skiller

Well apart from at least one other person on this thread saying so, and at least one link on this thread to a website saying so...

Even the Wiki page for Polar Coordinates says so. (Although, I will agree that Wiki is not always Gospel.)

17. Mar 21, 2013

### LCKurtz

Plot $r = \sin (3\theta)$ for $\theta$ from $0$ to $\pi$ and you will see an nice 3 leaved rose with one leaf plotted with $r<0$.

18. Mar 21, 2013

### SammyS

Staff Emeritus
I don't blame you.

It's clear that this argument concerning polar coordinates is not helping you understand how to solve this problem you have been given.

Try looking at the case of r = -2 as follows.

In Cartesian coordinates, the point you're working with is $\displaystyle \ (x,\, y)=\left(1,\,-\sqrt{3}\right)\ .$

In general, x = r cos(θ) and y = r sin(θ).

For this point, you then have $\displaystyle \ 1=(-2)\cos(\theta)\ \text{ and }\ -\sqrt{3}=(-2)\sin(\theta) \ .$

This gives you $\displaystyle \ \cos(\theta)=-\,\frac{1}{2}\ \text{ and }\ \sin(\theta) =\frac{\sqrt{3}}{2} \ .$

Thus the angle, θ, is in the second quadrant.

19. Mar 22, 2013

### eumyang

Precalculus by Larson (8th ed.) refers to r as a "directed distance from O to P" (p. 777) and it states that "another way to obtain multiple representations of a point is to use negative values for r" (p. 778).

Precalculus: Graphical, Numerical, Algebraic by Demana/Waits/Foley/Kennedy (7th ed.) states that "r is the directed distance from O to P" and that "if r < 0 then P is on the terminal side of θ + π" (both on p. 534)

That's two references, and I think it's safe to say that you are mistaken.

20. Mar 22, 2013

### Staff: Mentor

Actually, Wiki is a little schizophrenic on this. First the say "The radial distance ρ is the Euclidean distance from the z axis to the point P." This can only be positive. But then they say something about allowing negative values of ρ to locate a point.

In a 40 year career as an engineer/mathematician, I have never seen negative values of ρ being used in practice. So maybe negative values of ρ are only found in Ivory Tower land.