1. Sep 6, 2013

### PsychonautQQ

1. The problem statement, all variables and given/known data
I have to turn this homework in online... I just want someone to check my work
Convert from Cartesian coordinates to Polar coordinates
(-1,-sqrt(3))
if r > 0 and if r < 0.

2. Relevant equations

3. The attempt at a solution
if r > 0 then I believe the answer is
(r,∅) = (2,4.2)
if r < 0 then i'm a bit confused, is the only difference I make the 2 negative?
so
(r,∅) = (-2,4.2)

2. Sep 6, 2013

### eumyang

I guess you can't write the angle in terms of pi in the online system?
No.

Consider another example. The point (1, 1) (rectangular coordinates) is (sqrt 2, π/4) in polar coordinates. Imagine yourself standing at the origin. (sqrt 2, π/4) would mean that you would turn to face the π/4 "direction" and walk sqrt 2 units forward.

If r is negative, then you would walk backwards while still facing the direction indicated by θ. So if you look at (-sqrt 2, π/4), you would turn and face the π/4 "direction" and walk sqrt 2 units backwards. You see how this is not the same as (1, 1) in rectangular coordinates? In fact,
(-sqrt 2, π/4) (polar) = (-1, -1) (rectangular).

So, if you want r to be negative, what to do? You have to add/subtract π radians, to turn and face the opposite direction. If you look at (-sqrt 2, 5π/4), you face the 5π/4 "direction" and walk sqrt 2 units backwards. You end up at (1, 1) (rectangular).

3. Sep 7, 2013

### vanhees71

What do you mean by "if $r$ is negative"? By definition in polar coordinates $r>0$.

Further you didn't give the relevant equations, which indeed help to solve the problem. If you have given Cartesian coordinates $(x,y)$ of a vector you want to find $r$ and $\varphi$ such that
$$(x,y)=r(\cos \varphi,\sin \varphi).$$
From this you get
$$r=\sqrt{x^2+y^2}$$
and
$$\varphi=\mathrm{sign} (y) \arccos \left (\frac{x}{r} \right ),$$
where the range of the angle is $\varphi \in [-\pi,\pi]$. You can always add integer multiples of $2 \pi$ to this $\varphi$.

NB: Sometimes you find sloppy formulas involving $\tan \varphi$=y/x. This is dangerous, because you have to think in addition, in which quadrant the original vector is located and then adapt the result accordingly by hand!

4. Sep 9, 2013

### eumyang

According to Larson's Precalculus (8th ed., p. 777-778), r is the directed distance from O (the origin) to P (the point), and it proceeds to show examples with a negative r. Paul's Online Math Notes also state that "we also allow r to be negative."

5. Sep 9, 2013

### Enigman

Well, r can be negative, as 'r' is defined as directed distance ie. wrt angle in question. Just verified it with Thomas' Calculus 11th ed.

6. Sep 9, 2013

### vanhees71

These seem to be very strange books then! I've never heard about defining $r<0$ for plane or spherical polar coordinates ever, and it's not necessary!

7. Sep 9, 2013

### CAF123

I read it in my book too, but I would agree I have never had to define r < 0. Nor in any polar coordinates application to physics problems.

8. Sep 9, 2013

### Enigman

Its a pretty redundant concept as far as physics is concerned, though its useful to use it when finding intersection of two curves when only one representation of a point fits the equations, you will often get negative r then.

9. Sep 9, 2013

### LCKurtz

If you don't consider $r<0$ when you plot it, you don't get all three leaves of the 3-leaved rose $r=\sin(3\theta),~0 \le \theta \le\ pi$.