Polar Coordinates inverse Radius

[PsychonautQQ]

Homework Statement

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.

Homework Equations

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)

 

[eumyang]

if r > 0 then I believe the answer is
(r,∅) = (2,4.2)

I guess you can't write the angle in terms of pi in the online system?

if r < 0 then I'm a bit confused, is the only difference I make the 2 negative?
so
(r,∅) = (-2,4.2)

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).

Now fix your second point.

 

[vanhees71]

What do you mean by "if r is negative"? By definition in polar coordinates r&gt;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!

 

[eumyang]

What do you mean by "if r is negative"? By definition in polar coordinates r&gt;0.

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."

 

[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.

 

[vanhees71]

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

 

[CAF123]

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

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.

 

[Enigman]

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

Adding any number of full turns (360°) to the angular coordinate does not change the corresponding direction. Also, a negative radial coordinate is best interpreted as the corresponding positive distance measured in the opposite direction. Therefore, the same point can be expressed with an infinite number of different polar coordinates (r, φ ± n×360°) or (−r, φ ± (2n + 1)180°), where n is any integer.[10] Moreover, the pole itself can be expressed as (0, φ) for any angle φ.[11]
Where a unique representation is needed for any point, it is usual to limit r to non-negative numbers (r ≥ 0) and φ to the interval [0, 360°) or (−180°, 180°] (in radians, [0, 2π) or (−π, π]).[12] One must also choose a unique azimuth for the pole, e.g., φ = 0

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.

 

[LCKurtz]

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