convert rectangular coordinate (-3,0) to polar coordinate

Question

I'm trying to convert (-3,0) to polar coordinate.

I can get

r=$\sqrt {(-3)^2 +(0)^2}$ =3,

but when computing for the angle

$\theta$=$\tan^{-1} (\frac {0}{-3})$=0

but the answer for the angle is $\pi$, I don't understand why the answer is $\pi$ instead of 0 degree, can anyone explain this? TQ

Answer 1

Well, draw a picture showing the location of the point $(-3,0)$, and ask yourself what direction you'd need to go (from the origin) to get to it. Using an angle of $\theta = 0$ will send you along the positive $x$-axis, which is obviously wrong.

You can't get the polar angle from the inverse tangent function alone. You need to find an angle $\theta$ such that $r\cos\theta = x$ and $r\sin\theta=y$. In programming terms, you need to use the ATAN2 function to do this; the ATAN function is not sufficient. As you have already seen, there are two angles $\theta$ that give $\tan\theta = 0$, and the ATAN function won't tell you which of the two is correct.

Answer 2

To understand it first, plot $(-3, 0)$ on the Cartesian plane, and observe that it is $3$ units to the left of the origin. In polar coordinates, points with $r > 0$ and $\theta = 0$ are to the right of the origin; one must swing $180$ degrees—that is, $\pi$ radians—counterclockwise to end up on the left side of the origin. (Alternatively, one can use negative $r$ and $\theta = 0$ to get on the left side; ironically, one obtains $(-3, 0)$ as also valid polar coordinates for the same point.)

Your problem in finding this out algorithmically stems from $\tan^{-1}$ having a limited range, from $-\pi/2$ to $\pi/2$. In order to determine the proper answer algorithmically, follow lab bhattacharjee's link to atan2, a four-quadrant version of arctangent.

Answer 3

$ r = 3 $

$ \theta = \tan^{-1} 0 + \pi $

We have to add $\pi$, as the position of the tip of radius vector lies on the other side of origin for negative radius.

Answer 4

Since the result of the $\arctan$ function only ranges from $-\pi$ to $\pi$, you can't get the full circle of angles necessary to represent all points in the plane.

There are at least two approaches to getting the correct argument, both using $r=\sqrt{x^2+y^2}$

One can case on the sign of $x$: $$ \newcommand{sgn}{\operatorname{sgn}} \theta=\left\{\begin{array}{} \frac\pi2\sgn(y)&\text{if }x=0\\ \arctan\left(\dfrac yx\right)&\text{if }x\gt0\\ \arctan\left(\dfrac yx\right)+\pi&\text{if }x\lt0\text{ and }y\gt0\\ \arctan\left(\dfrac yx\right)-\pi&\text{if }x\lt0\text{ and }y\lt0\\ \pi&\text{if }x\lt0\text{ and }y=0 \end{array}\right. $$ One can use $$ \theta=\left\{\begin{array}{} 0&\text{if }r+x=0\text{ and }x=0\\ \pi&\text{if }r+x=0\text{ and }x\ne0\\ 2\arctan\left(\frac{y}{r+x}\right)&\text{if }r+x\ne0 \end{array}\right. $$ Both of these give $r=3$ and $\theta=\pi$ for $(-3,0)$.