# Find polar coordinates

• Dec 10th 2009, 08:24 PM
mmattson07
Find polar coordinates
Hey I am not sure how to find the second coordinate of these polar coordinates.

a) Find polar coordinates (r, θ) of the point (4, -4), where r > 0 and 0 ≤ θ ≤ 2π.

So $r= \sqrt{(4)^2+(4)^2}$
$= \sqrt{32}$
$=4\sqrt{2}$

I tried doing
θ= $tan^{-1}(\frac{y}{x})$
θ= $tan^{-1}(\frac{-4}{4})$
θ= $tan^{-1}(-1)$
θ= $\frac{-\pi}{4}$

But this is incorrect?

For r < 0 and 0 ≤ θ ≤ 2π.

$r=-4\sqrt{2}$

How do I find theta?
• Dec 10th 2009, 08:41 PM
Prove It
Quote:

Originally Posted by mmattson07
Hey I am not sure how to find the second coordinate of these polar coordinates.

a) Find polar coordinates (r, θ) of the point (4, -4), where r > 0 and 0 ≤ θ ≤ 2π.

So $r= \sqrt{(4)^2+(4)^2}$
$= \sqrt{32}$
$=4\sqrt{2}$

I tried doing
θ= $tan^{-1}(\frac{y}{x})$
θ= $tan^{-1}(\frac{-4}{4})$
θ= $tan^{-1}(-1)$
θ= $\frac{-\pi}{4}$

But this is incorrect?

For r < 0 and 0 ≤ θ ≤ 2π.

$r=-4\sqrt{2}$

How do I find theta?

Your calculation of $r$ is correct.

To find $\theta$, first note that the point $(4, -4)$ is in the fourth quadrant. So you would need to find the focus angle, i.e. for the point $(4, 4)$, and then subtract it from $2\pi$.
So

$\theta = 2\pi -\arctan{\frac{4}{4}}$

$= 2\pi -\arctan{1}$

$= 2\pi -\frac{\pi}{4}$

$= \frac{7\pi}{4}$.
• Dec 10th 2009, 08:58 PM
mmattson07
I see. So how would that differ for r<0?
• Dec 10th 2009, 09:53 PM
Prove It
Can you ever possibly have a negative radius?
• Dec 10th 2009, 10:12 PM
mmattson07
Perhaps not. I just added pi to account for -r.
• Dec 11th 2009, 08:01 AM
Prove It
$r$ represents the distance between a point and the origin.

A distance, in other words, a length, can never be negative.

I don't see why you would add $\pi$.
• Dec 11th 2009, 08:37 AM
HallsofIvy
Quote:

Originally Posted by mmattson07
Hey I am not sure how to find the second coordinate of these polar coordinates.

a) Find polar coordinates (r, θ) of the point (4, -4), where r > 0 and 0 ≤ θ ≤ 2π.

So $r= \sqrt{(4)^2+(4)^2}$
$= \sqrt{32}$
$=4\sqrt{2}$

I tried doing
θ= $tan^{-1}(\frac{y}{x})$
θ= $tan^{-1}(\frac{-4}{4})$
θ= $tan^{-1}(-1)$
θ= $\frac{-\pi}{4}$

But this is incorrect?

That's incorrect only because you specifically said " $0\le \theta\le 2\pi$" and $-\pi/4$ is less than 0. Since a complete circle is $2\pi$, adding $2\pi$, $-\pi/4+ 2\pi= -\pi/4+ 8\pi/4= 7\pi/4$ gives you exactly the same point and is now in the correct interval: $7\pi/4$ is larger than 0 and less than $2\pi$.

Quote:

For r < 0 and 0 ≤ θ ≤ 2π.

$r=-4\sqrt{2}$

How do I find theta?