# Find polar coordinates

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• Dec 10th 2009, 07:24 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 $\displaystyle r= \sqrt{(4)^2+(4)^2}$
$\displaystyle = \sqrt{32}$
$\displaystyle =4\sqrt{2}$

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

But this is incorrect?

For r < 0 and 0 ≤ θ ≤ 2π.

$\displaystyle r=-4\sqrt{2}$

How do I find theta?
• Dec 10th 2009, 07:41 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 $\displaystyle r= \sqrt{(4)^2+(4)^2}$
$\displaystyle = \sqrt{32}$
$\displaystyle =4\sqrt{2}$

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

But this is incorrect?

For r < 0 and 0 ≤ θ ≤ 2π.

$\displaystyle r=-4\sqrt{2}$

How do I find theta?

Your calculation of $\displaystyle r$ is correct.

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

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

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

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

$\displaystyle = \frac{7\pi}{4}$.
• Dec 10th 2009, 07:58 PM
mmattson07
I see. So how would that differ for r<0?
• Dec 10th 2009, 08:53 PM
Prove It
Can you ever possibly have a negative radius?
• Dec 10th 2009, 09:12 PM
mmattson07
Perhaps not. I just added pi to account for -r.
• Dec 11th 2009, 07:01 AM
Prove It
$\displaystyle r$ 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 $\displaystyle \pi$.
• Dec 11th 2009, 07:37 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 $\displaystyle r= \sqrt{(4)^2+(4)^2}$
$\displaystyle = \sqrt{32}$
$\displaystyle =4\sqrt{2}$

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

But this is incorrect?

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

Quote:

For r < 0 and 0 ≤ θ ≤ 2π.

$\displaystyle r=-4\sqrt{2}$

How do I find theta?