Question

Convert (-2/sqrt[3] , 2) to polar coordinates

Answer 1

\[|r|=\sqrt{x^2+y^2}; \theta = \arctan(y/x); x= r \cos(\theta), y = r \sin(\theta)\]
Find theta and r, and you're done.

Answer 2

I know that much.
\[\theta = \tan^-1 (2/(-2/\sqrt{3})\]
theta= - sqrt[3] right?

Answer 3

\[(x,y)=(\frac{-2}{\sqrt{3}},2)\]Ok, we need to translate this to \[(r,\theta)\]we use\[r=\sqrt{(x^2+y^2)}\]and \[\theta=\tan^{-1} (\frac{x}{y})\]so,\[r=\sqrt{\frac{4}{3}+4}=\sqrt{\frac{16}{3}}=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}\]and,\[\theta=\tan^{-1} (\frac{-1}{\sqrt{2}})\approx-35=325\]My answer is in degrees not radians. Your point lies at \[(r,\theta)=(\frac{4\sqrt{3}}{3},325)\]

Answer 4

theta = arctan(-sqrt(3)) = 5pi/3

Answer 5

why did you put (-1/root(2)) instead???

Answer 6

he did x/y, it should be y/x.

Answer 7

r=sqrt{(4/3)+4}=sqrt{16}{3}=4/sqrt{3}=4sqrt{3}/3

Answer 8

so (x,y) -> (4sqrt(3)/3, 5 pi/ 3)

Answer 9

Which is the exact answer :D

Answer 10

My teacher said the answer should b 2pi/3. Because i cant used a calculator to do this stuff, how are you getting those answers. If i have tan^-1 (y/x) it equals -sqrt[3]. And none of what you guys have

Answer 11

Wait, actually, the angle is wrong, I got the wrong quadrant, the arctangent is periodic so you need to add pi or subtract it as needed, in this case, subtract it. So the angle should be 2pi/3 or 120 degrees.

Answer 12

Notice, my answer can't be write because it puts the point in the fourth quadrant, but obviously that can't be right because the x is negative, not the y.