Find polar coordinates (r, θ) of the point.

[Gundown64]

Homework Statement

The Cartesian coordinates of a point are given. (3,-5)

(i) Find polar coordinates (r, θ) of the point, where
r > 0 and 0 ≤ θ < 2π.
(ii) Find polar coordinates (r, θ) of the point, where
r < 0 and 0 ≤ θ < 2π.

Homework Equations

r^2=x^2+y^2
tanθ=(y/x) → θ=arctan(y/x)

The Attempt at a Solution

r=√(9+25)=√(34)

θ=arctan(-5/3)

The problem must be in exact terms which typically involves pi (in the problems I have worked at least). Radians and degrees are not allowed as an answer. What is the value for theta? I can't figure out where to go from there. Any help is appreciated. Thanks!

 

[Office_Shredder]

The problem must be in exact terms which typically involves pi (in the problems I have worked at least). Radians and degrees are not allowed as an answer.

If radians and degrees are not allowed as an answer, what units IS your angle supposed to be in? Are you sure you don't mean you simply aren't supposed to submit a decimal approximation to the answer? In this case arctan(-5/3) is the best you can do

 

[Gundown64]

If radians and degrees are not allowed as an answer, what units IS your angle supposed to be in? Are you sure you don't mean you simply aren't supposed to submit a decimal approximation to the answer? In this case arctan(-5/3) is the best you can do

Yes, sorry, that is what I mean. Decimal approximation is not allowed as far as I know and I'm not even aware that leaving it as arctan(-5/3) is allowed. It does not specify, but all the other problems have had pi in the theta value so I have no idea how it is to be submitted. I guess that is more of my own problem than something you guys can help with, but I am unsure on what to do.

 

[SammyS]

[itex]\displaystyle -\frac{\pi}{2}<\arctan\left(-\frac{5}{3}\right)<0[/itex]

You need to get your answer into the correct quadrant.