Question

Please help me people
Given any Cartesian coordinates, (x,y), there are polar coordinates (r,θ) with −π/2<θ≤π/2.
Find polar coordinates with −π/2<θ≤π/2 for the following Cartesian coordinates:
(a) If (x,y)=(3,3) then (r,θ)=( , )
(b) If (x,y)=(13,6) then (r,θ)=( , )
(c) If (x,y)=(−9,−6) then (r,θ)=( , ),
(d) If (x,y)=(20,2) then (r,θ)=( , ),
(e) If (x,y)=(−4,6) then (r,θ)=( , ),
(f) If (x,y)=(0,6) then (r,θ)=( , )

Answer 1

use the following conversion formulas.
\[\Large r=\sqrt{x^2+y^2}\]
\[\Large \theta=\tan^{-1} (\frac{y}{x})\]

for example for first point
x=3 ,y=3
\[\Large r=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}\]
\[\Large \theta=\tan^{-1}(\frac{3}{3})=45^0=\frac{\pi}{4}\]

Answer 2

i know how to get the first one but the second one is confusing

Answer 3

never mind i got it