Respuesta :
Answer:
(a) [tex](3, -\frac{4\pi}{3})[/tex]
(b) [tex](-3, \frac{5\pi}{3})[/tex]
(c)[tex](3, \frac{8\pi}{3})[/tex]
Step-by-step explanation:
All polar coordinates of point (r, θ ) are
[tex](r,\theta+2n\pi)[/tex] and [tex](-r,\theta+(2n+1)\pi)[/tex]
where, θ is in radian and n is an integer.
The given point is [tex](3, \frac{2\pi}{3})[/tex]. So, all polar coordinates of point are
[tex](3, \frac{2\pi}{3}+2n\pi)[/tex] and [tex](-3, \frac{2\pi}{3}+(2n+1)\pi)[/tex]
(a) [tex]r>0,-2\pi\leq \theta <0[/tex]
Substitute n=-1 in [tex](3, \frac{2\pi}{3}+2n\pi)[/tex], to find the point for which [tex]r>0,-2\pi\leq \theta <0[/tex].
[tex](3, \frac{2\pi}{3}+2(-1)\pi)[/tex]
[tex](3, -\frac{4\pi}{3})[/tex]
Therefore, the required point is [tex](3, -\frac{4\pi}{3})[/tex].
(b) [tex]r<0,0\leq \theta <2\pi[/tex]
Substitute n=0 in [tex](-3, \frac{2\pi}{3}+(2n+1)\pi)[/tex], to find the point for which [tex]r>0,-2\pi\leq \theta <0[/tex].
[tex](-3, \frac{2\pi}{3}+(2(0)+1)\pi)[/tex]
[tex](-3, \frac{2\pi}{3}+\pi)[/tex]
[tex](-3, \frac{5\pi}{3})[/tex]
Therefore, the required point is [tex](-3, \frac{5\pi}{3})[/tex].
(c) [tex]r>0,2\pi \leq \theta <4\pi[/tex]
Substitute n=1 in [tex](3, \frac{2\pi}{3}+2n\pi)[/tex], to find the point for which [tex]r>0,2\pi \leq \theta <4\pi[/tex].
[tex](3, \frac{2\pi}{3}+2(1)\pi)[/tex]
[tex](3, \frac{2\pi}{3}+2\pi)[/tex]
[tex](3, \frac{8\pi}{3})[/tex]
Therefore, the required point is [tex](3, \frac{8\pi}{3})[/tex].
