Answer:
Part a)
(i) The polar coordinates are [tex](6\sqrt{2},\frac{7\pi}{4})[/tex]
(ii) The polar coordinates are [tex](-6\sqrt{2},\frac{3\pi}{4})[/tex]
Part b)
(i)The polar coordinates are [tex](\sqrt{10},0.60\pi)[/tex]
(ii)The polar coordinates are [tex](-\sqrt{10},1.60\pi)[/tex]
Step-by-step explanation:
Part a) we have
(6,-6)
(i) Find polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π
step 1
Find r
[tex]r^{2}=x^{2} + y^{2}[/tex]
substitute
[tex]r^{2}=(6)^{2} + (-6)^{2}[/tex]
[tex]r^{2}=72[/tex]
[tex]r=6\sqrt{2}[/tex]
step 2
Find angle theta
[tex]tan(\theta)=\frac{y}{x}[/tex]
substitute
[tex]tan(\theta)=\frac{6}{6}=1[/tex]
[tex]\theta=arctan(1)=\frac{\pi}{4}[/tex]
Remember that the point (6,-6) lies on IV quadrant
so
The angle theta is equal to
[tex]2\pi-\frac{\pi}{4}=\frac{7\pi}{4}[/tex]
therefore
The polar coordinates are [tex](6\sqrt{2},\frac{7\pi}{4})[/tex]
(ii) Find polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π.
For r< 0 the point lies on II Quadrant
so
the angle theta is equal to
[tex]\pi-\frac{\pi}{4}=\frac{3\pi}{4}[/tex]
therefore
The polar coordinates are [tex](-6\sqrt{2},\frac{3\pi}{4})[/tex]
Part b) we have
(-1,3)
(i) Find polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π
step 1
Find r
[tex]r^{2}=x^{2} + y^{2}[/tex]
substitute
[tex]r^{2}=(-1)^{2} + (3)^{2}[/tex]
[tex]r^{2}=10[/tex]
[tex]r=\sqrt{10}[/tex]
step 2
Find angle theta
[tex]tan(\theta)=\frac{y}{x}[/tex]
substitute
[tex]tan(\theta)=\frac{3}{1}=3[/tex]
[tex]\theta=arctan(3)=0.40\pi[/tex]
Remember that the point (-1,3) lies on II quadrant
so
The angle theta is equal to
[tex]\pi-0.40\pi=0.60\pi[/tex]
therefore
The polar coordinates are [tex](\sqrt{10},0.60\pi)[/tex]
(ii) Find polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π.
For r< 0 the point lies on IV Quadrant
so
the angle theta is equal to
[tex]2\pi-0.40\pi=1.60\pi[/tex]
therefore
The polar coordinates are [tex](-\sqrt{10},1.60\pi)[/tex]
